For a triangle with angles A , B , C A,B,C A , B , C a , b , c a,b,c a , b , c
sin β‘ A a = sin β‘ B b = sin β‘ C c . \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}. a sin A β = b sin B β = c sin C β . This is the Law of Sines . It is useful when a triangle has an angle-side opposite pair, especially in the following cases:
ASA or AAS , when two angles and one side are known.SSA , when two sides and a non-included angle are known.
The area of a triangle can also be written using two sides and the included angle:
K = 1 2 a b sin β‘ C = 1 2 b c sin β‘ A = 1 2 a c sin β‘ B . K=\frac12ab\sin C=\frac12bc\sin A=\frac12ac\sin B. K = 2 1 β ab sin C = 2 1 β b c sin A = 2 1 β a c sin B .
Proof (Area formula) Drop an altitude from angle B B B A C AC A C
Since side c c c A A A A A A
sin β‘ A = h c . \sin A=\frac{h}{c}. sin A = c h β . Thus,
h = c sin β‘ A . h=c\sin A. h = c sin A . Using the basic triangle area formula,
K = 1 2 ( base ) ( height ) , K=\frac12(\text{base})(\text{height}), K = 2 1 β ( base ) ( height ) , with base b b b h h h
K = 1 2 b ( c sin β‘ A ) = 1 2 b c sin β‘ A . K=\frac12 b(c\sin A)=\frac12bc\sin A. K = 2 1 β b ( c sin A ) = 2 1 β b c sin A . By dropping different altitudes, the same reasoning gives
K = 1 2 a b sin β‘ C = 1 2 a c sin β‘ B . K=\frac12ab\sin C=\frac12ac\sin B. K = 2 1 β ab sin C = 2 1 β a c sin B .
Proof (Law of Sines). Since the area of the same triangle can be written three ways,
1 2 b c sin β‘ A = 1 2 a c sin β‘ B = 1 2 a b sin β‘ C . \frac12bc\sin A=\frac12ac\sin B=\frac12ab\sin C. 2 1 β b c sin A = 2 1 β a c sin B = 2 1 β ab sin C . Divide each expression by 1 2 a b c \frac12abc 2 1 β ab c
sin β‘ A a = sin β‘ B b = sin β‘ C c . \frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}. a sin A β = b sin B β = c sin C β . This is the Law of Sines.
The Law of Sines also has an extended version, which is not particularly important for AP Precalculus.
Extension. Prove the extended Law of Sines: a sin β‘ A = b sin β‘ B = c sin β‘ C = 2 R . \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=2R. s i n A a β = s i n B b β = s i n C c β = 2 R .
The SSA case is called the ambiguous case because the given information may create no triangle, one triangle, or two triangles.
Suppose angle A A A a a a b b b a a a A A A
h = b sin β‘ A . h=b\sin A. h = b sin A . Then:
Condition NumberΒ ofΒ triangles a < h 0 h β€ a < b 2 a β₯ b 1 \begin{array}{c|c}
\text{Condition} & \text{Number of triangles}\\
\hline
a<h & 0\\
h\le a<b & 2\\
a\ge b & 1
\end{array} Condition a < h h β€ a < b a β₯ b β NumberΒ ofΒ triangles 0 2 1 β β Warning
When using the Law of Sines, remember that sin β‘ ΞΈ = sin β‘ ( 180 β β ΞΈ ) \sin \theta=\sin(180^\circ-\theta) sin ΞΈ = sin ( 18 0 β β ΞΈ )
Example. Suppose A = 45 β A=45^\circ A = 4 5 β a = 12 a=12 a = 12 b = 15 b=15 b = 15 B B B b b b
First find the height:
h = b sin β‘ A = 15 sin β‘ 45 β = 15 2 2 β 10.61. h=b\sin A=15\sin45^\circ=\frac{15\sqrt2}{2}\approx 10.61. h = b sin A = 15 sin 4 5 β = 2 15 2 β β β 10.61. Since
h < a < b , h<a<b, h < a < b , there are two possible triangles.
Using Law of Sines,
sin β‘ B 15 = sin β‘ 45 β 12 . \frac{\sin B}{15}=\frac{\sin45^\circ}{12}. 15 sin B β = 12 sin 4 5 β β . So
sin β‘ B = 15 sin β‘ 45 β 12 β 0.884. \sin B=\frac{15\sin45^\circ}{12}\approx 0.884. sin B = 12 15 sin 4 5 β β β 0.884. This gives one possible angle
B β 62.1 β , B\approx 62.1^\circ, B β 62. 1 β , and the second possible angle
180 β β 62.1 β = 117.9 β . 180^\circ-62.1^\circ=117.9^\circ. 18 0 β β 62. 1 β = 117. 9 β . Both angles create valid triangles because each leaves a positive value for angle C C C
The Law of Cosines relates all three sides of a triangle to one angle:
a 2 = b 2 + c 2 β 2 b c cos β‘ A . a^2=b^2+c^2-2bc\cos A. a 2 = b 2 + c 2 β 2 b c cos A . Similarly,
b 2 = a 2 + c 2 β 2 a c cos β‘ B b^2=a^2+c^2-2ac\cos B b 2 = a 2 + c 2 β 2 a c cos B and
c 2 = a 2 + b 2 β 2 a b cos β‘ C . c^2=a^2+b^2-2ab\cos C. c 2 = a 2 + b 2 β 2 ab cos C . Law of Cosines is useful for:
SSS , when all three sides are known.SAS , when two sides and the included angle are known.
To find an angle from three side lengths, rearrange the formula:
cos β‘ A = b 2 + c 2 β a 2 2 b c . \cos A=\frac{b^2+c^2-a^2}{2bc}. cos A = 2 b c b 2 + c 2 β a 2 β .
Proof (Law of Cosines). Place the triangle on the coordinate plane so that angle A A A b b b x x x
Then the endpoint of side b b b
C = ( b , 0 ) . C=(b,0). C = ( b , 0 ) . The endpoint of side c c c
B = ( c cos β‘ A , c sin β‘ A ) . B=(c\cos A,c\sin A). B = ( c cos A , c sin A ) . Side a a a B B B C C C
a 2 = ( c cos β‘ A β b ) 2 + ( c sin β‘ A β 0 ) 2 . a^2=(c\cos A-b)^2+(c\sin A-0)^2. a 2 = ( c cos A β b ) 2 + ( c sin A β 0 ) 2 . Expand:
a 2 = c 2 cos β‘ 2 A β 2 b c cos β‘ A + b 2 + c 2 sin β‘ 2 A . a^2=c^2\cos^2A-2bc\cos A+b^2+c^2\sin^2A. a 2 = c 2 cos 2 A β 2 b c cos A + b 2 + c 2 sin 2 A . Group the trig terms:
a 2 = b 2 + c 2 ( cos β‘ 2 A + sin β‘ 2 A ) β 2 b c cos β‘ A . a^2=b^2+c^2(\cos^2A+\sin^2A)-2bc\cos A. a 2 = b 2 + c 2 ( cos 2 A + sin 2 A ) β 2 b c cos A . Since
cos β‘ 2 A + sin β‘ 2 A = 1 , \cos^2A+\sin^2A=1, cos 2 A + sin 2 A = 1 , we get
a 2 = b 2 + c 2 β 2 b c cos β‘ A . a^2=b^2+c^2-2bc\cos A. a 2 = b 2 + c 2 β 2 b c cos A .
Example. If b = 7 b=7 b = 7 c = 10 c=10 c = 10 A = 60 β A=60^\circ A = 6 0 β a a a
By Law of Cosines,
a 2 = 7 2 + 10 2 β 2 ( 7 ) ( 10 ) cos β‘ 60 β . a^2=7^2+10^2-2(7)(10)\cos60^\circ. a 2 = 7 2 + 1 0 2 β 2 ( 7 ) ( 10 ) cos 6 0 β . Since cos β‘ 60 β = 1 2 \cos60^\circ=\frac12 cos 6 0 β = 2 1 β
a 2 = 49 + 100 β 70 = 79. a^2=49+100-70=79. a 2 = 49 + 100 β 70 = 79. Thus,
a = 79 . a=\sqrt{79}. a = 79 β .
A scalar is a quantity described by one number, such as temperature, length, or mass.
A vector is a quantity described by both magnitude and direction, such as force or velocity.
Geometrically, a vector is drawn as a directed line segment. The starting point is the initial point , and the ending point is the terminal point .
The magnitude of vector P Q β \overrightarrow{PQ} P Q β
β£ P Q β β£ . \left\lvert \overrightarrow{PQ}\right \rvert. β P Q β β . Two vectors are equal if they have the same magnitude and the same direction, even if they are drawn in different locations. Basically, they are the same vector just translated.
If a vector has magnitude β£ v β£ \lvert \mathbf v \rvert β£ v β£ ΞΈ \theta ΞΈ
v x = β£ v β£ cos β‘ ΞΈ v_x=\lvert \mathbf v \rvert\cos\theta v x β = β£ v β£ cos ΞΈ and
v y = β£ v β£ sin β‘ ΞΈ . v_y=\lvert \mathbf v\rvert \sin\theta. v y β = β£ v β£ sin ΞΈ . Thus,
v = β¨ β£ v β£ cos β‘ ΞΈ , β£ v β£ sin β‘ ΞΈ β© . \mathbf v=\langle \lvert \mathbf v \rvert \cos\theta, \lvert \mathbf v \rvert\sin\theta\rangle. v = β¨β£ v β£ cos ΞΈ , β£ v β£ sin ΞΈ β© .
Proof (Component Definition). Draw a vector v \mathbf v v ΞΈ \theta ΞΈ β£ v β£ \lvert \mathbf v \rvert β£ v β£
By right triangle trigonometry,
cos β‘ ΞΈ = v x β£ v β£ \cos\theta=\frac{v_x}{\lvert \mathbf v \rvert} cos ΞΈ = β£ v β£ v x β β and
sin β‘ ΞΈ = v y β£ v β£ . \sin\theta=\frac{v_y}{\lvert \mathbf v \rvert}. sin ΞΈ = β£ v β£ v y β β . Solving these equations gives
v x = β£ v β£ cos β‘ ΞΈ v_x=\lvert \mathbf v \rvert\cos\theta v x β = β£ v β£ cos ΞΈ and
v y = β£ v β£ sin β‘ ΞΈ . v_y=\lvert \mathbf v \rvert\sin\theta. v y β = β£ v β£ sin ΞΈ .
For a vector
v = β¨ a , b β© , \mathbf v=\langle a,b\rangle, v = β¨ a , b β© , the magnitude is
β£ v β£ = a 2 + b 2 . \lvert \mathbf v \rvert=\sqrt{a^2+b^2}. β£ v β£ = a 2 + b 2 β . The direction angle satisfies
tan β‘ ΞΈ = b a , \tan\theta=\frac{b}{a}, tan ΞΈ = a b β , but the quadrant of the vector must be considered.
x y v = h a; b i a b Β΅ ( a; b )
Example. A force has magnitude 6.34 6.34 6.34 175 β 175^\circ 17 5 β
The components of the vector are
F x = 6.34 cos β‘ 175 β β β 6.32 F_x=6.34\cos175^\circ\approx -6.32 F x β = 6.34 cos 17 5 β β β 6.32 and
F y = 6.34 sin β‘ 175 β β 0.55. F_y=6.34\sin175^\circ\approx 0.55. F y β = 6.34 sin 17 5 β β 0.55. So the vector is approximately
β¨ β 6.32 , 0.55 β© . \langle -6.32,0.55\rangle. β¨ β 6.32 , 0.55 β© . The negative x x x 175 β 175^\circ 17 5 β
Vectors can be added geometrically by placing the initial point of one vector at the terminal point of the other. The resulting vector is called the resultant .
Vector addition is commutative:
u + v = v + u . \mathbf u+\mathbf v=\mathbf v+\mathbf u. u + v = v + u . For vectors written in component form,
u = β¨ u 1 , u 2 β© , v = β¨ v 1 , v 2 β© , \mathbf u=\langle u_1,u_2\rangle,\qquad
\mathbf v=\langle v_1,v_2\rangle, u = β¨ u 1 β , u 2 β β© , v = β¨ v 1 β , v 2 β β© , their sum is
u + v = β¨ u 1 + v 1 , u 2 + v 2 β© . \mathbf u+\mathbf v=\langle u_1+v_1,u_2+v_2\rangle. u + v = β¨ u 1 β + v 1 β , u 2 β + v 2 β β© . To understand vector addition, remember that a vector β¨ u 1 , u 2 β© \langle u_1,u_2\rangle β¨ u 1 β , u 2 β β© u 1 u_1 u 1 β u 2 u_2 u 2 β
Adding
u = β¨ u 1 , u 2 β© \mathbf u=\langle u_1,u_2\rangle u = β¨ u 1 β , u 2 β β© and
v = β¨ v 1 , v 2 β© \mathbf v=\langle v_1,v_2\rangle v = β¨ v 1 β , v 2 β β© means doing both moves. Horizontally, the total movement is
u 1 + v 1 . u_1+v_1. u 1 β + v 1 β . Vertically, the total movement is
u 2 + v 2 . u_2+v_2. u 2 β + v 2 β . So,
u + v = β¨ u 1 + v 1 , u 2 + v 2 β© . \mathbf u+\mathbf v=\langle u_1+v_1,u_2+v_2\rangle. u + v = β¨ u 1 β + v 1 β , u 2 β + v 2 β β© . If
u = β¨ u 1 , u 2 β© , \mathbf u=\langle u_1,u_2\rangle, u = β¨ u 1 β , u 2 β β© , then
k u = β¨ k u 1 , k u 2 β© . k\mathbf u=\langle ku_1,ku_2\rangle. k u = β¨ k u 1 β , k u 2 β β© . The magnitude changes by a factor of β£ k β£ \lvert k \rvert β£ k β£
β£ k u β£ = β£ k β£ β£ u β£ . \lvert k\mathbf u \rvert = \lvert k \rvert \lvert \mathbf u \rvert. β£ k u β£ = β£ k β£ β£ u β£ . If k > 0 k>0 k > 0 k < 0 k<0 k < 0
A unit vector has magnitude 1 1 1
The standard unit vectors are
i = β¨ 1 , 0 β© \mathbf i=\langle 1,0\rangle i = β¨ 1 , 0 β© and
j = β¨ 0 , 1 β© . \mathbf j=\langle 0,1\rangle. j = β¨ 0 , 1 β© . So
β¨ a , b β© = a i + b j . \langle a,b\rangle=a\mathbf i+b\mathbf j. β¨ a , b β© = a i + b j . To find a unit vector in the direction of a nonzero vector v \mathbf v v
u = v β£ v β£ . \mathbf u=\frac{\mathbf v}{\lvert \mathbf v \rvert}. u = β£ v β£ v β . For
A = β¨ x 1 , y 1 β© \mathbf A=\langle x_1,y_1\rangle A = β¨ x 1 β , y 1 β β© and
B = β¨ x 2 , y 2 β© , \mathbf B=\langle x_2,y_2\rangle, B = β¨ x 2 β , y 2 β β© , the dot product is
A β
B = x 1 x 2 + y 1 y 2 . \mathbf A\cdot \mathbf B=x_1x_2+y_1y_2. A β
B = x 1 β x 2 β + y 1 β y 2 β . The dot product can also be written as
A β
B = β£ A β£ β£ B β£ cos β‘ ΞΈ , \mathbf A\cdot \mathbf B=\lvert \mathbf A \rvert \lvert \mathbf B \rvert \cos\theta, A β
B = β£ A β£ β£ B β£ cos ΞΈ , where ΞΈ \theta ΞΈ
Therefore,
cos β‘ ΞΈ = A β
B β£ A β£ β£ B β£ . \cos\theta=\frac{\mathbf A\cdot \mathbf B}{\lvert \mathbf A \rvert \lvert \mathbf B \rvert}. cos ΞΈ = β£ A β£ β£ B β£ A β
B β . The dot product is a method of multiplication of two vectors, and always returns a scalar . Another form of vector multiplication is the cross product, which will not be taught here.
Proof (Dot product formula). Let
A = β¨ x 1 , y 1 β© , B = β¨ x 2 , y 2 β© . \mathbf A=\langle x_1,y_1\rangle,\qquad \mathbf B=\langle x_2,y_2\rangle. A = β¨ x 1 β , y 1 β β© , B = β¨ x 2 β , y 2 β β© . The vector between their terminal points is
A β B . \mathbf A-\mathbf B. A β B . Using the magnitude formula,
β£ A β B β£ 2 = ( x 1 β x 2 ) 2 + ( y 1 β y 2 ) 2 . \lvert\mathbf A-\mathbf B\rvert^2=(x_1-x_2)^2+(y_1-y_2)^2. β£ A β B β£ 2 = ( x 1 β β x 2 β ) 2 + ( y 1 β β y 2 β ) 2 . Expanding gives
β£ A β B β£ 2 = x 1 2 + y 1 2 + x 2 2 + y 2 2 β 2 ( x 1 x 2 + y 1 y 2 ) . \lvert\mathbf A-\mathbf B\rvert^2=x_1^2+y_1^2+x_2^2+y_2^2-2(x_1x_2+y_1y_2). β£ A β B β£ 2 = x 1 2 β + y 1 2 β + x 2 2 β + y 2 2 β β 2 ( x 1 β x 2 β + y 1 β y 2 β ) . Since
β£ A β£ 2 = x 1 2 + y 1 2 , β£ B β£ 2 = x 2 2 + y 2 2 , \lvert\mathbf A\rvert^2=x_1^2+y_1^2,\qquad \lvert\mathbf B\rvert^2=x_2^2+y_2^2, β£ A β£ 2 = x 1 2 β + y 1 2 β , β£ B β£ 2 = x 2 2 β + y 2 2 β , this becomes
β£ A β B β£ 2 = β£ A β£ 2 + β£ B β£ 2 β 2 ( A β
B ) . \lvert\mathbf A-\mathbf B\rvert^2=\lvert\mathbf A\rvert^2+\lvert\mathbf B\rvert^2-2(\mathbf A\cdot \mathbf B). β£ A β B β£ 2 = β£ A β£ 2 + β£ B β£ 2 β 2 ( A β
B ) . Now use the Law of Cosines on the triangle formed by A \mathbf A A B \mathbf B B A β B \mathbf A-\mathbf B A β B
β£ A β B β£ 2 = β£ A β£ 2 + β£ B β£ 2 β 2 β£ A β£ β£ B β£ cos β‘ ΞΈ . \lvert\mathbf A-\mathbf B\rvert^2=\lvert\mathbf A\rvert^2+\lvert\mathbf B\rvert^2-2\lvert\mathbf A\rvert\lvert\mathbf B\rvert\cos\theta. β£ A β B β£ 2 = β£ A β£ 2 + β£ B β£ 2 β 2 β£ A β£ β£ B β£ cos ΞΈ . Comparing the two equations,
A β
B = β£ A β£ β£ B β£ cos β‘ ΞΈ . \mathbf A\cdot \mathbf B=\lvert\mathbf A\rvert\lvert\mathbf B\rvert\cos\theta. A β
B = β£ A β£ β£ B β£ cos ΞΈ .
Useful vector formulas that follow from the dot product include
v β
v = β£ v β£ 2 , \mathbf v\cdot \mathbf v=\lvert\mathbf v\rvert^2, v β
v = β£ v β£ 2 , so a vector dotted with itself gives the square of its magnitude. Also,
β£ v β£ = v β
v . \lvert\mathbf v\rvert=\sqrt{\mathbf v\cdot \mathbf v}. β£ v β£ = v β
v β . For two nonzero vectors A \mathbf A A B \mathbf B B
A β₯ B ifΒ andΒ onlyΒ if A β
B = 0. \mathbf A\perp \mathbf B\quad \text{if and only if}\quad \mathbf A\cdot \mathbf B=0. A β₯ B ifΒ andΒ onlyΒ if A β
B = 0. The scalar projection of A \mathbf A A B \mathbf B B
comp β‘ B A = A β
B β£ B β£ , \operatorname{comp}_{\mathbf B}\mathbf A=\frac{\mathbf A\cdot \mathbf B}{\lvert\mathbf B\rvert}, comp B β A = β£ B β£ A β
B β , and the vector projection of A \mathbf A A B \mathbf B B
proj β‘ B A = A β
B β£ B β£ 2 B . \operatorname{proj}_{\mathbf B}\mathbf A=\frac{\mathbf A\cdot \mathbf B}{\lvert \mathbf B \rvert^2}\mathbf B. proj B β A = β£ B β£ 2 A β
B β B .
Example. Find the angle between
A = β¨ 3 , 4 β© \mathbf A=\langle 3,4\rangle A = β¨ 3 , 4 β© and
B = β¨ 5 , 0 β© . \mathbf B=\langle 5,0\rangle. B = β¨ 5 , 0 β© . First,
A β
B = 3 ( 5 ) + 4 ( 0 ) = 15. \mathbf A\cdot \mathbf B=3(5)+4(0)=15. A β
B = 3 ( 5 ) + 4 ( 0 ) = 15. Also,
β£ A β£ = 5 , β£ B β£ = 5. \lvert\mathbf A\rvert=5,\qquad \lvert\mathbf B\rvert=5. β£ A β£ = 5 , β£ B β£ = 5. So
cos β‘ ΞΈ = 15 5 β
5 = 3 5 . \cos\theta=\frac{15}{5\cdot5}=\frac35. cos ΞΈ = 5 β
5 15 β = 5 3 β . Therefore,
ΞΈ = cos β‘ β 1 ( 3 5 ) β 53.1 β . \theta=\cos^{-1}\left(\frac35\right)\approx 53.1^\circ. ΞΈ = cos β 1 ( 5 3 β ) β 53. 1 β .
In a parametric equation, both x x x y y y t t t
x = f ( t ) , y = g ( t ) . x=f(t),\qquad y=g(t). x = f ( t ) , y = g ( t ) . The variable t t t parameter .
Parametric equations describe both:
the set of points on a curve,
the direction the curve is traced as t t t
To eliminate the parameter, solve one equation for t t t
For example, if
x = 3 t , x=3t, x = 3 t , then
t = x 3 . t=\frac{x}{3}. t = 3 x β . Substituting this into the equation for y y y x x x y y y
The rectangular equation may not show the full behavior of the parametric curve, because the parameter can restrict the domain or determine the direction of travel. Always indicate the allowed domain for t t t
Example. Consider
x = 3 t , y = 1 2 t 2 . x=3t,\qquad y=\frac12t^2. x = 3 t , y = 2 1 β t 2 . From the first equation,
t = x 3 . t=\frac{x}{3}. t = 3 x β . Substitute into the equation for y y y
y = 1 2 ( x 3 ) 2 . y=\frac12\left(\frac{x}{3}\right)^2. y = 2 1 β ( 3 x β ) 2 . So
y = x 2 18 . y=\frac{x^2}{18}. y = 18 x 2 β . This is a parabola. However, if the parameter is restricted, such as 0 β€ t β€ 2 0\le t\le 2 0 β€ t β€ 2
0 β€ x β€ 6. 0\le x\le 6. 0 β€ x β€ 6. 3 6 1 2 x y
The identity
sin β‘ 2 t + cos β‘ 2 t = 1 \sin^2 t+\cos^2 t=1 sin 2 t + cos 2 t = 1 is often used to eliminate a parameter from trig parametrizations.
For a circle centered at the origin with radius a a a
x = a cos β‘ t , y = a sin β‘ t . x=a\cos t,\qquad y=a\sin t. x = a cos t , y = a sin t . Eliminating t t t
( x a ) 2 + ( y a ) 2 = 1 , \left(\frac{x}{a}\right)^2+\left(\frac{y}{a}\right)^2=1, ( a x β ) 2 + ( a y β ) 2 = 1 , or
x 2 + y 2 = a 2 . x^2+y^2=a^2. x 2 + y 2 = a 2 . For an ellipse centered at the origin:
x = a cos β‘ t , y = b sin β‘ t . x=a\cos t,\qquad y=b\sin t. x = a cos t , y = b sin t . Eliminating t t t
x 2 a 2 + y 2 b 2 = 1. \frac{x^2}{a^2}+\frac{y^2}{b^2}=1. a 2 x 2 β + b 2 y 2 β = 1. Switching sine and cosine, or making one coefficient negative, can change where the graph starts and which direction it is traced, but the rectangular equation may stay the same.
Problem-solving strategy
Isolate the two trig expressions (e.g. cos β‘ t = x a \cos t = \frac{x}{a} cos t = a x β
Use a version of the Pythagoren Identity (e.g. cos β‘ 2 t + sin β‘ 2 t = 1 \cos^2 t + \sin^2 t = 1 cos 2 t + sin 2 t = 1
State any domain restrictions on the graph and indicate orientation/direction.
Example. For
x = β 3 sin β‘ t , y = 4 cos β‘ t , x=-3\sin t,\qquad y=4\cos t, x = β 3 sin t , y = 4 cos t , divide each equation by its coefficient:
x β 3 = sin β‘ t , y 4 = cos β‘ t . \frac{x}{-3}=\sin t,\qquad \frac{y}{4}=\cos t. β 3 x β = sin t , 4 y β = cos t . Now square and add:
( x β 3 ) 2 + ( y 4 ) 2 = sin β‘ 2 t + cos β‘ 2 t . \left(\frac{x}{-3}\right)^2+\left(\frac{y}{4}\right)^2=\sin^2t+\cos^2t. ( β 3 x β ) 2 + ( 4 y β ) 2 = sin 2 t + cos 2 t . So
x 2 9 + y 2 16 = 1. \frac{x^2}{9}+\frac{y^2}{16}=1. 9 x 2 β + 16 y 2 β = 1. The graph is an ellipse centered at the origin.
Problem-solving strategy
Choose values of t t t
Calculate the corresponding x x x y y y
Plot the ordered pairs ( x , y ) (x,y) ( x , y )
Use arrows to show the direction of motion as t t t
An alternative is to solve for the combined equation and then graph that equation on the allowed domain and then draw the direction arrows.
In rectangular coordinates, a point is written as
( x , y ) . (x,y). ( x , y ) . In polar coordinates, a point is written as
( r , ΞΈ ) , (r,\theta), ( r , ΞΈ ) , where:
r r r ΞΈ \theta ΞΈ
The pole is the origin, and the polar axis is the positive x x x
p olar axis y r Β‘ r Β΅ p ole ( r ; Β΅ ) ( Β‘ r ; Β΅ ) negativ e radius plots opp osite Every polar point has infinitely many representations.
For any integer k k k
( r , ΞΈ ) = ( r , ΞΈ + 2 k Ο ) . (r,\theta)=(r,\theta+2k\pi). ( r , ΞΈ ) = ( r , ΞΈ + 2 k Ο ) . A point can also be represented using a negative radius:
( r , ΞΈ ) = ( β r , ΞΈ + ( 2 k + 1 ) Ο ) . (r,\theta)=(-r,\theta+(2k+1)\pi). ( r , ΞΈ ) = ( β r , ΞΈ + ( 2 k + 1 ) Ο ) . A negative value of r r r ΞΈ \theta ΞΈ
Example. Find two points that are the same point as ( 4 , Ο 3 ) \left(4,\frac{\pi}{3}\right) ( 4 , 3 Ο β )
The polar points
( 4 , Ο 3 ) , ( 4 , Ο 3 + 2 Ο ) , ( β 4 , 4 Ο 3 ) \left(4,\frac{\pi}{3}\right),\qquad
\left(4,\frac{\pi}{3}+2\pi\right),\qquad
\left(-4,\frac{4\pi}{3}\right) ( 4 , 3 Ο β ) , ( 4 , 3 Ο β + 2 Ο ) , ( β 4 , 3 4 Ο β ) all represent the same point.
The first two use the same radius and coterminal angles. The third uses a negative radius, so it points in the opposite direction from 4 Ο 3 \frac{4\pi}{3} 3 4 Ο β Ο 3 \frac{\pi}{3} 3 Ο β
The main conversion formulas are
x = r cos β‘ ΞΈ x=r\cos\theta x = r cos ΞΈ and
y = r sin β‘ ΞΈ . y=r\sin\theta. y = r sin ΞΈ . Also,
x 2 + y 2 = r 2 x^2+y^2=r^2 x 2 + y 2 = r 2 and
tan β‘ ΞΈ = y x . \tan\theta=\frac{y}{x}. tan ΞΈ = x y β . When converting from rectangular to polar form, use
r = x 2 + y 2 r=\sqrt{x^2+y^2} r = x 2 + y 2 β and choose ΞΈ \theta ΞΈ
Proof (Conversion formulas). Prove the conversion formulas above.
A polar point ( r , ΞΈ ) (r,\theta) ( r , ΞΈ ) x x x y y y r r r
By right triangle trig,
cos β‘ ΞΈ = x r \cos\theta=\frac{x}{r} cos ΞΈ = r x β and
sin β‘ ΞΈ = y r . \sin\theta=\frac{y}{r}. sin ΞΈ = r y β . Multiplying by r r r
x = r cos β‘ ΞΈ , y = r sin β‘ ΞΈ . x=r\cos\theta,\qquad y=r\sin\theta. x = r cos ΞΈ , y = r sin ΞΈ . The Pythagorean Theorem gives
x 2 + y 2 = r 2 . x^2+y^2=r^2. x 2 + y 2 = r 2 . Similarly, since tan β‘ ΞΈ \tan\theta tan ΞΈ
tan β‘ ΞΈ = y x . \tan\theta=\frac{y}{x}. tan ΞΈ = x y β .
Example. Convert
( 2 , 7 Ο 6 ) \left(2,\frac{7\pi}{6}\right) ( 2 , 6 7 Ο β ) to rectangular coordinates.
Use
x = r cos β‘ ΞΈ , y = r sin β‘ ΞΈ . x=r\cos\theta,\qquad y=r\sin\theta. x = r cos ΞΈ , y = r sin ΞΈ . Then
x = 2 cos β‘ ( 7 Ο 6 ) = 2 ( β 3 2 ) = β 3 x=2\cos\left(\frac{7\pi}{6}\right)=2\left(-\frac{\sqrt3}{2}\right)=-\sqrt3 x = 2 cos ( 6 7 Ο β ) = 2 ( β 2 3 β β ) = β 3 β and
y = 2 sin β‘ ( 7 Ο 6 ) = 2 ( β 1 2 ) = β 1. y=2\sin\left(\frac{7\pi}{6}\right)=2\left(-\frac12\right)=-1. y = 2 sin ( 6 7 Ο β ) = 2 ( β 2 1 β ) = β 1. So the rectangular point is
( β 3 , β 1 ) . (-\sqrt3,-1). ( β 3 β , β 1 ) .
To convert a polar equation into rectangular form, use:
x = r cos β‘ ΞΈ , y = r sin β‘ ΞΈ , r 2 = x 2 + y 2 . x=r\cos\theta,\qquad y=r\sin\theta,\qquad r^2=x^2+y^2. x = r cos ΞΈ , y = r sin ΞΈ , r 2 = x 2 + y 2 . Sometimes it helps to multiply both sides of a polar equation by r r r r cos β‘ ΞΈ r\cos\theta r cos ΞΈ r sin β‘ ΞΈ r\sin\theta r sin ΞΈ
The two most common substitutions are:
r cos β‘ ΞΈ = x r\cos\theta=x r cos ΞΈ = x and
r sin β‘ ΞΈ = y . r\sin\theta=y. r sin ΞΈ = y .
Example. Convert
r sin β‘ ( ΞΈ + Ο 4 ) = 6 r\sin\left(\theta+\frac{\pi}{4}\right)=6 r sin ( ΞΈ + 4 Ο β ) = 6 to rectangular form.
Use the angle-sum identity:
sin β‘ ( ΞΈ + Ο 4 ) = sin β‘ ΞΈ cos β‘ Ο 4 + cos β‘ ΞΈ sin β‘ Ο 4 . \sin\left(\theta+\frac{\pi}{4}\right)
=\sin\theta\cos\frac{\pi}{4}+\cos\theta\sin\frac{\pi}{4}. sin ( ΞΈ + 4 Ο β ) = sin ΞΈ cos 4 Ο β + cos ΞΈ sin 4 Ο β . So
r ( sin β‘ ΞΈ β
2 2 + cos β‘ ΞΈ β
2 2 ) = 6. r\left(\sin\theta\cdot\frac{\sqrt2}{2}+\cos\theta\cdot\frac{\sqrt2}{2}\right)=6. r ( sin ΞΈ β
2 2 β β + cos ΞΈ β
2 2 β β ) = 6. Distribute r r r
2 2 r sin β‘ ΞΈ + 2 2 r cos β‘ ΞΈ = 6. \frac{\sqrt2}{2}r\sin\theta+\frac{\sqrt2}{2}r\cos\theta=6. 2 2 β β r sin ΞΈ + 2 2 β β r cos ΞΈ = 6. Substitute r sin β‘ ΞΈ = y r\sin\theta=y r sin ΞΈ = y r cos β‘ ΞΈ = x r\cos\theta=x r cos ΞΈ = x
2 2 y + 2 2 x = 6. \frac{\sqrt2}{2}y+\frac{\sqrt2}{2}x=6. 2 2 β β y + 2 2 β β x = 6. Multiply by 2 \sqrt2 2 β
x + y = 6 2 . x+y=6\sqrt2. x + y = 6 2 β .
If two points are written in polar form as
P ( r 1 , ΞΈ 1 ) P(r_1,\theta_1) P ( r 1 β , ΞΈ 1 β ) and
Q ( r 2 , ΞΈ 2 ) , Q(r_2,\theta_2), Q ( r 2 β , ΞΈ 2 β ) , then the distance between them is
d = r 1 2 + r 2 2 β 2 r 1 r 2 cos β‘ ( ΞΈ 2 β ΞΈ 1 ) . d=\sqrt{r_1^2+r_2^2-2r_1r_2\cos(\theta_2-\theta_1)}. d = r 1 2 β + r 2 2 β β 2 r 1 β r 2 β cos ( ΞΈ 2 β β ΞΈ 1 β ) β . This comes from the Law of Cosines.
Proof (Distance formula in polar). Draw the two polar points from the pole. Their distances from the pole are r 1 r_1 r 1 β r 2 r_2 r 2 β
The angle between the two segments is
ΞΈ 2 β ΞΈ 1 . \theta_2-\theta_1. ΞΈ 2 β β ΞΈ 1 β . The distance d d d
d 2 = r 1 2 + r 2 2 β 2 r 1 r 2 cos β‘ ( ΞΈ 2 β ΞΈ 1 ) . d^2=r_1^2+r_2^2-2r_1r_2\cos(\theta_2-\theta_1). d 2 = r 1 2 β + r 2 2 β β 2 r 1 β r 2 β cos ( ΞΈ 2 β β ΞΈ 1 β ) . Taking the square root gives
d = r 1 2 + r 2 2 β 2 r 1 r 2 cos β‘ ( ΞΈ 2 β ΞΈ 1 ) . d=\sqrt{r_1^2+r_2^2-2r_1r_2\cos(\theta_2-\theta_1)}. d = r 1 2 β + r 2 2 β β 2 r 1 β r 2 β cos ( ΞΈ 2 β β ΞΈ 1 β ) β .
A circle with center ( r 0 , ΞΈ 0 ) (r_0,\theta_0) ( r 0 β , ΞΈ 0 β ) a a a
a 2 = r 2 + r 0 2 β 2 r r 0 cos β‘ ( ΞΈ β ΞΈ 0 ) . a^2=r^2+r_0^2-2rr_0\cos(\theta-\theta_0). a 2 = r 2 + r 0 2 β β 2 r r 0 β cos ( ΞΈ β ΞΈ 0 β ) . This is the polar distance formula applied to a moving point ( r , ΞΈ ) (r,\theta) ( r , ΞΈ ) ( r 0 , ΞΈ 0 ) (r_0,\theta_0) ( r 0 β , ΞΈ 0 β )
Proof (Circle equation in polar). A point ( r , ΞΈ ) (r,\theta) ( r , ΞΈ ) ( r 0 , ΞΈ 0 ) (r_0,\theta_0) ( r 0 β , ΞΈ 0 β ) a a a a a a
Using the polar distance formula,
a = r 2 + r 0 2 β 2 r r 0 cos β‘ ( ΞΈ β ΞΈ 0 ) . a=\sqrt{r^2+r_0^2-2rr_0\cos(\theta-\theta_0)}. a = r 2 + r 0 2 β β 2 r r 0 β cos ( ΞΈ β ΞΈ 0 β ) β . Squaring both sides gives
a 2 = r 2 + r 0 2 β 2 r r 0 cos β‘ ( ΞΈ β ΞΈ 0 ) . a^2=r^2+r_0^2-2rr_0\cos(\theta-\theta_0). a 2 = r 2 + r 0 2 β β 2 r r 0 β cos ( ΞΈ β ΞΈ 0 β ) .
Alternatively, you can write the circle equation (centered at the origin) as r = a r=a r = a a a a
A line through the pole has the form
ΞΈ = ΞΈ 0 . \theta=\theta_0. ΞΈ = ΞΈ 0 β . A line not passing through the pole can be written as
d = r cos β‘ ( ΞΈ β Ξ± ) , d=r\cos(\theta-\alpha), d = r cos ( ΞΈ β Ξ± ) , where ( d , Ξ± ) (d,\alpha) ( d , Ξ± )
Proof (Polar line equation). Let ( d , Ξ± ) (d,\alpha) ( d , Ξ± )
For any point ( r , ΞΈ ) (r,\theta) ( r , ΞΈ ) ( d , Ξ± ) (d,\alpha) ( d , Ξ± ) ( r , ΞΈ ) (r,\theta) ( r , ΞΈ )
The angle between the segment of length r r r d d d
ΞΈ β Ξ± . \theta-\alpha. ΞΈ β Ξ± . Since d d d r r r
cos β‘ ( ΞΈ β Ξ± ) = d r . \cos(\theta-\alpha)=\frac{d}{r}. cos ( ΞΈ β Ξ± ) = r d β . Multiplying by r r r
d = r cos β‘ ( ΞΈ β Ξ± ) . d=r\cos(\theta-\alpha). d = r cos ( ΞΈ β Ξ± ) .
Example. A line is tangent to the circle
x 2 + y 2 = 36 x^2+y^2=36 x 2 + y 2 = 36 at the point
( β 3 , β 3 3 ) . (-3,-3\sqrt3). ( β 3 , β 3 3 β ) . The circle has radius 6 6 6 6 6 6 4 Ο 3 \frac{4\pi}{3} 3 4 Ο β
( 6 , 4 Ο 3 ) . \left(6,\frac{4\pi}{3}\right). ( 6 , 3 4 Ο β ) . Using the polar line formula,
6 = r cos β‘ ( ΞΈ β 4 Ο 3 ) . 6=r\cos\left(\theta-\frac{4\pi}{3}\right). 6 = r cos ( ΞΈ β 3 4 Ο β ) .
A polar curve is written as
r = f ( ΞΈ ) . r=f(\theta). r = f ( ΞΈ ) . Problem-solving strategy
Make a table of ΞΈ \theta ΞΈ r r r r r r ΞΈ \theta ΞΈ
Plot each polar point.
Watch for negative r r r
Use symmetry and periodicity to complete the curve.
However, it is also very helpful to know some of the most common types of polar curves.
An equation like
r = a ΞΈ r=a\theta r = a ΞΈ creates a spiral. As ΞΈ \theta ΞΈ r r r
Example. Graph r = ΞΈ Ο , r=\frac{\theta}{\pi}, r = Ο ΞΈ β ,
The radius grows as ΞΈ \theta ΞΈ
ΞΈ 0 Ο 2 Ο 3 Ο 2 2 Ο r 0 1 2 1 3 2 2 \begin{array}{c|ccccc}
\theta & 0 & \frac{\pi}{2} & \pi & \frac{3\pi}{2} & 2\pi\\
\hline
r & 0 & \frac12 & 1 & \frac32 & 2
\end{array} ΞΈ r β 0 0 β 2 Ο β 2 1 β β Ο 1 β 2 3 Ο β 2 3 β β 2 Ο 2 β β So the graph spirals outward from the pole.
A graph of the function is shown below:
Β‘ 2 Β‘ 1 1 2 Β‘ 2 Β‘ 1 1 2 x y
Equations of the form
r = a + b sin β‘ ΞΈ r=a+b\sin\theta r = a + b sin ΞΈ or
r = a + b cos β‘ ΞΈ r=a+b\cos\theta r = a + b cos ΞΈ create limacons.
If
β£ a β£ = β£ b β£ , \lvert a \rvert = \lvert b \rvert, β£ a β£ = β£ b β£ , the graph is a cardioid .
If
β£ a β£ < β£ b β£ \lvert a \rvert < \lvert b \rvert β£ a β£ < β£ b β£ the graph has an inner loop.
If the equation uses sine, the main symmetry is usually vertical. If the equation uses cosine, the main symmetry is usually horizontal.
A list of common limacon shapes is shown below (note that a circle is technically a limacon as well):
j a j > j b j : dimple/no lo op j a j = j b j : cardioid j a j < j b j : inner lo op
Example. Graph r = 2 + 4 cos β‘ ΞΈ . r=2+4\cos\theta. r = 2 + 4 cos ΞΈ .
Here a = 2 a=2 a = 2 b = 4 b=4 b = 4
β£ a β£ < β£ b β£ , \lvert a \rvert < \lvert b \rvert, β£ a β£ < β£ b β£ , the graph is a limacon with an inner loop.
Since the equation uses cosine, the graph has symmetry across the polar axis.
A graph of the function is shown below:
Β‘ 2 2 4 6 Β‘ 4 Β‘ 2 2 4 x y
Rose curves have the form
r = a sin β‘ ( n ΞΈ ) r=a\sin(n\theta) r = a sin ( n ΞΈ ) or
r = a cos β‘ ( n ΞΈ ) . r=a\cos(n\theta). r = a cos ( n ΞΈ ) . The number of petals depends on n n n
n NumberΒ ofΒ petals odd n even 2 n \begin{array}{c|c}
n & \text{Number of petals}\\
\hline
\text{odd} & n\\
\text{even} & 2n
\end{array} n odd even β NumberΒ ofΒ petals n 2 n β β The value β£ a β£ \lvert a \rvert β£ a β£
Proof (Rose petal count formula). The helper graph
y = a sin β‘ ( n ΞΈ ) y=a\sin(n\theta) y = a sin ( n ΞΈ ) has period
2 Ο n . \frac{2\pi}{n}. n 2 Ο β . For odd n n n n n n
For even n n n 2 n 2n 2 n
Example. Draw the rose curve r = 4 sin β‘ ( 3 ΞΈ ) r=4\sin(3\theta) r = 4 sin ( 3 ΞΈ )
For this curve, the value of n n n 3 3 3 3 3 3 4 4 4
To draw the rose, start at ΞΈ = 0 \theta=0 ΞΈ = 0 0 , 0 0,0 0 , 0
A graph of the function is shown below:
Β‘ 4 Β‘ 2 2 4 Β‘ 4 Β‘ 2 2 4 x y
Lemniscates are figure-eight shaped curves. Common forms include
r 2 = a 2 cos β‘ ( 2 ΞΈ ) r^2=a^2\cos(2\theta) r 2 = a 2 cos ( 2 ΞΈ ) and
r 2 = a 2 sin β‘ ( 2 ΞΈ ) . r^2=a^2\sin(2\theta). r 2 = a 2 sin ( 2 ΞΈ ) . Because r 2 r^2 r 2
Example. Graph r 2 = 9 sin β‘ ( 2 ΞΈ ) r^2=9\sin(2\theta) r 2 = 9 sin ( 2 ΞΈ )
For this graph, the maximum value of r 2 r^2 r 2 9 9 9 r r r 3 3 3
The graph is a lemniscate. Since it uses sin β‘ ( 2 ΞΈ ) \sin(2\theta) sin ( 2 ΞΈ )
A graph of the lemniscate is shown below:
Β‘ 3 Β‘ 2 Β‘ 1 1 2 3 Β‘ 3 Β‘ 2 Β‘ 1 1 2 3 x y
A complex number has the form
z = a + b i , z=a+bi, z = a + bi , where a a a b b b
i = β 1 . i=\sqrt{-1}. i = β 1 β . The number a a a real part , and b b b imaginary part .
The powers of i i i
i 1 = i , i 2 = β 1 , i 3 = β i , i 4 = 1. i^1=i,\qquad i^2=-1,\qquad i^3=-i,\qquad i^4=1. i 1 = i , i 2 = β 1 , i 3 = β i , i 4 = 1. After that, the pattern repeats every four powers.
Complex numbers can be graphed on the complex plane . The horizontal axis is the real axis, and the vertical axis is the imaginary axis.
The complex number
z = a + b i z=a+bi z = a + bi corresponds to the point
( a , b ) . (a,b). ( a , b ) . The magnitude, or modulus , of z z z
β£ z β£ = a 2 + b 2 . \lvert z\rvert=\sqrt{a^2+b^2}. β£ z β£ = a 2 + b 2 β . This is the distance from the origin to the point ( a , b ) (a,b) ( a , b )
The complex conjugate of
z = a + b i z=a+bi z = a + bi is defined as
z βΎ = a β b i . \overline z=a-bi. z = a β bi . Multiplying a complex number by its conjugate gives
z z βΎ = ( a + b i ) ( a β b i ) = a 2 + b 2 = β£ z β£ 2 . z\overline z=(a+bi)(a-bi)=a^2+b^2=\lvert z\rvert^2. z z = ( a + bi ) ( a β bi ) = a 2 + b 2 = β£ z β£ 2 . To add or subtract complex numbers, combine real parts with real parts and imaginary parts with imaginary parts like vector components:
( a + b i ) + ( c + d i ) = ( a + c ) + ( b + d ) i (a+bi)+(c+di)=(a+c)+(b+d)i ( a + bi ) + ( c + d i ) = ( a + c ) + ( b + d ) i and
( a + b i ) β ( c + d i ) = ( a β c ) + ( b β d ) i . (a+bi)-(c+di)=(a-c)+(b-d)i. ( a + bi ) β ( c + d i ) = ( a β c ) + ( b β d ) i . To multiply, distribute and use i 2 = β 1 i^2=-1 i 2 = β 1
( a + b i ) ( c + d i ) = a c + a d i + b c i + b d i 2 . (a+bi)(c+di)=ac+adi+bci+bdi^2. ( a + bi ) ( c + d i ) = a c + a d i + b c i + b d i 2 . Since i 2 = β 1 i^2=-1 i 2 = β 1
( a + b i ) ( c + d i ) = ( a c β b d ) + ( a d + b c ) i . (a+bi)(c+di)=(ac-bd)+(ad+bc)i. ( a + bi ) ( c + d i ) = ( a c β b d ) + ( a d + b c ) i . To divide complex numbers, multiply the numerator and denominator by the conjugate of the denominator, kind of like rationalizing the denominator.
Example. Simplify
3 + 2 i 1 β 4 i . \frac{3+2i}{1-4i}. 1 β 4 i 3 + 2 i β . Multiply by the conjugate of the denominator:
3 + 2 i 1 β 4 i β
1 + 4 i 1 + 4 i . \frac{3+2i}{1-4i}\cdot \frac{1+4i}{1+4i}. 1 β 4 i 3 + 2 i β β
1 + 4 i 1 + 4 i β . The numerator is
( 3 + 2 i ) ( 1 + 4 i ) = 3 + 12 i + 2 i + 8 i 2 . (3+2i)(1+4i)=3+12i+2i+8i^2. ( 3 + 2 i ) ( 1 + 4 i ) = 3 + 12 i + 2 i + 8 i 2 . Since i 2 = β 1 i^2=-1 i 2 = β 1
3 + 14 i + 8 i 2 = β 5 + 14 i . 3+14i+8i^2=-5+14i. 3 + 14 i + 8 i 2 = β 5 + 14 i . The denominator is
( 1 β 4 i ) ( 1 + 4 i ) = 1 2 + 4 2 = 17. (1-4i)(1+4i)=1^2+4^2=17. ( 1 β 4 i ) ( 1 + 4 i ) = 1 2 + 4 2 = 17. So
3 + 2 i 1 β 4 i = β 5 17 + 14 17 i . \frac{3+2i}{1-4i}=-\frac5{17}+\frac{14}{17}i. 1 β 4 i 3 + 2 i β = β 17 5 β + 17 14 β i .
Since a + b i a+bi a + bi ( a , b ) (a,b) ( a , b )
Let
r = β£ z β£ = a 2 + b 2 . r=\lvert z\rvert=\sqrt{a^2+b^2}. r = β£ z β£ = a 2 + b 2 β . Let ΞΈ \theta ΞΈ
a = r cos β‘ ΞΈ a=r\cos\theta a = r cos ΞΈ and
b = r sin β‘ ΞΈ . b=r\sin\theta. b = r sin ΞΈ . Therefore,
z = a + b i = r cos β‘ ΞΈ + i r sin β‘ ΞΈ . z=a+bi=r\cos\theta+ir\sin\theta. z = a + bi = r cos ΞΈ + i r sin ΞΈ . Factoring out r r r polar form :
z = r ( cos β‘ ΞΈ + i sin β‘ ΞΈ ) . z=r(\cos\theta+i\sin\theta). z = r ( cos ΞΈ + i sin ΞΈ ) . This is sometimes abbreviated as
z = r cis β‘ ΞΈ , z=r\operatorname{cis}\theta, z = r cis ΞΈ , where
cis β‘ ΞΈ = cos β‘ ΞΈ + i sin β‘ ΞΈ . \operatorname{cis}\theta=\cos\theta+i\sin\theta. cis ΞΈ = cos ΞΈ + i sin ΞΈ . Suppose
z 1 = r 1 ( cos β‘ Ξ± + i sin β‘ Ξ± ) z_1=r_1(\cos\alpha+i\sin\alpha) z 1 β = r 1 β ( cos Ξ± + i sin Ξ± ) and
z 2 = r 2 ( cos β‘ Ξ² + i sin β‘ Ξ² ) . z_2=r_2(\cos\beta+i\sin\beta). z 2 β = r 2 β ( cos Ξ² + i sin Ξ² ) . Then
z 1 z 2 = r 1 r 2 ( cos β‘ ( Ξ± + Ξ² ) + i sin β‘ ( Ξ± + Ξ² ) ) . z_1z_2=r_1r_2\left(\cos(\alpha+\beta)+i\sin(\alpha+\beta)\right). z 1 β z 2 β = r 1 β r 2 β ( cos ( Ξ± + Ξ² ) + i sin ( Ξ± + Ξ² ) ) . In words, when multiplying complex numbers in polar form, multiply the moduli and add the angles.
For division,
z 1 z 2 = r 1 r 2 ( cos β‘ ( Ξ± β Ξ² ) + i sin β‘ ( Ξ± β Ξ² ) ) , \frac{z_1}{z_2}
=\frac{r_1}{r_2}\left(\cos(\alpha-\beta)+i\sin(\alpha-\beta)\right), z 2 β z 1 β β = r 2 β r 1 β β ( cos ( Ξ± β Ξ² ) + i sin ( Ξ± β Ξ² ) ) , where z 2 β 0 z_2\ne 0 z 2 β ξ = 0
In words, when dividing, divide the moduli and subtract the angles.
These two operations show why polar form for complex numbers is sometimes preferred.
Proof (Polar multiplication formula). Multiply directly:
z 1 z 2 = r 1 r 2 ( cos β‘ Ξ± + i sin β‘ Ξ± ) ( cos β‘ Ξ² + i sin β‘ Ξ² ) . z_1z_2=r_1r_2(\cos\alpha+i\sin\alpha)(\cos\beta+i\sin\beta). z 1 β z 2 β = r 1 β r 2 β ( cos Ξ± + i sin Ξ± ) ( cos Ξ² + i sin Ξ² ) . Distribute:
z 1 z 2 = r 1 r 2 ( cos β‘ Ξ± cos β‘ Ξ² + i cos β‘ Ξ± sin β‘ Ξ² + i sin β‘ Ξ± cos β‘ Ξ² + i 2 sin β‘ Ξ± sin β‘ Ξ² ) . z_1z_2=r_1r_2(\cos\alpha\cos\beta+i\cos\alpha\sin\beta+i\sin\alpha\cos\beta+i^2\sin\alpha\sin\beta). z 1 β z 2 β = r 1 β r 2 β ( cos Ξ± cos Ξ² + i cos Ξ± sin Ξ² + i sin Ξ± cos Ξ² + i 2 sin Ξ± sin Ξ² ) . Since i 2 = β 1 i^2=-1 i 2 = β 1
z 1 z 2 = r 1 r 2 ( ( cos β‘ Ξ± cos β‘ Ξ² β sin β‘ Ξ± sin β‘ Ξ² ) + i ( sin β‘ Ξ± cos β‘ Ξ² + cos β‘ Ξ± sin β‘ Ξ² ) ) . z_1z_2=r_1r_2\left((\cos\alpha\cos\beta-\sin\alpha\sin\beta)+i(\sin\alpha\cos\beta+\cos\alpha\sin\beta)\right). z 1 β z 2 β = r 1 β r 2 β ( ( cos Ξ± cos Ξ² β sin Ξ± sin Ξ² ) + i ( sin Ξ± cos Ξ² + cos Ξ± sin Ξ² ) ) . Using the angle addition identities,
cos β‘ ( Ξ± + Ξ² ) = cos β‘ Ξ± cos β‘ Ξ² β sin β‘ Ξ± sin β‘ Ξ² \cos(\alpha+\beta)=\cos\alpha\cos\beta-\sin\alpha\sin\beta cos ( Ξ± + Ξ² ) = cos Ξ± cos Ξ² β sin Ξ± sin Ξ² and
sin β‘ ( Ξ± + Ξ² ) = sin β‘ Ξ± cos β‘ Ξ² + cos β‘ Ξ± sin β‘ Ξ² , \sin(\alpha+\beta)=\sin\alpha\cos\beta+\cos\alpha\sin\beta, sin ( Ξ± + Ξ² ) = sin Ξ± cos Ξ² + cos Ξ± sin Ξ² , we get
z 1 z 2 = r 1 r 2 ( cos β‘ ( Ξ± + Ξ² ) + i sin β‘ ( Ξ± + Ξ² ) ) . z_1z_2=r_1r_2\left(\cos(\alpha+\beta)+i\sin(\alpha+\beta)\right). z 1 β z 2 β = r 1 β r 2 β ( cos ( Ξ± + Ξ² ) + i sin ( Ξ± + Ξ² ) ) .
Extension. Derive the division formula for two complex numbers.
Example. Multiply
z 1 = 3 ( cos β‘ Ο 6 + i sin β‘ Ο 6 ) z_1=3\left(\cos\frac{\pi}{6}+i\sin\frac{\pi}{6}\right) z 1 β = 3 ( cos 6 Ο β + i sin 6 Ο β ) and
z 2 = 2 ( cos β‘ Ο 3 + i sin β‘ Ο 3 ) . z_2=2\left(\cos\frac{\pi}{3}+i\sin\frac{\pi}{3}\right). z 2 β = 2 ( cos 3 Ο β + i sin 3 Ο β ) . Multiply the moduli and add the angles:
z 1 z 2 = 6 ( cos β‘ Ο 2 + i sin β‘ Ο 2 ) = 6 i . z_1z_2=6\left(\cos\frac{\pi}{2}+i\sin\frac{\pi}{2}\right)=6i. z 1 β z 2 β = 6 ( cos 2 Ο β + i sin 2 Ο β ) = 6 i .
Eulerβs formula states that
e i ΞΈ = cos β‘ ΞΈ + i sin β‘ ΞΈ e^{i\theta}=\cos\theta+i\sin\theta e i ΞΈ = cos ΞΈ + i sin ΞΈ A proof of this does require calculus/linear algebra and wonβt be discussed here.
Therefore, the polar form
z = r ( cos β‘ ΞΈ + i sin β‘ ΞΈ ) z=r(\cos\theta+i\sin\theta) z = r ( cos ΞΈ + i sin ΞΈ ) can also be written in exponential form :
z = r e i ΞΈ . z=re^{i\theta}. z = r e i ΞΈ . This means the following forms are equivalent:
a + b i = r ( cos β‘ ΞΈ + i sin β‘ ΞΈ ) = r cis β‘ ΞΈ = r e i ΞΈ . a+bi=r(\cos\theta+i\sin\theta)=r\operatorname{cis}\theta=re^{i\theta}. a + bi = r ( cos ΞΈ + i sin ΞΈ ) = r cis ΞΈ = r e i ΞΈ . The exponential form is especially useful because it makes multiplication, division, and powers look like normal exponent rules.
For example,
( r 1 e i Ξ± ) ( r 2 e i Ξ² ) = r 1 r 2 e i ( Ξ± + Ξ² ) . (r_1e^{i\alpha})(r_2e^{i\beta})=r_1r_2e^{i(\alpha+\beta)}. ( r 1 β e i Ξ± ) ( r 2 β e i Ξ² ) = r 1 β r 2 β e i ( Ξ± + Ξ² ) . For division,
r 1 e i Ξ± r 2 e i Ξ² = r 1 r 2 e i ( Ξ± β Ξ² ) . \frac{r_1e^{i\alpha}}{r_2e^{i\beta}}=\frac{r_1}{r_2}e^{i(\alpha-\beta)}. r 2 β e i Ξ² r 1 β e i Ξ± β = r 2 β r 1 β β e i ( Ξ± β Ξ² ) . For powers,
( r e i ΞΈ ) n = r n e i n ΞΈ . (re^{i\theta})^n=r^ne^{in\theta}. ( r e i ΞΈ ) n = r n e in ΞΈ . These are the same rules as polar form, just written more compactly.
Example. Write
z = β 2 + 2 i z=-2+2i z = β 2 + 2 i in exponential form.
First find the modulus:
r = ( β 2 ) 2 + 2 2 = 2 2 . r=\sqrt{(-2)^2+2^2}=2\sqrt2. r = ( β 2 ) 2 + 2 2 β = 2 2 β . The point ( β 2 , 2 ) (-2,2) ( β 2 , 2 ) Ο 4 \frac{\pi}{4} 4 Ο β
ΞΈ = 3 Ο 4 . \theta=\frac{3\pi}{4}. ΞΈ = 4 3 Ο β . So
z = 2 2 e i 3 Ο / 4 . z=2\sqrt2 e^{i3\pi/4}. z = 2 2 β e i 3 Ο /4 .
Example. Write
z = β 1 + 3 i z=-1+\sqrt3i z = β 1 + 3 β i in polar form.
First find the modulus:
r = ( β 1 ) 2 + ( 3 ) 2 = 2. r=\sqrt{(-1)^2+(\sqrt3)^2}=2. r = ( β 1 ) 2 + ( 3 β ) 2 β = 2. The point ( β 1 , 3 ) (-1,\sqrt3) ( β 1 , 3 β ) Ο 3 \frac{\pi}{3} 3 Ο β
ΞΈ = 2 Ο 3 . \theta=\frac{2\pi}{3}. ΞΈ = 3 2 Ο β . Thus,
z = 2 ( cos β‘ 2 Ο 3 + i sin β‘ 2 Ο 3 ) . z=2\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right). z = 2 ( cos 3 2 Ο β + i sin 3 2 Ο β ) .
Key info De Moivre's Theorem
If z = r ( cos β‘ ΞΈ + i sin β‘ ΞΈ ) z=r(\cos\theta+i\sin\theta) z = r ( cos ΞΈ + i sin ΞΈ ) n n n z n = r n ( cos β‘ ( n ΞΈ ) + i sin β‘ ( n ΞΈ ) ) z^n=r^n(\cos(n\theta)+i\sin(n\theta)) z n = r n ( cos ( n ΞΈ ) + i sin ( n ΞΈ ))
Proof (De Moivreβs Theorem). Multiplying complex numbers in polar form multiplies their moduli and adds their angles.
If
z = r ( cos β‘ ΞΈ + i sin β‘ ΞΈ ) , z=r(\cos\theta+i\sin\theta), z = r ( cos ΞΈ + i sin ΞΈ ) , then
z 2 = r β
r β
( cos β‘ ( ΞΈ + ΞΈ ) + i sin β‘ ( ΞΈ + ΞΈ ) ) = r 2 ( cos β‘ ( 2 ΞΈ ) + i sin β‘ ( 2 ΞΈ ) ) . z^2=r\cdot r \cdot (\cos(\theta + \theta) + i\sin(\theta + \theta)) = r^2(\cos(2\theta)+i\sin(2\theta)). z 2 = r β
r β
( cos ( ΞΈ + ΞΈ ) + i sin ( ΞΈ + ΞΈ )) = r 2 ( cos ( 2 ΞΈ ) + i sin ( 2 ΞΈ )) . Multiplying by another copy of z z z
z 3 = r 3 ( cos β‘ ( 3 ΞΈ ) + i sin β‘ ( 3 ΞΈ ) ) . z^3=r^3(\cos(3\theta)+i\sin(3\theta)). z 3 = r 3 ( cos ( 3 ΞΈ ) + i sin ( 3 ΞΈ )) . Repeating this process n n n
z n = r n ( cos β‘ ( n ΞΈ ) + i sin β‘ ( n ΞΈ ) ) , z^n=r^n(\cos(n\theta)+i\sin(n\theta)), z n = r n ( cos ( n ΞΈ ) + i sin ( n ΞΈ )) , which is De Moivreβs Theorem. To reverse De Moivreβs Theorem (e.g. solve for roots of a complex number), just swap n n n 1 n \frac{1}{n} n 1 β
Example. Find
( 1 + i ) 6 . (1+i)^6. ( 1 + i ) 6 . First write 1 + i 1+i 1 + i
1 + i = 2 ( cos β‘ Ο 4 + i sin β‘ Ο 4 ) . 1+i=\sqrt2\left(\cos\frac{\pi}{4}+i\sin\frac{\pi}{4}\right). 1 + i = 2 β ( cos 4 Ο β + i sin 4 Ο β ) . By De Moivreβs Theorem,
( 1 + i ) 6 = ( 2 ) 6 ( cos β‘ 6 Ο 4 + i sin β‘ 6 Ο 4 ) . (1+i)^6=(\sqrt2)^6\left(\cos\frac{6\pi}{4}+i\sin\frac{6\pi}{4}\right). ( 1 + i ) 6 = ( 2 β ) 6 ( cos 4 6 Ο β + i sin 4 6 Ο β ) . Since
( 2 ) 6 = 8 (\sqrt2)^6=8 ( 2 β ) 6 = 8 and
6 Ο 4 = 3 Ο 2 , \frac{6\pi}{4}=\frac{3\pi}{2}, 4 6 Ο β = 2 3 Ο β , we get
( 1 + i ) 6 = 8 ( cos β‘ 3 Ο 2 + i sin β‘ 3 Ο 2 ) = β 8 i . (1+i)^6=8\left(\cos\frac{3\pi}{2}+i\sin\frac{3\pi}{2}\right)=-8i. ( 1 + i ) 6 = 8 ( cos 2 3 Ο β + i sin 2 3 Ο β ) = β 8 i .
To solve
w n = z , w^n=z, w n = z , write z z z
z = r ( cos β‘ ΞΈ + i sin β‘ ΞΈ ) . z=r(\cos\theta+i\sin\theta). z = r ( cos ΞΈ + i sin ΞΈ ) . Since angles repeat every 2 Ο 2\pi 2 Ο
z = r ( cos β‘ ( ΞΈ + 2 k Ο ) + i sin β‘ ( ΞΈ + 2 k Ο ) ) . z=r(\cos(\theta+2k\pi)+i\sin(\theta+2k\pi)). z = r ( cos ( ΞΈ + 2 k Ο ) + i sin ( ΞΈ + 2 k Ο )) . The n n n
w k = r n ( cos β‘ ΞΈ + 2 k Ο n + i sin β‘ ΞΈ + 2 k Ο n ) , w_k=\sqrt[n]{r}\left(\cos\frac{\theta+2k\pi}{n}+i\sin\frac{\theta+2k\pi}{n}\right), w k β = n r β ( cos n ΞΈ + 2 k Ο β + i sin n ΞΈ + 2 k Ο β ) , where
k = 0 , 1 , 2 , β¦ , n β 1. k=0,1,2,\ldots,n-1. k = 0 , 1 , 2 , β¦ , n β 1. This is just a classic example of De Moivreβs Theorem.
Example. Find the cube roots of 8 8 8
Write
8 = 8 ( cos β‘ 0 + i sin β‘ 0 ) . 8=8(\cos0+i\sin0). 8 = 8 ( cos 0 + i sin 0 ) . The cube roots are
w k = 2 ( cos β‘ 0 + 2 k Ο 3 + i sin β‘ 0 + 2 k Ο 3 ) , w_k=2\left(\cos\frac{0+2k\pi}{3}+i\sin\frac{0+2k\pi}{3}\right), w k β = 2 ( cos 3 0 + 2 k Ο β + i sin 3 0 + 2 k Ο β ) , where k = 0 , 1 , 2 k=0,1,2 k = 0 , 1 , 2
So the roots are
2 , 2, 2 , 2 ( cos β‘ 2 Ο 3 + i sin β‘ 2 Ο 3 ) = β 1 + 3 i , 2\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right)=-1+\sqrt3i, 2 ( cos 3 2 Ο β + i sin 3 2 Ο β ) = β 1 + 3 β i , and
2 ( cos β‘ 4 Ο 3 + i sin β‘ 4 Ο 3 ) = β 1 β 3 i . 2\left(\cos\frac{4\pi}{3}+i\sin\frac{4\pi}{3}\right)=-1-\sqrt3i. 2 ( cos 3 4 Ο β + i sin 3 4 Ο β ) = β 1 β 3 β i .
The n n n
z n = 1. z^n=1. z n = 1. Since
1 = cos β‘ 0 + i sin β‘ 0 , 1=\cos0+i\sin0, 1 = cos 0 + i sin 0 , and angles repeat every 2 Ο 2\pi 2 Ο
1 = cos β‘ ( 2 k Ο ) + i sin β‘ ( 2 k Ο ) 1=\cos(2k\pi)+i\sin(2k\pi) 1 = cos ( 2 k Ο ) + i sin ( 2 k Ο ) for any integer k k k
Using the complex-root formula, the n n n
z k = cos β‘ ( 2 k Ο n ) + i sin β‘ ( 2 k Ο n ) , z_k=\cos\left(\frac{2k\pi}{n}\right)+i\sin\left(\frac{2k\pi}{n}\right), z k β = cos ( n 2 k Ο β ) + i sin ( n 2 k Ο β ) , where
k = 0 , 1 , 2 , β¦ , n β 1. k=0,1,2,\ldots,n-1. k = 0 , 1 , 2 , β¦ , n β 1. In exponential form,
z k = e 2 k Ο i / n . z_k=e^{2k\pi i/n}. z k β = e 2 k Ο i / n . These points all lie on the unit circle because each one has modulus 1 1 1 n n n
The n n n
The angle between consecutive roots is
2 Ο n . \frac{2\pi}{n}. n 2 Ο β . So the roots form the vertices of a regular n n n
For example:
The 3 3 3
The 4 4 4
The 5 5 5
Example. Find the fourth roots of unity.
Use
z k = e 2 k Ο i / 4 z_k=e^{2k\pi i/4} z k β = e 2 k Ο i /4 for
k = 0 , 1 , 2 , 3. k=0,1,2,3. k = 0 , 1 , 2 , 3. Then the roots are
z 0 = e 0 = 1 , z_0=e^0=1, z 0 β = e 0 = 1 , z 1 = e Ο i / 2 = i , z_1=e^{\pi i/2}=i, z 1 β = e Ο i /2 = i , z 2 = e Ο i = β 1 , z_2=e^{\pi i}=-1, z 2 β = e Ο i = β 1 , and
z 3 = e 3 Ο i / 2 = β i . z_3=e^{3\pi i/2}=-i. z 3 β = e 3 Ο i /2 = β i . So the fourth roots of unity are
1 , Β i , Β β 1 , Β β i . 1,\ i,\ -1,\ -i. 1 , Β i , Β β 1 , Β β i . Geometrically, the solutions look like this:
real imaginary 1 i Β‘ 1 Β‘ i fourth ro ots of unit y on the unit circle Note that the rendering on the y y y
For n > 1 n>1 n > 1 n n n
1 + Ο + Ο 2 + β― + Ο n β 1 = 0. 1+\omega+\omega^2+\cdots+\omega^{n-1}=0. 1 + Ο + Ο 2 + β― + Ο n β 1 = 0. There are two ways to understand this.
Geometrically, the roots are evenly spaced on the unit circle, so their vectors balance perfectly around the origin.
Algebraically, use the finite geometric series formula (Unit 13 & 14: Additional Topics in Algebra ):
1 + Ο + Ο 2 + β― + Ο n β 1 = 1 β Ο n 1 β Ο . 1+\omega+\omega^2+\cdots+\omega^{n-1}
=\frac{1-\omega^n}{1-\omega}. 1 + Ο + Ο 2 + β― + Ο n β 1 = 1 β Ο 1 β Ο n β . Since Ο n = 1 \omega^n=1 Ο n = 1 Ο β 1 \omega\ne1 Ο ξ = 1
1 β Ο n 1 β Ο = 1 β 1 1 β Ο = 0. \frac{1-\omega^n}{1-\omega}
=\frac{1-1}{1-\omega}=0. 1 β Ο 1 β Ο n β = 1 β Ο 1 β 1 β = 0. Additionally, the product of the n n n ( β 1 ) n + 1 (-1)^{n+1} ( β 1 ) n + 1
Extension. Prove that the product of the n n n ( β 1 ) n + 1 (-1)^{n+1} ( β 1 ) n + 1
Roots of unity help organize roots of any nonzero complex number.
If w w w n n n z z z
w , w Ο , w Ο 2 , β¦ , w Ο n β 1 , w,\quad w\omega,\quad w\omega^2,\quad \ldots,\quad w\omega^{n-1}, w , w Ο , w Ο 2 , β¦ , w Ο n β 1 , where
Ο = e 2 Ο i / n . \omega=e^{2\pi i/n}. Ο = e 2 Ο i / n . Multiplying by Ο \omega Ο
2 Ο n \frac{2\pi}{n} n 2 Ο β without changing its magnitude.
So the n n n n n n 0 0 0 ( β 1 ) n + 1 β£ z β£ (-1)^{n+1}\lvert z \rvert ( β 1 ) n + 1 β£ z β£
Solve the triangle with A = 60 β A=60^\circ A = 6 0 β a = 9 a=9 a = 9 b = 10 b=10 b = 10
By the Law of Sines,
sin β‘ B 10 = sin β‘ 60 β 9 , \frac{\sin B}{10}=\frac{\sin 60^\circ}{9}, 10 sin B β = 9 sin 6 0 β β , so
sin β‘ B = 10 sin β‘ 60 β 9 = 5 3 9 . \sin B=\frac{10\sin 60^\circ}{9}=\frac{5\sqrt3}{9}. sin B = 9 10 sin 6 0 β β = 9 5 3 β β . This gives two possible angles:
B β 74.2 β or B β 105.8 β . B\approx 74.2^\circ\quad \text{or}\quad B\approx 105.8^\circ. B β 74. 2 β or B β 105. 8 β . If B β 74.2 β B\approx 74.2^\circ B β 74. 2 β
C β 45.8 β , c β 7.45 Β cm . C\approx 45.8^\circ,\qquad c\approx 7.45\text{ cm}. C β 45. 8 β , c β 7.45 Β cm . If B β 105.8 β B\approx 105.8^\circ B β 105. 8 β
C β 14.2 β , c β 2.55 Β cm . C\approx 14.2^\circ,\qquad c\approx 2.55\text{ cm}. C β 14. 2 β , c β 2.55 Β cm . Therefore, there are two possible triangles .
A plane is flying above the ocean. The angle of depression to a submarine is 24 β 24^\circ 2 4 β 17 β 17^\circ 1 7 β 5120 5120 5120
The plane-to-ship distance is 5120 5120 5120 17 β 17^\circ 1 7 β
h = 5120 sin β‘ 17 β β 1496.94 Β ft . h=5120\sin 17^\circ\approx 1496.94\text{ ft}. h = 5120 sin 1 7 β β 1496.94 Β ft . The horizontal distance from the plane to the ship is
5120 cos β‘ 17 β β 4896.28 Β ft . 5120\cos 17^\circ\approx 4896.28\text{ ft}. 5120 cos 1 7 β β 4896.28 Β ft . The horizontal distance from the plane to the submarine is
h tan β‘ 24 β β 3362.19 Β ft . \frac{h}{\tan 24^\circ}\approx 3362.19\text{ ft}. tan 2 4 β h β β 3362.19 Β ft . Assuming the ship and submarine are on the same side of the plane, their distance apart is
4896.28 β 3362.19 β 1534.09 Β ft . 4896.28-3362.19\approx 1534.09\text{ ft}. 4896.28 β 3362.19 β 1534.09 Β ft .
Two towns A A A B B B 1.4 1.4 1.4 B B B A A A S 22 β E S22^\circ E S 2 2 β E A A A S 43 β W S43^\circ W S 4 3 β W B B B
Place town A A A ( 0 , 0 ) (0,0) ( 0 , 0 ) B B B ( 1.4 , 0 ) (1.4,0) ( 1.4 , 0 ) S 22 β E S22^\circ E S 2 2 β E A A A S 43 β W S43^\circ W S 4 3 β W B B B
Solving the two bearing lines gives the signal at approximately
( 0.423 , β 1.047 ) . (0.423,-1.047). ( 0.423 , β 1.047 ) . Therefore,
A S β 1.13 Β miles , B S β 1.43 Β miles . AS\approx 1.13\text{ miles},\qquad BS\approx 1.43\text{ miles}. A S β 1.13 Β miles , B S β 1.43 Β miles .
Prove that for any triangle with side lengths a , b , c a,b,c a , b , c s = 1 2 ( a + b + c ) s=\frac12(a+b+c) s = 2 1 β ( a + b + c ) sin β‘ 2 ( C 2 ) = ( s β a ) ( s β b ) a b . \sin^2\left(\frac C2\right)=\frac{(s-a)(s-b)}{ab}. sin 2 ( 2 C β ) = ab ( s β a ) ( s β b ) β .
Use the half-angle identity:
sin β‘ 2 ( C 2 ) = 1 β cos β‘ C 2 . \sin^2\left(\frac C2\right)=\frac{1-\cos C}{2}. sin 2 ( 2 C β ) = 2 1 β cos C β . By the Law of Cosines,
cos β‘ C = a 2 + b 2 β c 2 2 a b . \cos C=\frac{a^2+b^2-c^2}{2ab}. cos C = 2 ab a 2 + b 2 β c 2 β . Substitute:
sin β‘ 2 ( C 2 ) = 1 2 ( 1 β a 2 + b 2 β c 2 2 a b ) = c 2 β ( a β b ) 2 4 a b . \sin^2\left(\frac C2\right)
=\frac{1}{2}\left(1-\frac{a^2+b^2-c^2}{2ab}\right)
=\frac{c^2-(a-b)^2}{4ab}. sin 2 ( 2 C β ) = 2 1 β ( 1 β 2 ab a 2 + b 2 β c 2 β ) = 4 ab c 2 β ( a β b ) 2 β . Factor the numerator:
c 2 β ( a β b ) 2 = ( c β a + b ) ( c + a β b ) . c^2-(a-b)^2=(c-a+b)(c+a-b). c 2 β ( a β b ) 2 = ( c β a + b ) ( c + a β b ) . Since s = 1 2 ( a + b + c ) s=\frac12(a+b+c) s = 2 1 β ( a + b + c )
c β a + b = 2 ( s β a ) , c + a β b = 2 ( s β b ) . c-a+b=2(s-a),\qquad c+a-b=2(s-b). c β a + b = 2 ( s β a ) , c + a β b = 2 ( s β b ) . Thus,
sin β‘ 2 ( C 2 ) = ( s β a ) ( s β b ) a b . \sin^2\left(\frac C2\right)=\frac{(s-a)(s-b)}{ab}. sin 2 ( 2 C β ) = ab ( s β a ) ( s β b ) β .
Let u = β¨ β 2 , 1 β© \mathbf u=\langle -2,1\rangle u = β¨ β 2 , 1 β© v = β¨ β 5 , 3 β© \mathbf v=\langle -5,3\rangle v = β¨ β 5 , 3 β©
( A ) (A) ( A ) 2 u β 3 v 2\mathbf u-3\mathbf v 2 u β 3 v
( B ) (B) ( B ) u \mathbf u u v \mathbf v v
( C ) (C) ( C ) u + v \mathbf u+\mathbf v u + v
We have
u = β¨ β 2 , 1 β© , v = β¨ β 5 , 3 β© . \mathbf u=\langle -2,1\rangle,\qquad \mathbf v=\langle -5,3\rangle. u = β¨ β 2 , 1 β© , v = β¨ β 5 , 3 β© . For part (A),
2 u β 3 v = β¨ β 4 , 2 β© β β¨ β 15 , 9 β© = β¨ 11 , β 7 β© . 2\mathbf u-3\mathbf v
=\langle -4,2\rangle-\langle -15,9\rangle
=\langle 11,-7\rangle. 2 u β 3 v = β¨ β 4 , 2 β© β β¨ β 15 , 9 β© = β¨ 11 , β 7 β© . For part (B),
u β
v = ( β 2 ) ( β 5 ) + ( 1 ) ( 3 ) = 13. \mathbf u\cdot \mathbf v=(-2)(-5)+(1)(3)=13. u β
v = ( β 2 ) ( β 5 ) + ( 1 ) ( 3 ) = 13. Also,
β£ u β£ = 5 , β£ v β£ = 34 . \lvert \mathbf u \rvert=\sqrt5,\qquad \lvert \mathbf v \rvert=\sqrt{34}. β£ u β£ = 5 β , β£ v β£ = 34 β . So
cos β‘ ΞΈ = 13 5 34 = 13 170 . \cos\theta=\frac{13}{\sqrt5\sqrt{34}}=\frac{13}{\sqrt{170}}. cos ΞΈ = 5 β 34 β 13 β = 170 β 13 β . Therefore,
ΞΈ β 4 β . \theta\approx 4^\circ. ΞΈ β 4 β . For part (C), since
u + v = β¨ β 7 , 4 β© , \mathbf u+\mathbf v=\langle -7,4\rangle, u + v = β¨ β 7 , 4 β© , a unit vector in that direction is
β¨ β 7 , 4 β© 65 = β¨ β 7 65 , 4 65 β© . \frac{\langle -7,4\rangle}{\sqrt{65}}
=\left\langle \frac{-7}{\sqrt{65}},\frac{4}{\sqrt{65}}\right\rangle. 65 β β¨ β 7 , 4 β© β = β¨ 65 β β 7 β , 65 β 4 β β© .
A plane is heading 40 β 40^\circ 4 0 β 120 120 120 10 10 10
Let east be positive x x x y y y 40 β 40^\circ 4 0 β 120 120 120
β¨ 120 sin β‘ 40 β , 120 cos β‘ 40 β β© . \langle 120\sin40^\circ,120\cos40^\circ\rangle. β¨ 120 sin 4 0 β , 120 cos 4 0 β β© . A wind directly from the east blows west, so its vector is
β¨ β 10 , 0 β© . \langle -10,0\rangle. β¨ β 10 , 0 β© . The ground-speed vector is
β¨ 120 sin β‘ 40 β β 10 , 120 cos β‘ 40 β β© β β¨ 67.13 , 91.93 β© . \langle 120\sin40^\circ-10,120\cos40^\circ\rangle
\approx \langle 67.13,91.93\rangle. β¨ 120 sin 4 0 β β 10 , 120 cos 4 0 β β© β β¨ 67.13 , 91.93 β© . The ground speed is
67.13 2 + 91.93 2 β 113.83 Β mph . \sqrt{67.13^2+91.93^2}\approx 113.83\text{ mph}. 67.1 3 2 + 91.9 3 2 β β 113.83 Β mph . The actual direction is about
tan β‘ β 1 ( 67.13 91.93 ) β 36.1 β \tan^{-1}\left(\frac{67.13}{91.93}\right)\approx 36.1^\circ tan β 1 ( 91.93 67.13 β ) β 36. 1 β east of north. Since the intended heading was 40 β 40^\circ 4 0 β
40 β β 36.1 β = 3.9 β 40^\circ-36.1^\circ=3.9^\circ 4 0 β β 36. 1 β = 3. 9 β west of the intended heading.
A force of 18 18 18 235 β 235^\circ 23 5 β x x x β¨ 12 , β 5 β© \langle 12,-5\rangle β¨ 12 , β 5 β©
The force components are
β¨ 18 cos β‘ 235 β , 18 sin β‘ 235 β β© β β¨ β 10.32 , β 14.74 β© . \langle 18\cos235^\circ,18\sin235^\circ\rangle
\approx \langle -10.32,-14.74\rangle. β¨ 18 cos 23 5 β , 18 sin 23 5 β β© β β¨ β 10.32 , β 14.74 β© . Add β¨ 12 , β 5 β© \langle 12,-5\rangle β¨ 12 , β 5 β©
β¨ β 10.32 , β 14.74 β© + β¨ 12 , β 5 β© β β¨ 1.68 , β 19.74 β© . \langle -10.32,-14.74\rangle+\langle 12,-5\rangle
\approx \langle 1.68,-19.74\rangle. β¨ β 10.32 , β 14.74 β© + β¨ 12 , β 5 β© β β¨ 1.68 , β 19.74 β© . The magnitude is
1.68 2 + ( β 19.74 ) 2 β 19.82. \sqrt{1.68^2+(-19.74)^2}\approx 19.82. 1.6 8 2 + ( β 19.74 ) 2 β β 19.82. The direction angle is approximately
274.9 β . 274.9^\circ. 274. 9 β .
A particle moves according to x ( t ) = 2 cos β‘ t β sin β‘ ( 2 t ) , y ( t ) = 6 sin β‘ t , x(t)=2\cos t-\sin(2t),\qquad y(t)=6\sin t, x ( t ) = 2 cos t β sin ( 2 t ) , y ( t ) = 6 sin t , 0 β€ t β€ 2 Ο 0\le t\le 2\pi 0 β€ t β€ 2 Ο
( A ) (A) ( A ) x x x
( B ) (B) ( B ) t = Ο 2 t=\frac{\pi}{2} t = 2 Ο β t = 7 Ο 6 t=\frac{7\pi}{6} t = 6 7 Ο β
( C ) (C) ( C ) t t t
The particle has
x ( t ) = 2 cos β‘ t β sin β‘ ( 2 t ) , y ( t ) = 6 sin β‘ t . x(t)=2\cos t-\sin(2t),\qquad y(t)=6\sin t. x ( t ) = 2 cos t β sin ( 2 t ) , y ( t ) = 6 sin t . For part (A), use sin β‘ ( 2 t ) = 2 sin β‘ t cos β‘ t \sin(2t)=2\sin t\cos t sin ( 2 t ) = 2 sin t cos t
x ( t ) = 2 cos β‘ t β 2 sin β‘ t cos β‘ t = 2 cos β‘ t ( 1 β sin β‘ t ) . x(t)=2\cos t-2\sin t\cos t=2\cos t(1-\sin t). x ( t ) = 2 cos t β 2 sin t cos t = 2 cos t ( 1 β sin t ) . So x ( t ) = 0 x(t)=0 x ( t ) = 0
cos β‘ t = 0 or sin β‘ t = 1. \cos t=0\quad \text{or}\quad \sin t=1. cos t = 0 or sin t = 1. On 0 β€ t β€ 2 Ο 0\le t\le 2\pi 0 β€ t β€ 2 Ο
t = Ο 2 , 3 Ο 2 . t=\frac{\pi}{2},\frac{3\pi}{2}. t = 2 Ο β , 2 3 Ο β . The exact x x x
( 0 , 6 ) and ( 0 , β 6 ) . (0,6)\quad \text{and}\quad (0,-6). ( 0 , 6 ) and ( 0 , β 6 ) . For part (B), when t = Ο 2 t=\frac{\pi}{2} t = 2 Ο β
( x , y ) = ( 0 , 6 ) . (x,y)=(0,6). ( x , y ) = ( 0 , 6 ) . When t = 7 Ο 6 t=\frac{7\pi}{6} t = 6 7 Ο β
x = 2 cos β‘ 7 Ο 6 β sin β‘ 7 Ο 3 = β 3 β 3 2 = β 3 3 2 , x=2\cos\frac{7\pi}{6}-\sin\frac{7\pi}{3}
=-\sqrt3-\frac{\sqrt3}{2}
=-\frac{3\sqrt3}{2}, x = 2 cos 6 7 Ο β β sin 3 7 Ο β = β 3 β β 2 3 β β = β 2 3 3 β β , and
y = 6 sin β‘ 7 Ο 6 = β 3. y=6\sin\frac{7\pi}{6}=-3. y = 6 sin 6 7 Ο β = β 3. So the position is
( β 3 3 2 , β 3 ) . \left(-\frac{3\sqrt3}{2},-3\right). ( β 2 3 3 β β , β 3 ) . For part (C), the distance from the origin is
d ( t ) = x ( t ) 2 + y ( t ) 2 . d(t)=\sqrt{x(t)^2+y(t)^2}. d ( t ) = x ( t ) 2 + y ( t ) 2 β . Therefore,
d ( t ) = ( 2 cos β‘ t β sin β‘ ( 2 t ) ) 2 + 36 sin β‘ 2 t . d(t)=\sqrt{(2\cos t-\sin(2t))^2+36\sin^2t}. d ( t ) = ( 2 cos t β sin ( 2 t ) ) 2 + 36 sin 2 t β .
Eliminate the parameter and describe the curve, including any domain restrictions and orientation: x = e t x=e^t x = e t y = e 2 t β 3 y=e^{2t}-3 y = e 2 t β 3 β ln β‘ 2 β€ t β€ ln β‘ 3. -\ln2\le t\le \ln3. β ln 2 β€ t β€ ln 3.
Since
x = e t , x=e^t, x = e t , we have
t = ln β‘ x . t=\ln x. t = ln x . Then
y = e 2 t β 3 = ( e t ) 2 β 3 = x 2 β 3. y=e^{2t}-3=(e^t)^2-3=x^2-3. y = e 2 t β 3 = ( e t ) 2 β 3 = x 2 β 3. Because
β ln β‘ 2 β€ t β€ ln β‘ 3 , -\ln2\le t\le \ln3, β ln 2 β€ t β€ ln 3 , the domain is
1 2 β€ x β€ 3. \frac12\le x\le 3. 2 1 β β€ x β€ 3. The curve is the parabola
y = x 2 β 3 y=x^2-3 y = x 2 β 3 restricted to 1 2 β€ x β€ 3 \frac12\le x\le 3 2 1 β β€ x β€ 3 t t t x = e t x=e^t x = e t
( 1 2 , β 11 4 ) \left(\frac12,-\frac{11}{4}\right) ( 2 1 β , β 4 11 β ) to
( 3 , 6 ) . (3,6). ( 3 , 6 ) .
The curve x = 4 cos β‘ t x=4\cos t x = 4 cos t y = β 2 sin β‘ t y=-2\sin t y = β 2 sin t 0 β€ t β€ 2 Ο 0\le t\le 2\pi 0 β€ t β€ 2 Ο
From
x = 4 cos β‘ t , x=4\cos t, x = 4 cos t , we get
cos β‘ t = x 4 . \cos t=\frac{x}{4}. cos t = 4 x β . From
y = β 2 sin β‘ t , y=-2\sin t, y = β 2 sin t , we get
sin β‘ t = β y 2 . \sin t=-\frac{y}{2}. sin t = β 2 y β . Using cos β‘ 2 t + sin β‘ 2 t = 1 \cos^2t+\sin^2t=1 cos 2 t + sin 2 t = 1
x 2 16 + y 2 4 = 1. \frac{x^2}{16}+\frac{y^2}{4}=1. 16 x 2 β + 4 y 2 β = 1. This is an ellipse. At t = 0 t=0 t = 0
( 4 , 0 ) . (4,0). ( 4 , 0 ) . For small positive t t t y = β 2 sin β‘ t y=-2\sin t y = β 2 sin t ( 4 , 0 ) (4,0) ( 4 , 0 ) clockwise .
Classify each polar curve as a cardioid, limacon, rose curve, lemniscate, circle, line, or spiral. For rose curves, state the number of petals. Graph each of the curves.
( A ) (A) ( A ) r = 3 β 3 sin β‘ ΞΈ r=3-3\sin\theta r = 3 β 3 sin ΞΈ
( B ) (B) ( B ) r = 2 + 5 cos β‘ ΞΈ r=2+5\cos\theta r = 2 + 5 cos ΞΈ
( C ) (C) ( C ) r = 4 sin β‘ ( 3 ΞΈ ) r=4\sin(3\theta) r = 4 sin ( 3 ΞΈ )
( D ) (D) ( D ) r 2 = 25 sin β‘ ( 2 ΞΈ ) r^2=25\sin(2\theta) r 2 = 25 sin ( 2 ΞΈ )
( A ) (A) ( A )
r = 3 β 3 sin β‘ ΞΈ r=3-3\sin\theta r = 3 β 3 sin ΞΈ is a cardioid because it has the form r = a β a sin β‘ ΞΈ r=a-a\sin\theta r = a β a sin ΞΈ ΞΈ = Ο 2 \theta=\frac{\pi}{2} ΞΈ = 2 Ο β
Β‘ 6 Β‘ 3 3 6 Β‘ 6 Β‘ 3 x y ( B ) (B) ( B )
r = 2 + 5 cos β‘ ΞΈ r=2+5\cos\theta r = 2 + 5 cos ΞΈ is a limacon with an inner loop because β£ 2 β£ < β£ 5 β£ \lvert 2\rvert<\lvert 5\rvert β£ 2 β£ < β£ 5 β£
Β‘ 4 Β‘ 2 2 4 6 Β‘ 4 Β‘ 2 2 4 x y ( C ) (C) ( C )
r = 4 sin β‘ ( 3 ΞΈ ) r=4\sin(3\theta) r = 4 sin ( 3 ΞΈ ) is a rose curve. Since 3 3 3 3 3 3
Β‘ 4 Β‘ 2 2 4 Β‘ 4 Β‘ 2 2 4 x y ( D ) (D) ( D )
r 2 = 25 sin β‘ ( 2 ΞΈ ) r^2=25\sin(2\theta) r 2 = 25 sin ( 2 ΞΈ ) is a lemniscate. Since sin β‘ ( 2 ΞΈ ) β₯ 0 \sin(2\theta)\ge0 sin ( 2 ΞΈ ) β₯ 0 y = x y=x y = x
Β‘ 4 Β‘ 2 2 4 Β‘ 4 Β‘ 2 2 4 x y
Graph both r = 2 + 2 cos β‘ ΞΈ r=2+2\cos\theta r = 2 + 2 cos ΞΈ r = 2 β 2 cos β‘ ΞΈ r=2-2\cos\theta r = 2 β 2 cos ΞΈ
The curves are
r = 2 + 2 cos β‘ ΞΈ r=2+2\cos\theta r = 2 + 2 cos ΞΈ and
r = 2 β 2 cos β‘ ΞΈ . r=2-2\cos\theta. r = 2 β 2 cos ΞΈ . Both are cardioids. The first opens to the right, and the second opens to the left.
Β‘ 4 Β‘ 2 2 4 Β‘ 2 2 x y To find intersections with the same angle, set the equations equal:
2 + 2 cos β‘ ΞΈ = 2 β 2 cos β‘ ΞΈ . 2+2\cos\theta=2-2\cos\theta. 2 + 2 cos ΞΈ = 2 β 2 cos ΞΈ . Then
4 cos β‘ ΞΈ = 0 , 4\cos\theta=0, 4 cos ΞΈ = 0 , so
cos β‘ ΞΈ = 0. \cos\theta=0. cos ΞΈ = 0. Thus,
ΞΈ = Ο 2 , 3 Ο 2 . \theta=\frac{\pi}{2},\frac{3\pi}{2}. ΞΈ = 2 Ο β , 2 3 Ο β . These give the points
( 0 , 2 ) and ( 0 , β 2 ) . (0,2)\quad \text{and}\quad (0,-2). ( 0 , 2 ) and ( 0 , β 2 ) . Both cardioids also pass through the pole: r = 0 r=0 r = 0 ΞΈ = Ο \theta=\pi ΞΈ = Ο r = 0 r=0 r = 0 ΞΈ = 0 \theta=0 ΞΈ = 0
3 \boxed{3} 3 β intersection points.
Martina and Carl are part of a team that is studying weather patterns.
The team is about to launch a weather balloon to collect data. Martinaβs
rope is 7.8 7.8 7.8 36.0 β 36.0\circ 36.0 β 5.9 5.9 5.9
Let d d d 7.8 7.8 7.8 5.9 5.9 5.9 36.0 β 36.0^\circ 36. 0 β
5.9 2 = 7.8 2 + d 2 β 2 ( 7.8 ) ( d ) cos β‘ ( 36.0 β ) . 5.9^2=7.8^2+d^2-2(7.8)(d)\cos(36.0^\circ). 5. 9 2 = 7. 8 2 + d 2 β 2 ( 7.8 ) ( d ) cos ( 36. 0 β ) . Move everything to one side:
d 2 β 15.6 cos β‘ ( 36.0 β ) d + ( 7.8 2 β 5.9 2 ) = 0. d^2-15.6\cos(36.0^\circ)d+(7.8^2-5.9^2)=0. d 2 β 15.6 cos ( 36. 0 β ) d + ( 7. 8 2 β 5. 9 2 ) = 0. Using cos β‘ ( 36.0 β ) β 0.8090 \cos(36.0^\circ)\approx0.8090 cos ( 36. 0 β ) β 0.8090
d 2 β 12.6207 d + 25.03 = 0. d^2-12.6207d+25.03=0. d 2 β 12.6207 d + 25.03 = 0. Apply the quadratic formula:
d = 12.6207 Β± 12.6207 2 β 4 ( 25.03 ) 2 . d=\frac{12.6207\pm\sqrt{12.6207^2-4(25.03)}}{2}. d = 2 12.6207 Β± 12.620 7 2 β 4 ( 25.03 ) β β . This gives
d β 10.0 or d β 2.6. d\approx10.0
\qquad\text{or}\qquad
d\approx2.6. d β 10.0 or d β 2.6. So there are two possible distances unless the diagram or context says which side Carl is on:
10.0 Β mΒ orΒ 2.6 Β m . \boxed{10.0\text{ m or }2.6\text{ m}}. 10.0 Β mΒ orΒ 2.6 Β m β .
Let z 1 = β 2 + 2 3 i , z 2 = 1 β i . z_1=-2+2\sqrt3i,\qquad z_2=1-i. z 1 β = β 2 + 2 3 β i , z 2 β = 1 β i .
( A ) (A) ( A )
( B ) (B) ( B )
( C ) (C) ( C ) z 1 z 2 z_1z_2 z 1 β z 2 β
For part (A), for
z 1 = β 2 + 2 3 i , z_1=-2+2\sqrt3i, z 1 β = β 2 + 2 3 β i , the magnitude is
β£ z 1 β£ = ( β 2 ) 2 + ( 2 3 ) 2 = 4. \lvert z_1 \rvert=\sqrt{(-2)^2+(2\sqrt3)^2}=4. β£ z 1 β β£ = ( β 2 ) 2 + ( 2 3 β ) 2 β = 4. Since z 1 z_1 z 1 β
ΞΈ 1 = 120 β = 2 Ο 3 . \theta_1=120^\circ=\frac{2\pi}{3}. ΞΈ 1 β = 12 0 β = 3 2 Ο β . So
z 1 = 4 ( cos β‘ 2 Ο 3 + i sin β‘ 2 Ο 3 ) . z_1=4\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right). z 1 β = 4 ( cos 3 2 Ο β + i sin 3 2 Ο β ) . For
z 2 = 1 β i , z_2=1-i, z 2 β = 1 β i , the magnitude is 2 \sqrt2 2 β β 45 β = β Ο 4 -45^\circ=-\frac{\pi}{4} β 4 5 β = β 4 Ο β
z 2 = 2 ( cos β‘ ( β Ο 4 ) + i sin β‘ ( β Ο 4 ) ) . z_2=\sqrt2\left(\cos\left(-\frac{\pi}{4}\right)+i\sin\left(-\frac{\pi}{4}\right)\right). z 2 β = 2 β ( cos ( β 4 Ο β ) + i sin ( β 4 Ο β ) ) . For part (B), in exponential form,
z 1 = 4 e 2 Ο i / 3 , z 2 = 2 e β Ο i / 4 . z_1=4e^{2\pi i/3},\qquad z_2=\sqrt2e^{-\pi i/4}. z 1 β = 4 e 2 Ο i /3 , z 2 β = 2 β e β Ο i /4 . For part (C), multiply magnitudes and add angles:
z 1 z 2 = 4 2 e i ( 2 Ο / 3 β Ο / 4 ) = 4 2 e 5 Ο i / 12 . z_1z_2=4\sqrt2e^{i(2\pi/3-\pi/4)}=4\sqrt2e^{5\pi i/12}. z 1 β z 2 β = 4 2 β e i ( 2 Ο /3 β Ο /4 ) = 4 2 β e 5 Ο i /12 . In polar form,
z 1 z 2 = 4 2 ( cos β‘ 5 Ο 12 + i sin β‘ 5 Ο 12 ) . z_1z_2=4\sqrt2\left(\cos\frac{5\pi}{12}+i\sin\frac{5\pi}{12}\right). z 1 β z 2 β = 4 2 β ( cos 12 5 Ο β + i sin 12 5 Ο β ) . In rectangular form,
z 1 z 2 = ( 2 3 β 2 ) + ( 2 3 + 2 ) i . z_1z_2=(2\sqrt3-2)+(2\sqrt3+2)i. z 1 β z 2 β = ( 2 3 β β 2 ) + ( 2 3 β + 2 ) i .
Use De Moivreβs Theorem to find all fourth roots of 16 ( cos β‘ 2 Ο 3 + i sin β‘ 2 Ο 3 ) . 16\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right). 16 ( cos 3 2 Ο β + i sin 3 2 Ο β ) .
Write the complex number as
16 ( cos β‘ 2 Ο 3 + i sin β‘ 2 Ο 3 ) . 16\left(\cos\frac{2\pi}{3}+i\sin\frac{2\pi}{3}\right). 16 ( cos 3 2 Ο β + i sin 3 2 Ο β ) . The fourth roots have magnitude
16 4 = 2. \sqrt[4]{16}=2. 4 16 β = 2. Their angles are
2 Ο 3 + 2 Ο k 4 = Ο 6 + k Ο 2 , \frac{\frac{2\pi}{3}+2\pi k}{4}
=\frac{\pi}{6}+\frac{k\pi}{2}, 4 3 2 Ο β + 2 Ο k β = 6 Ο β + 2 k Ο β , where k = 0 , 1 , 2 , 3 k=0,1,2,3 k = 0 , 1 , 2 , 3
2 cis β‘ Ο 6 , 2 cis β‘ 2 Ο 3 , 2 cis β‘ 7 Ο 6 , 2 cis β‘ 5 Ο 3 . 2\operatorname{cis}\frac{\pi}{6},\quad
2\operatorname{cis}\frac{2\pi}{3},\quad
2\operatorname{cis}\frac{7\pi}{6},\quad
2\operatorname{cis}\frac{5\pi}{3}. 2 cis 6 Ο β , 2 cis 3 2 Ο β , 2 cis 6 7 Ο β , 2 cis 3 5 Ο β . In rectangular form, these are
3 + i , β 1 + 3 i , β 3 β i , 1 β 3 i . \sqrt3+i,\quad -1+\sqrt3i,\quad -\sqrt3-i,\quad 1-\sqrt3i. 3 β + i , β 1 + 3 β i , β 3 β β i , 1 β 3 β i .
(Bonus, Brahmaguptaβs and Bretschneiderβs formulas)
Let A B C D ABCD A B C D a , b , c , d a,b,c,d a , b , c , d
s = 1 2 ( a + b + c + d ) . s=\frac12(a+b+c+d). s = 2 1 β ( a + b + c + d ) . The goal is to prove the area of a cyclic quadrilateral is:
K = ( s β a ) ( s β b ) ( s β c ) ( s β d ) , K=\sqrt{(s-a)(s-b)(s-c)(s-d)}, K = ( s β a ) ( s β b ) ( s β c ) ( s β d ) β , where K K K
( A ) (A) ( A ) A C AC A C β A B C = B \angle ABC=B β A B C = B β A D C = D \angle ADC=D β A D C = D D = 180 β β B D=180^\circ-B D = 18 0 β β B K = 1 2 ( a b + c d ) sin β‘ B . K=\frac12(ab+cd)\sin B. K = 2 1 β ( ab + c d ) sin B .
( B ) (B) ( B ) A B C ABC A B C A D C ADC A D C 2 a b cos β‘ B = a 2 + b 2 β c 2 β d 2 + 2 c d cos β‘ D . 2ab\cos B=a^2+b^2-c^2-d^2+2cd\cos D. 2 ab cos B = a 2 + b 2 β c 2 β d 2 + 2 c d cos D .
( C ) (C) ( C ) cos β‘ D = β cos β‘ B \cos D=-\cos B cos D = β cos B cos β‘ B \cos B cos B K = 1 2 ( a b + c d ) sin β‘ B K=\frac12(ab+cd)\sin B K = 2 1 β ( ab + c d ) sin B sin β‘ 2 B = 1 β cos β‘ 2 B \sin^2B=1-\cos^2B sin 2 B = 1 β cos 2 B K 2 = ( s β a ) ( s β b ) ( s β c ) ( s β d ) . K^2=(s-a)(s-b)(s-c)(s-d). K 2 = ( s β a ) ( s β b ) ( s β c ) ( s β d ) .
( D ) (D) ( D ) A B C D ABCD A B C D B B B D D D ( 4 K ) 2 + ( a 2 + b 2 β c 2 β d 2 ) 2 = 4 ( a 2 b 2 + c 2 d 2 β 2 a b c d cos β‘ ( B + D ) ) . (4K)^2+(a^2+b^2-c^2-d^2)^2=4(a^2b^2+c^2d^2-2abcd\cos(B+D)). ( 4 K ) 2 + ( a 2 + b 2 β c 2 β d 2 ) 2 = 4 ( a 2 b 2 + c 2 d 2 β 2 ab c d cos ( B + D )) .
( E ) (E) ( E ) Bretschneiderβs formula : K 2 = ( s β a ) ( s β b ) ( s β c ) ( s β d ) β a b c d cos β‘ 2 ( B + D 2 ) . K^2=(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\left(\frac{B+D}{2}\right). K 2 = ( s β a ) ( s β b ) ( s β c ) ( s β d ) β ab c d cos 2 ( 2 B + D β ) .
( F ) (F) ( F ) A B C D ABCD A B C D
For part (A), draw diagonal A C AC A C A B C ABC A B C A D C ADC A D C K = 1 2 a b sin β‘ C K=\frac12ab\sin C K = 2 1 β ab sin C
[ A B C ] = 1 2 a b sin β‘ B [ABC]=\frac12ab\sin B [ A B C ] = 2 1 β ab sin B and
[ A D C ] = 1 2 c d sin β‘ D . [ADC]=\frac12cd\sin D. [ A D C ] = 2 1 β c d sin D . Therefore,
K = 1 2 a b sin β‘ B + 1 2 c d sin β‘ D . K=\frac12ab\sin B+\frac12cd\sin D. K = 2 1 β ab sin B + 2 1 β c d sin D . Since A B C D ABCD A B C D
B + D = 180 β . B+D=180^\circ. B + D = 18 0 β . So
D = 180 β β B D=180^\circ-B D = 18 0 β β B and
sin β‘ D = sin β‘ ( 180 β β B ) = sin β‘ B . \sin D=\sin(180^\circ-B)=\sin B. sin D = sin ( 18 0 β β B ) = sin B . Therefore,
K = 1 2 a b sin β‘ B + 1 2 c d sin β‘ B = 1 2 ( a b + c d ) sin β‘ B . K=\frac12ab\sin B+\frac12cd\sin B
=\frac12(ab+cd)\sin B. K = 2 1 β ab sin B + 2 1 β c d sin B = 2 1 β ( ab + c d ) sin B . For part (B), apply the Law of Cosines to triangle A B C ABC A B C
A C 2 = a 2 + b 2 β 2 a b cos β‘ B . AC^2=a^2+b^2-2ab\cos B. A C 2 = a 2 + b 2 β 2 ab cos B . Apply the Law of Cosines to triangle A D C ADC A D C
A C 2 = c 2 + d 2 β 2 c d cos β‘ D . AC^2=c^2+d^2-2cd\cos D. A C 2 = c 2 + d 2 β 2 c d cos D . Set these equal:
a 2 + b 2 β 2 a b cos β‘ B = c 2 + d 2 β 2 c d cos β‘ D . a^2+b^2-2ab\cos B=c^2+d^2-2cd\cos D. a 2 + b 2 β 2 ab cos B = c 2 + d 2 β 2 c d cos D . Rearranging gives
2 a b cos β‘ B = a 2 + b 2 β c 2 β d 2 + 2 c d cos β‘ D . 2ab\cos B=a^2+b^2-c^2-d^2+2cd\cos D. 2 ab cos B = a 2 + b 2 β c 2 β d 2 + 2 c d cos D . For part (C), since D = 180 β β B D=180^\circ-B D = 18 0 β β B
cos β‘ D = β cos β‘ B . \cos D=-\cos B. cos D = β cos B . Substitute into the result from part ( C ) (C) ( C )
2 a b cos β‘ B = a 2 + b 2 β c 2 β d 2 β 2 c d cos β‘ B . 2ab\cos B=a^2+b^2-c^2-d^2-2cd\cos B. 2 ab cos B = a 2 + b 2 β c 2 β d 2 β 2 c d cos B . So
2 ( a b + c d ) cos β‘ B = a 2 + b 2 β c 2 β d 2 , 2(ab+cd)\cos B=a^2+b^2-c^2-d^2, 2 ( ab + c d ) cos B = a 2 + b 2 β c 2 β d 2 , and
cos β‘ B = a 2 + b 2 β c 2 β d 2 2 ( a b + c d ) . \cos B=\frac{a^2+b^2-c^2-d^2}{2(ab+cd)}. cos B = 2 ( ab + c d ) a 2 + b 2 β c 2 β d 2 β . From part ( B ) (B) ( B )
K = 1 2 ( a b + c d ) sin β‘ B . K=\frac12(ab+cd)\sin B. K = 2 1 β ( ab + c d ) sin B . Square both sides:
K 2 = 1 4 ( a b + c d ) 2 sin β‘ 2 B . K^2=\frac14(ab+cd)^2\sin^2B. K 2 = 4 1 β ( ab + c d ) 2 sin 2 B . Use sin β‘ 2 B = 1 β cos β‘ 2 B \sin^2B=1-\cos^2B sin 2 B = 1 β cos 2 B
K 2 = 1 4 ( a b + c d ) 2 ( 1 β ( a 2 + b 2 β c 2 β d 2 2 ( a b + c d ) ) 2 ) . K^2=\frac14(ab+cd)^2\left(1-\left(\frac{a^2+b^2-c^2-d^2}{2(ab+cd)}\right)^2\right). K 2 = 4 1 β ( ab + c d ) 2 ( 1 β ( 2 ( ab + c d ) a 2 + b 2 β c 2 β d 2 β ) 2 ) . Simplifying gives
K 2 = 4 ( a b + c d ) 2 β ( a 2 + b 2 β c 2 β d 2 ) 2 16 . K^2=\frac{4(ab+cd)^2-(a^2+b^2-c^2-d^2)^2}{16}. K 2 = 16 4 ( ab + c d ) 2 β ( a 2 + b 2 β c 2 β d 2 ) 2 β . This factors as
K 2 = ( a + b + c β d ) ( a + b β c + d ) ( a β b + c + d ) ( β a + b + c + d ) 16 . K^2=\frac{(a+b+c-d)(a+b-c+d)(a-b+c+d)(-a+b+c+d)}{16}. K 2 = 16 ( a + b + c β d ) ( a + b β c + d ) ( a β b + c + d ) ( β a + b + c + d ) β . Since
s = 1 2 ( a + b + c + d ) , s=\frac12(a+b+c+d), s = 2 1 β ( a + b + c + d ) , the four factors are
2 ( s β d ) , 2 ( s β c ) , 2 ( s β b ) , 2 ( s β a ) . 2(s-d),\quad 2(s-c),\quad 2(s-b),\quad 2(s-a). 2 ( s β d ) , 2 ( s β c ) , 2 ( s β b ) , 2 ( s β a ) . Therefore,
K 2 = ( s β a ) ( s β b ) ( s β c ) ( s β d ) , K^2=(s-a)(s-b)(s-c)(s-d), K 2 = ( s β a ) ( s β b ) ( s β c ) ( s β d ) , so
K = ( s β a ) ( s β b ) ( s β c ) ( s β d ) . K=\sqrt{(s-a)(s-b)(s-c)(s-d)}. K = ( s β a ) ( s β b ) ( s β c ) ( s β d ) β . For part (D), for a general quadrilateral, the area is still
K = 1 2 a b sin β‘ B + 1 2 c d sin β‘ D , K=\frac12ab\sin B+\frac12cd\sin D, K = 2 1 β ab sin B + 2 1 β c d sin D , so
4 K = 2 a b sin β‘ B + 2 c d sin β‘ D . 4K=2ab\sin B+2cd\sin D. 4 K = 2 ab sin B + 2 c d sin D . Also, using the shared diagonal A C AC A C
a 2 + b 2 β 2 a b cos β‘ B = c 2 + d 2 β 2 c d cos β‘ D . a^2+b^2-2ab\cos B=c^2+d^2-2cd\cos D. a 2 + b 2 β 2 ab cos B = c 2 + d 2 β 2 c d cos D . Thus,
a 2 + b 2 β c 2 β d 2 = 2 a b cos β‘ B β 2 c d cos β‘ D . a^2+b^2-c^2-d^2=2ab\cos B-2cd\cos D. a 2 + b 2 β c 2 β d 2 = 2 ab cos B β 2 c d cos D . Now compute
( 4 K ) 2 + ( a 2 + b 2 β c 2 β d 2 ) 2 . (4K)^2+(a^2+b^2-c^2-d^2)^2. ( 4 K ) 2 + ( a 2 + b 2 β c 2 β d 2 ) 2 . Substitute the two expressions above:
( 2 a b sin β‘ B + 2 c d sin β‘ D ) 2 + ( 2 a b cos β‘ B β 2 c d cos β‘ D ) 2 . (2ab\sin B+2cd\sin D)^2+(2ab\cos B-2cd\cos D)^2. ( 2 ab sin B + 2 c d sin D ) 2 + ( 2 ab cos B β 2 c d cos D ) 2 . Expand:
4 a 2 b 2 sin β‘ 2 B + 8 a b c d sin β‘ B sin β‘ D + 4 c 2 d 2 sin β‘ 2 D 4a^2b^2\sin^2B+8abcd\sin B\sin D+4c^2d^2\sin^2D 4 a 2 b 2 sin 2 B + 8 ab c d sin B sin D + 4 c 2 d 2 sin 2 D + 4 a 2 b 2 cos β‘ 2 B β 8 a b c d cos β‘ B cos β‘ D + 4 c 2 d 2 cos β‘ 2 D . +4a^2b^2\cos^2B-8abcd\cos B\cos D+4c^2d^2\cos^2D. + 4 a 2 b 2 cos 2 B β 8 ab c d cos B cos D + 4 c 2 d 2 cos 2 D . Group terms:
4 a 2 b 2 ( sin β‘ 2 B + cos β‘ 2 B ) + 4 c 2 d 2 ( sin β‘ 2 D + cos β‘ 2 D ) 4a^2b^2(\sin^2B+\cos^2B)+4c^2d^2(\sin^2D+\cos^2D) 4 a 2 b 2 ( sin 2 B + cos 2 B ) + 4 c 2 d 2 ( sin 2 D + cos 2 D ) + 8 a b c d ( sin β‘ B sin β‘ D β cos β‘ B cos β‘ D ) . +8abcd(\sin B\sin D-\cos B\cos D). + 8 ab c d ( sin B sin D β cos B cos D ) . Since
sin β‘ 2 x + cos β‘ 2 x = 1 \sin^2x+\cos^2x=1 sin 2 x + cos 2 x = 1 and
cos β‘ ( B + D ) = cos β‘ B cos β‘ D β sin β‘ B sin β‘ D , \cos(B+D)=\cos B\cos D-\sin B\sin D, cos ( B + D ) = cos B cos D β sin B sin D , this becomes
4 a 2 b 2 + 4 c 2 d 2 β 8 a b c d cos β‘ ( B + D ) . 4a^2b^2+4c^2d^2-8abcd\cos(B+D). 4 a 2 b 2 + 4 c 2 d 2 β 8 ab c d cos ( B + D ) . Therefore,
( 4 K ) 2 + ( a 2 + b 2 β c 2 β d 2 ) 2 = 4 ( a 2 b 2 + c 2 d 2 β 2 a b c d cos β‘ ( B + D ) ) . (4K)^2+(a^2+b^2-c^2-d^2)^2=4(a^2b^2+c^2d^2-2abcd\cos(B+D)). ( 4 K ) 2 + ( a 2 + b 2 β c 2 β d 2 ) 2 = 4 ( a 2 b 2 + c 2 d 2 β 2 ab c d cos ( B + D )) . For part (E), rearrange the result from part (D):
16 K 2 = 4 ( a 2 b 2 + c 2 d 2 β 2 a b c d cos β‘ ( B + D ) ) β ( a 2 + b 2 β c 2 β d 2 ) 2 . 16K^2=4(a^2b^2+c^2d^2-2abcd\cos(B+D))-(a^2+b^2-c^2-d^2)^2. 16 K 2 = 4 ( a 2 b 2 + c 2 d 2 β 2 ab c d cos ( B + D )) β ( a 2 + b 2 β c 2 β d 2 ) 2 . Use
cos β‘ ( B + D ) = 2 cos β‘ 2 ( B + D 2 ) β 1. \cos(B+D)=2\cos^2\left(\frac{B+D}{2}\right)-1. cos ( B + D ) = 2 cos 2 ( 2 B + D β ) β 1. Then
β 2 a b c d cos β‘ ( B + D ) = β 4 a b c d cos β‘ 2 ( B + D 2 ) + 2 a b c d . -2abcd\cos(B+D)
=-4abcd\cos^2\left(\frac{B+D}{2}\right)+2abcd. β 2 ab c d cos ( B + D ) = β 4 ab c d cos 2 ( 2 B + D β ) + 2 ab c d . So
16 K 2 = 4 ( a 2 b 2 + c 2 d 2 + 2 a b c d ) β ( a 2 + b 2 β c 2 β d 2 ) 2 16K^2=4(a^2b^2+c^2d^2+2abcd)-(a^2+b^2-c^2-d^2)^2 16 K 2 = 4 ( a 2 b 2 + c 2 d 2 + 2 ab c d ) β ( a 2 + b 2 β c 2 β d 2 ) 2 β 16 a b c d cos β‘ 2 ( B + D 2 ) . -16abcd\cos^2\left(\frac{B+D}{2}\right). β 16 ab c d cos 2 ( 2 B + D β ) . Since
4 ( a 2 b 2 + c 2 d 2 + 2 a b c d ) = 4 ( a b + c d ) 2 , 4(a^2b^2+c^2d^2+2abcd)=4(ab+cd)^2, 4 ( a 2 b 2 + c 2 d 2 + 2 ab c d ) = 4 ( ab + c d ) 2 , we get
16 K 2 = 4 ( a b + c d ) 2 β ( a 2 + b 2 β c 2 β d 2 ) 2 β 16 a b c d cos β‘ 2 ( B + D 2 ) . 16K^2=4(ab+cd)^2-(a^2+b^2-c^2-d^2)^2
-16abcd\cos^2\left(\frac{B+D}{2}\right). 16 K 2 = 4 ( ab + c d ) 2 β ( a 2 + b 2 β c 2 β d 2 ) 2 β 16 ab c d cos 2 ( 2 B + D β ) . From part ( D ) (D) ( D )
16 ( s β a ) ( s β b ) ( s β c ) ( s β d ) . 16(s-a)(s-b)(s-c)(s-d). 16 ( s β a ) ( s β b ) ( s β c ) ( s β d ) . Therefore,
16 K 2 = 16 ( s β a ) ( s β b ) ( s β c ) ( s β d ) β 16 a b c d cos β‘ 2 ( B + D 2 ) . 16K^2=16(s-a)(s-b)(s-c)(s-d)
-16abcd\cos^2\left(\frac{B+D}{2}\right). 16 K 2 = 16 ( s β a ) ( s β b ) ( s β c ) ( s β d ) β 16 ab c d cos 2 ( 2 B + D β ) . Divide by 16 16 16
K 2 = ( s β a ) ( s β b ) ( s β c ) ( s β d ) β a b c d cos β‘ 2 ( B + D 2 ) . K^2=(s-a)(s-b)(s-c)(s-d)-abcd\cos^2\left(\frac{B+D}{2}\right). K 2 = ( s β a ) ( s β b ) ( s β c ) ( s β d ) β ab c d cos 2 ( 2 B + D β ) . This is Bretschneiderβs formula.
For part (F), if A B C D ABCD A B C D
B + D = 180 β . B+D=180^\circ. B + D = 18 0 β . Thus,
B + D 2 = 90 β , \frac{B+D}{2}=90^\circ, 2 B + D β = 9 0 β , so
cos β‘ 2 ( B + D 2 ) = cos β‘ 2 ( 90 β ) = 0. \cos^2\left(\frac{B+D}{2}\right)=\cos^2(90^\circ)=0. cos 2 ( 2 B + D β ) = cos 2 ( 9 0 β ) = 0. Bretschneiderβs formula becomes
K 2 = ( s β a ) ( s β b ) ( s β c ) ( s β d ) , K^2=(s-a)(s-b)(s-c)(s-d), K 2 = ( s β a ) ( s β b ) ( s β c ) ( s β d ) , which is Brahmaguptaβs formula.