The parent sine and cosine functions are periodic, meaning that their values repeat over regular intervals.
For every integer k k k
sin β‘ ( t + 2 Ο k ) = sin β‘ t \sin(t+2\pi k)=\sin t sin ( t + 2 Ο k ) = sin t and
cos β‘ ( t + 2 Ο k ) = cos β‘ t . \cos(t+2\pi k)=\cos t. cos ( t + 2 Ο k ) = cos t . So both y = sin β‘ x y=\sin x y = sin x y = cos β‘ x y=\cos x y = cos x period 2 Ο 2\pi 2 Ο
For
y = sin β‘ x , y=\sin x, y = sin x , the key facts are:
Domain: ( β β , β ) (-\infty,\infty) ( β β , β )
Range: [ β 1 , 1 ] [-1,1] [ β 1 , 1 ]
Period: 2 Ο 2\pi 2 Ο
Amplitude: 1 1 1
x x x x = k Ο x=k\pi x = k Ο k β Z k\in\mathbb Z k β Z
One cycle of sine can be tracked with the five key points:
x x x 0 0 0 Ο 2 \frac{\pi}{2} 2 Ο β Ο \pi Ο 3 Ο 2 \frac{3\pi}{2} 2 3 Ο β 2 Ο 2\pi 2 Ο sin β‘ x \sin x sin x 0 0 0 1 1 1 0 0 0 β 1 -1 β 1 0 0 0
The sine graph starts at the midline, reaches a maximum after one fourth of a period, returns to the midline after half a period, reaches a minimum after three fourths of a period, and returns to the midline after one full period.
For
y = cos β‘ x , y=\cos x, y = cos x , the key facts are:
Domain: ( β β , β ) (-\infty,\infty) ( β β , β )
Range: [ β 1 , 1 ] [-1,1] [ β 1 , 1 ]
Period: 2 Ο 2\pi 2 Ο
Amplitude: 1 1 1
x x x x = Ο 2 + k Ο x=\frac{\pi}{2}+k\pi x = 2 Ο β + k Ο k β Z k\in\mathbb Z k β Z
One cycle of cosine can be tracked with the five key points:
x x x 0 0 0 Ο 2 \frac{\pi}{2} 2 Ο β Ο \pi Ο 3 Ο 2 \frac{3\pi}{2} 2 3 Ο β 2 Ο 2\pi 2 Ο cos β‘ x \cos x cos x 1 1 1 0 0 0 β 1 -1 β 1 0 0 0 1 1 1
The cosine graph starts at a maximum, crosses the midline after one fourth of a period, reaches a minimum after half a period, crosses the midline again after three fourths of a period, and returns to a maximum after one full period.
A transformed sine or cosine function often has the form
y = A sin β‘ ( B ( x β C ) ) + D y=A\sin(B(x-C))+D y = A sin ( B ( x β C )) + D or
y = A cos β‘ ( B ( x β C ) ) + D . y=A\cos(B(x-C))+D. y = A cos ( B ( x β C )) + D . The constants control the shape of the graph:
Key info Amplitude, period, phase shift, midline
β£ A β£ \lvert A \rvert β£ A β£ amplitude .If A < 0 A<0 A < 0 x x x
2 Ο β£ B β£ \frac{2\pi}{\lvert B \rvert} β£ B β£ 2 Ο β period .C C C phase shift .D D D vertical shift , so the midline is y = D y=D y = D The range is [ D β β£ A β£ , D + β£ A β£ ] [D-\lvert A \rvert,D+ \lvert A \rvert] [ D β β£ A β£ , D + β£ A β£]
If the function is written as
y = A sin β‘ ( B x β C ) + D y=A\sin(Bx-C)+D y = A sin ( B x β C ) + D or
y = A cos β‘ ( B x β C ) + D , y=A\cos(Bx-C)+D, y = A cos ( B x β C ) + D , factor the inside first:
B x β C = B ( x β C B ) . Bx-C=B\left(x-\frac{C}{B}\right). B x β C = B ( x β B C β ) . So the phase shift is
C B . \frac{C}{B}. B C β . All the transformations learned in Unit 4 apply here as well.
To graph one full period:
Problem-solving strategy
Find the amplitude β£ A β£ \lvert A \rvert β£ A β£
Find the period 2 Ο β£ B β£ \frac{2\pi}{\lvert B \rvert} β£ B β£ 2 Ο β
Find the phase shift and starting point.
Divide the period into four equal increments.
Plot the five key points, then apply any reflection and vertical shift.
Repeat your graph for the amount of periods specified.
The increment between consecutive key points is
1 4 β
2 Ο β£ B β£ = Ο 2 β£ B β£ . \frac14\cdot\frac{2\pi}{\lvert B \rvert}
=\frac{\pi}{2\lvert B \rvert}. 4 1 β β
β£ B β£ 2 Ο β = 2 β£ B β£ Ο β . For example, if the period is 2 Ο 3 \frac{2\pi}{3} 3 2 Ο β
1 4 β
2 Ο 3 = Ο 6 . \frac14\cdot\frac{2\pi}{3}
=\frac{\pi}{6}. 4 1 β β
3 2 Ο β = 6 Ο β .
Example. Find the amplitude, period, phase shift, midline, range, and key-point increment for y = β 2 cos β‘ ( 3 ( x β Ο 4 ) ) + 1. y=-2\cos\left(3\left(x-\frac{\pi}{4}\right)\right)+1. y = β 2 cos ( 3 ( x β 4 Ο β ) ) + 1.
Here
A = β 2 , B = 3 , C = Ο 4 , D = 1. A=-2,\qquad B=3,\qquad C=\frac{\pi}{4},\qquad D=1. A = β 2 , B = 3 , C = 4 Ο β , D = 1. So the amplitude is
β£ A β£ = 2. \lvert A\rvert=2. β£ A β£ = 2. The period is
2 Ο β£ B β£ = 2 Ο 3 . \frac{2\pi}{\lvert B\rvert}
=\frac{2\pi}{3}. β£ B β£ 2 Ο β = 3 2 Ο β . The phase shift is right Ο 4 \frac{\pi}{4} 4 Ο β
y = 1. y=1. y = 1. The range is
[ 1 β 2 , 1 + 2 ] = [ β 1 , 3 ] . [1-2,1+2]=[-1,3]. [ 1 β 2 , 1 + 2 ] = [ β 1 , 3 ] . Since one full period is 2 Ο 3 \frac{2\pi}{3} 3 2 Ο β
1 4 β
2 Ο 3 = Ο 6 . \frac14\cdot\frac{2\pi}{3}
=\frac{\pi}{6}. 4 1 β β
3 2 Ο β = 6 Ο β . Since A < 0 A<0 A < 0
A graph with many key points is shown below:
Β‘ 2 ΒΌ 3 Β‘ ΒΌ 3 ΒΌ 3 2 ΒΌ 3 ΒΌ 4 ΒΌ 3 Β‘ 1 1 3 x y
Graphs are useful for understanding why trigonometric equations often have more than one solution.
For example, solving
sin β‘ x = a \sin x=a sin x = a on [ 0 , 2 Ο ) [0,2\pi) [ 0 , 2 Ο ) a a a β 1 < a < 1 -1<a<1 β 1 < a < 1 a β 0 a\ne 0 a ξ = 0
Similarly,
cos β‘ x = a \cos x=a cos x = a usually has two solutions on [ 0 , 2 Ο ) [0,2\pi) [ 0 , 2 Ο ) β 1 < a < 1 -1<a<1 β 1 < a < 1 a β 0 a\ne 0 a ξ = 0
Use the unit circle to find the reference angle, then use the sign of sine or cosine to choose the correct quadrants.
For all real solutions, add the period:
x = solution + 2 Ο k , k β Z . x=\text{solution}+2\pi k,\qquad k\in\mathbb Z. x = solution + 2 Ο k , k β Z .
Example. Solve on [ 0 , 4 Ο ) [0,4\pi) [ 0 , 4 Ο ) sin β‘ x = 1 2 . \sin x=\frac12. sin x = 2 1 β .
The reference angle is
Ο 6 , \frac{\pi}{6}, 6 Ο β , because
sin β‘ ( Ο 6 ) = 1 2 . \sin\left(\frac{\pi}{6}\right)=\frac12. sin ( 6 Ο β ) = 2 1 β . Sine is positive in Quadrants I and II, so the two standard solutions are
x = Ο 6 , 5 Ο 6 . x=\frac{\pi}{6},\frac{5\pi}{6}. x = 6 Ο β , 6 5 Ο β . Since we repeat over 2 periods, we add on 2 Ο 2\pi 2 Ο
x = Ο 6 , 5 Ο 6 , 13 Ο 6 , 17 Ο 6 . x=\frac{\pi}{6},\frac{5\pi}{6},\frac{13\pi}{6},\frac{17\pi}{6}. x = 6 Ο β , 6 5 Ο β , 6 13 Ο β , 6 17 Ο β . Graphically, the horizontal line y = 1 2 y=\frac12 y = 2 1 β
The remaining trig graphs come from tangent, cotangent, secant, and cosecant.
Since
tan β‘ x = sin β‘ x cos β‘ x , \tan x=\frac{\sin x}{\cos x}, tan x = cos x sin x β , tangent is undefined wherever cos β‘ x = 0 \cos x=0 cos x = 0
For
y = tan β‘ x , y=\tan x, y = tan x , the key facts are:
Domain: all real numbers except x = Ο 2 + k Ο x=\frac{\pi}{2}+k\pi x = 2 Ο β + k Ο
Range: ( β β , β ) (-\infty,\infty) ( β β , β )
Period: Ο \pi Ο
Vertical asymptotes: x = Ο 2 + k Ο x=\frac{\pi}{2}+k\pi x = 2 Ο β + k Ο
x x x x = k Ο x=k\pi x = k Ο
A transformed tangent function has the form
y = A tan β‘ ( B ( x β C ) ) + D . y=A\tan(B(x-C))+D. y = A tan ( B ( x β C )) + D . Its period is
Ο β£ B β£ . \frac{\pi}{\lvert B \rvert}. β£ B β£ Ο β . The distance from the center point to each neighboring vertical asymptote is one half of the period.
Example. Graph y = 3 tan β‘ ( 2 ( x + Ο 8 ) ) β 4 , y=3\tan\left(2\left(x+\frac{\pi}{8}\right)\right)-4, y = 3 tan ( 2 ( x + 8 Ο β ) ) β 4 ,
the period is
Ο 2 . \frac{\pi}{2}. 2 Ο β . The center of a tangent branch happens where the inside angle equals 0 0 0
2 ( x + Ο 8 ) = 0 β x = β Ο 8 . 2\left(x+\frac{\pi}{8}\right)=0
\quad\Rightarrow\quad
x=-\frac{\pi}{8}. 2 ( x + 8 Ο β ) = 0 β x = β 8 Ο β . At this point, tan β‘ ( 0 ) = 0 \tan(0)=0 tan ( 0 ) = 0
y = 3 ( 0 ) β 4 = β 4. y=3(0)-4=-4. y = 3 ( 0 ) β 4 = β 4. The center point is
( β Ο 8 , β 4 ) . \left(-\frac{\pi}{8},-4\right). ( β 8 Ο β , β 4 ) . Since half the period is Ο 4 \frac{\pi}{4} 4 Ο β
x = β Ο 8 β Ο 4 = β 3 Ο 8 and x = β Ο 8 + Ο 4 = Ο 8 . x=-\frac{\pi}{8}-\frac{\pi}{4}=-\frac{3\pi}{8}
\qquad\text{and}\qquad
x=-\frac{\pi}{8}+\frac{\pi}{4}=\frac{\pi}{8}. x = β 8 Ο β β 4 Ο β = β 8 3 Ο β and x = β 8 Ο β + 4 Ο β = 8 Ο β . A graph with many key points is shown below:
Β‘ ΒΌ 8 Β‘ 4 x y
Since
cot β‘ x = cos β‘ x sin β‘ x , \cot x=\frac{\cos x}{\sin x}, cot x = sin x cos x β , cotangent is undefined wherever sin β‘ x = 0 \sin x=0 sin x = 0
For
y = cot β‘ x , y=\cot x, y = cot x , the key facts are:
Domain: all real numbers except x = k Ο x=k\pi x = k Ο
Range: ( β β , β ) (-\infty,\infty) ( β β , β )
Period: Ο \pi Ο
Vertical asymptotes: x = k Ο x=k\pi x = k Ο
x x x x = Ο 2 + k Ο x=\frac{\pi}{2}+k\pi x = 2 Ο β + k Ο
A transformed cotangent function
y = A cot β‘ ( B ( x β C ) ) + D y=A\cot(B(x-C))+D y = A cot ( B ( x β C )) + D has period
Ο β£ B β£ . \frac{\pi}{\lvert B \rvert}. β£ B β£ Ο β . Since
sec β‘ x = 1 cos β‘ x , \sec x=\frac1{\cos x}, sec x = cos x 1 β , secant is undefined wherever cos β‘ x = 0 \cos x=0 cos x = 0
For
y = sec β‘ x , y=\sec x, y = sec x , the key facts are:
Domain: all real numbers except x = Ο 2 + k Ο x=\frac{\pi}{2}+k\pi x = 2 Ο β + k Ο
Range: ( β β , β 1 ] βͺ [ 1 , β ) (-\infty,-1]\cup[1,\infty) ( β β , β 1 ] βͺ [ 1 , β )
Period: 2 Ο 2\pi 2 Ο
Vertical asymptotes: x = Ο 2 + k Ο x=\frac{\pi}{2}+k\pi x = 2 Ο β + k Ο
The graph of secant follows the reciprocal of cosine. Where cosine has a maximum of 1 1 1 1 1 1 β 1 -1 β 1 β 1 -1 β 1 0 0 0
For
y = A sec β‘ ( B ( x β C ) ) + D , y=A\sec(B(x-C))+D, y = A sec ( B ( x β C )) + D , the period is
2 Ο β£ B β£ . \frac{2\pi}{\lvert B \rvert}. β£ B β£ 2 Ο β . Since
csc β‘ x = 1 sin β‘ x , \csc x=\frac1{\sin x}, csc x = sin x 1 β , cosecant is undefined wherever sin β‘ x = 0 \sin x=0 sin x = 0
For
y = csc β‘ x , y=\csc x, y = csc x , the key facts are:
Domain: all real numbers except x = k Ο x=k\pi x = k Ο
Range: ( β β , β 1 ] βͺ [ 1 , β ) (-\infty,-1]\cup[1,\infty) ( β β , β 1 ] βͺ [ 1 , β )
Period: 2 Ο 2\pi 2 Ο
Vertical asymptotes: x = k Ο x=k\pi x = k Ο
The graph of cosecant follows the reciprocal of sine. Where sine has a maximum of 1 1 1 1 1 1 β 1 -1 β 1 β 1 -1 β 1 0 0 0
For
y = A csc β‘ ( B ( x β C ) ) + D , y=A\csc(B(x-C))+D, y = A csc ( B ( x β C )) + D , the period is
2 Ο β£ B β£ . \frac{2\pi}{\lvert B \rvert}. β£ B β£ 2 Ο β . The angle addition and subtraction formulas let us rewrite trig functions of sums and differences of angles.
Proof (Angle addition formulas). Let two points on the unit circle be
P = ( cos β‘ A , sin β‘ A ) P=(\cos A,\sin A) P = ( cos A , sin A ) and
Q = ( cos β‘ B , sin β‘ B ) . Q=(\cos B,\sin B). Q = ( cos B , sin B ) . The distance between them depends only on the angle between them, which is A β B A-B A β B
P Q 2 = ( cos β‘ A β cos β‘ B ) 2 + ( sin β‘ A β sin β‘ B ) 2 . PQ^2=(\cos A-\cos B)^2+(\sin A-\sin B)^2. P Q 2 = ( cos A β cos B ) 2 + ( sin A β sin B ) 2 . Expanding gives
P Q 2 = 2 β 2 ( cos β‘ A cos β‘ B + sin β‘ A sin β‘ B ) . PQ^2=2-2(\cos A\cos B+\sin A\sin B). P Q 2 = 2 β 2 ( cos A cos B + sin A sin B ) . The same chord length can also be written using the angle difference:
P Q 2 = ( 1 β cos β‘ ( A β B ) ) 2 + sin β‘ 2 ( A β B ) = 2 β 2 cos β‘ ( A β B ) . PQ^2=(1-\cos(A-B))^2+\sin^2(A-B)=2-2\cos(A-B). P Q 2 = ( 1 β cos ( A β B ) ) 2 + sin 2 ( A β B ) = 2 β 2 cos ( A β B ) . Set the two expressions equal:
2 β 2 ( cos β‘ A cos β‘ B + sin β‘ A sin β‘ B ) = 2 β 2 cos β‘ ( A β B ) . 2-2(\cos A\cos B+\sin A\sin B)=2-2\cos(A-B). 2 β 2 ( cos A cos B + sin A sin B ) = 2 β 2 cos ( A β B ) . Therefore
cos β‘ ( A β B ) = cos β‘ A cos β‘ B + sin β‘ A sin β‘ B . \cos(A-B)=\cos A\cos B+\sin A\sin B. cos ( A β B ) = cos A cos B + sin A sin B . The other addition and subtraction formulas follow from this identity, even/odd identities, and cofunction relationships. For example, a similar proof can be done for sine addition/subtraction, and the proof is left to the reader as an exercise.
sin β‘ ( A + B ) = sin β‘ A cos β‘ B + cos β‘ A sin β‘ B \sin(A+B)=\sin A\cos B+\cos A\sin B sin ( A + B ) = sin A cos B + cos A sin B sin β‘ ( A β B ) = sin β‘ A cos β‘ B β cos β‘ A sin β‘ B \sin(A-B)=\sin A\cos B-\cos A\sin B sin ( A β B ) = sin A cos B β cos A sin B cos β‘ ( A + B ) = cos β‘ A cos β‘ B β sin β‘ A sin β‘ B \cos(A+B)=\cos A\cos B-\sin A\sin B cos ( A + B ) = cos A cos B β sin A sin B cos β‘ ( A β B ) = cos β‘ A cos β‘ B + sin β‘ A sin β‘ B \cos(A-B)=\cos A\cos B+\sin A\sin B cos ( A β B ) = cos A cos B + sin A sin B tan β‘ ( A + B ) = tan β‘ A + tan β‘ B 1 β tan β‘ A tan β‘ B \tan(A+B)=\frac{\tan A+\tan B}{1-\tan A\tan B} tan ( A + B ) = 1 β tan A tan B tan A + tan B β tan β‘ ( A β B ) = tan β‘ A β tan β‘ B 1 + tan β‘ A tan β‘ B \tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B} tan ( A β B ) = 1 + tan A tan B tan A β tan B β
Proof (Tangent addition formula). Use the fact that tangent is sine divided by cosine: tan β‘ ( A + B ) = sin β‘ ( A + B ) cos β‘ ( A + B ) \tan(A+B)=\frac{\sin(A+B)}{\cos(A+B)} tan ( A + B ) = c o s ( A + B ) s i n ( A + B ) β
Substitute the addition formulas:
tan β‘ ( A + B ) = sin β‘ A cos β‘ B + cos β‘ A sin β‘ B cos β‘ A cos β‘ B β sin β‘ A sin β‘ B . \tan(A+B)
=\frac{\sin A\cos B+\cos A\sin B}{\cos A\cos B-\sin A\sin B}. tan ( A + B ) = cos A cos B β sin A sin B sin A cos B + cos A sin B β . Divide every term in the numerator and denominator by cos β‘ A cos β‘ B \cos A\cos B cos A cos B
tan β‘ ( A + B ) = sin β‘ A cos β‘ B cos β‘ A cos β‘ B + cos β‘ A sin β‘ B cos β‘ A cos β‘ B cos β‘ A cos β‘ B cos β‘ A cos β‘ B β sin β‘ A sin β‘ B cos β‘ A cos β‘ B . \tan(A+B)
=\frac{\frac{\sin A\cos B}{\cos A\cos B}+\frac{\cos A\sin B}{\cos A\cos B}}
{\frac{\cos A\cos B}{\cos A\cos B}-\frac{\sin A\sin B}{\cos A\cos B}}. tan ( A + B ) = c o s A c o s B c o s A c o s B β β c o s A c o s B s i n A s i n B β c o s A c o s B s i n A c o s B β + c o s A c o s B c o s A s i n B β β . This simplifies to
tan β‘ ( A + B ) = tan β‘ A + tan β‘ B 1 β tan β‘ A tan β‘ B . \tan(A+B)
=\frac{\tan A+\tan B}{1-\tan A\tan B}. tan ( A + B ) = 1 β tan A tan B tan A + tan B β . The subtraction formula follows the same way using sin β‘ ( A β B ) \sin(A-B) sin ( A β B ) cos β‘ ( A β B ) \cos(A-B) cos ( A β B )
These formulas are especially useful for finding exact trig values for angles that can be written as sums or differences of special angles, such as 75 β = 45 β + 30 β 75^\circ=45^\circ+30^\circ 7 5 β = 4 5 β + 3 0 β
Example. Find the exact value of sin β‘ 75 β \sin 75^\circ sin 7 5 β
Rewrite the angle as a sum of special angles:
75 β = 45 β + 30 β . 75^\circ=45^\circ+30^\circ. 7 5 β = 4 5 β + 3 0 β . Then use the sine addition formula:
sin β‘ ( 75 β ) = sin β‘ ( 45 β + 30 β ) = sin β‘ 45 β cos β‘ 30 β + cos β‘ 45 β sin β‘ 30 β . \sin(75^\circ)
=\sin(45^\circ+30^\circ)
=\sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ. sin ( 7 5 β ) = sin ( 4 5 β + 3 0 β ) = sin 4 5 β cos 3 0 β + cos 4 5 β sin 3 0 β . Substitute exact values:
sin β‘ 75 β = 2 2 β
3 2 + 2 2 β
1 2 6 + 2 4 . \sin75^\circ
=\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}
+\frac{\sqrt2}{2}\cdot\frac12
\frac{\sqrt6+\sqrt2}{4}. sin 7 5 β = 2 2 β β β
2 3 β β + 2 2 β β β
2 1 β 4 6 β + 2 β β .
Double-angle formulas are the addition formulas with the same angle used twice.
Proof. Start with the angle addition formula for sine:
sin β‘ ( A + B ) = sin β‘ A cos β‘ B + cos β‘ A sin β‘ B . \sin(A+B)=\sin A\cos B+\cos A\sin B. sin ( A + B ) = sin A cos B + cos A sin B . Let A = ΞΈ A=\theta A = ΞΈ B = ΞΈ B=\theta B = ΞΈ
sin β‘ ( 2 ΞΈ ) = sin β‘ ΞΈ cos β‘ ΞΈ + cos β‘ ΞΈ sin β‘ ΞΈ = 2 sin β‘ ΞΈ cos β‘ ΞΈ . \sin(2\theta)
=\sin\theta\cos\theta+\cos\theta\sin\theta
=2\sin\theta\cos\theta. sin ( 2 ΞΈ ) = sin ΞΈ cos ΞΈ + cos ΞΈ sin ΞΈ = 2 sin ΞΈ cos ΞΈ . So
sin β‘ ( 2 ΞΈ ) = 2 sin β‘ ΞΈ cos β‘ ΞΈ . \sin(2\theta)=2\sin\theta\cos\theta. sin ( 2 ΞΈ ) = 2 sin ΞΈ cos ΞΈ . The cosine double-angle formula comes from the cosine addition formula:
cos β‘ ( 2 ΞΈ ) = cos β‘ ( ΞΈ + ΞΈ ) = cos β‘ 2 ΞΈ β sin β‘ 2 ΞΈ . \cos(2\theta)
=\cos(\theta+\theta)
=\cos^2\theta-\sin^2\theta. cos ( 2 ΞΈ ) = cos ( ΞΈ + ΞΈ ) = cos 2 ΞΈ β sin 2 ΞΈ . The alternate forms come from the Pythagorean identity:
sin β‘ 2 ΞΈ + cos β‘ 2 ΞΈ = 1. \sin^2\theta+\cos^2\theta=1. sin 2 ΞΈ + cos 2 ΞΈ = 1. Since sin β‘ 2 ΞΈ = 1 β cos β‘ 2 ΞΈ \sin^2\theta=1-\cos^2\theta sin 2 ΞΈ = 1 β cos 2 ΞΈ
cos β‘ 2 ΞΈ β sin β‘ 2 ΞΈ = cos β‘ 2 ΞΈ β ( 1 β cos β‘ 2 ΞΈ ) = 2 cos β‘ 2 ΞΈ β 1. \cos^2\theta-\sin^2\theta
=\cos^2\theta-(1-\cos^2\theta)
=2\cos^2\theta-1. cos 2 ΞΈ β sin 2 ΞΈ = cos 2 ΞΈ β ( 1 β cos 2 ΞΈ ) = 2 cos 2 ΞΈ β 1. Since cos β‘ 2 ΞΈ = 1 β sin β‘ 2 ΞΈ \cos^2\theta=1-\sin^2\theta cos 2 ΞΈ = 1 β sin 2 ΞΈ
cos β‘ 2 ΞΈ β sin β‘ 2 ΞΈ = ( 1 β sin β‘ 2 ΞΈ ) β sin β‘ 2 ΞΈ = 1 β 2 sin β‘ 2 ΞΈ . \cos^2\theta-\sin^2\theta
=(1-\sin^2\theta)-\sin^2\theta
=1-2\sin^2\theta. cos 2 ΞΈ β sin 2 ΞΈ = ( 1 β sin 2 ΞΈ ) β sin 2 ΞΈ = 1 β 2 sin 2 ΞΈ . So
cos β‘ ( 2 ΞΈ ) = cos β‘ 2 ΞΈ β sin β‘ 2 ΞΈ = ( 1 β sin β‘ 2 ΞΈ ) β sin β‘ 2 ΞΈ = 1 β 2 sin β‘ 2 ΞΈ . . \cos(2\theta)=\cos^2\theta-\sin^2\theta=(1-\sin^2\theta)-\sin^2\theta=1-2\sin^2\theta.. cos ( 2 ΞΈ ) = cos 2 ΞΈ β sin 2 ΞΈ = ( 1 β sin 2 ΞΈ ) β sin 2 ΞΈ = 1 β 2 sin 2 ΞΈ .. Similarly, the tangent double angle formula can come from either dividing sine by cosine or using the tangent addition formula:
tan β‘ ( 2 ΞΈ ) = tan β‘ ( ΞΈ + ΞΈ ) = 2 tan β‘ A 1 β tan β‘ 2 A . \tan(2\theta)
=\tan(\theta+\theta)
=\frac{2\tan A}{1 - \tan^2 A}. tan ( 2 ΞΈ ) = tan ( ΞΈ + ΞΈ ) = 1 β tan 2 A 2 tan A β . tan β‘ ( 2 ΞΈ ) = 2 tan β‘ A 1 β tan β‘ 2 A . \tan(2\theta)=\frac{2\tan A}{1 - \tan^2 A}. tan ( 2 ΞΈ ) = 1 β tan 2 A 2 tan A β .
Power-reducing formulas come from the double-angle formulas for cosine.
cos β‘ 2 ΞΈ = 1 + cos β‘ ( 2 ΞΈ ) 2 \cos^2\theta=\frac{1+\cos(2\theta)}{2} cos 2 ΞΈ = 2 1 + cos ( 2 ΞΈ ) β sin β‘ 2 ΞΈ = 1 β cos β‘ ( 2 ΞΈ ) 2 \sin^2\theta=\frac{1-\cos(2\theta)}{2} sin 2 ΞΈ = 2 1 β cos ( 2 ΞΈ ) β These are useful when rewriting expressions with powers of sine or cosine. Later we will learn De Moivreβs Theorem to generalize power formulas.
Proof (Power reduction formulas). Prove the two power reduction formulas.
Use the cosine double-angle identities:
cos β‘ ( 2 ΞΈ ) = 2 cos β‘ 2 ΞΈ β 1. \cos(2\theta)=2\cos^2\theta-1. cos ( 2 ΞΈ ) = 2 cos 2 ΞΈ β 1. Solve for cos β‘ 2 ΞΈ \cos^2\theta cos 2 ΞΈ
2 cos β‘ 2 ΞΈ = 1 + cos β‘ ( 2 ΞΈ ) 2\cos^2\theta=1+\cos(2\theta) 2 cos 2 ΞΈ = 1 + cos ( 2 ΞΈ ) so
cos β‘ 2 ΞΈ = 1 + cos β‘ ( 2 ΞΈ ) 2 . \cos^2\theta=\frac{1+\cos(2\theta)}{2}. cos 2 ΞΈ = 2 1 + cos ( 2 ΞΈ ) β . Similarly, use
cos β‘ ( 2 ΞΈ ) = 1 β 2 sin β‘ 2 ΞΈ . \cos(2\theta)=1-2\sin^2\theta. cos ( 2 ΞΈ ) = 1 β 2 sin 2 ΞΈ . Solving for sin β‘ 2 ΞΈ \sin^2\theta sin 2 ΞΈ
sin β‘ 2 ΞΈ = 1 β cos β‘ ( 2 ΞΈ ) 2 . \sin^2\theta=\frac{1-\cos(2\theta)}{2}. sin 2 ΞΈ = 2 1 β cos ( 2 ΞΈ ) β .
The half-angle formulas come from replacing ΞΈ \theta ΞΈ ΞΈ 2 \frac{\theta}{2} 2 ΞΈ β
cos β‘ ( ΞΈ 2 ) = Β± 1 + cos β‘ ΞΈ 2 \cos\left(\frac{\theta}{2}\right)=\pm\sqrt{\frac{1+\cos\theta}{2}} cos ( 2 ΞΈ β ) = Β± 2 1 + cos ΞΈ β β sin β‘ ( ΞΈ 2 ) = Β± 1 β cos β‘ ΞΈ 2 \sin\left(\frac{\theta}{2}\right)=\pm\sqrt{\frac{1-\cos\theta}{2}} sin ( 2 ΞΈ β ) = Β± 2 1 β cos ΞΈ β β The sign depends on the quadrant of ΞΈ 2 \frac{\theta}{2} 2 ΞΈ β
Another useful half-angle identity is
tan β‘ ( ΞΈ 2 ) = sin β‘ ΞΈ 1 + cos β‘ ΞΈ \tan\left(\frac{\theta}{2}\right)=\frac{\sin\theta}{1+\cos\theta} tan ( 2 ΞΈ β ) = 1 + cos ΞΈ sin ΞΈ β which can also be written as
tan β‘ ( ΞΈ 2 ) = 1 β cos β‘ ΞΈ sin β‘ ΞΈ . \tan\left(\frac{\theta}{2}\right)=\frac{1-\cos\theta}{\sin\theta}. tan ( 2 ΞΈ β ) = sin ΞΈ 1 β cos ΞΈ β .
Extension. Prove the tangent half angle identity. As a bonus, try to solve it geometrically!
Example. Find the exact value of sin β‘ 105 β \sin 105^\circ sin 10 5 β
Since
105 β = 210 β 2 , 105^\circ=\frac{210^\circ}{2}, 10 5 β = 2 21 0 β β , use the half-angle formula:
sin β‘ ( ΞΈ 2 ) = Β± 1 β cos β‘ ΞΈ 2 . \sin\left(\frac{\theta}{2}\right)=\pm\sqrt{\frac{1-\cos\theta}{2}}. sin ( 2 ΞΈ β ) = Β± 2 1 β cos ΞΈ β β . Here ΞΈ = 210 β \theta=210^\circ ΞΈ = 21 0 β ΞΈ 2 = 105 β \frac{\theta}{2}=105^\circ 2 ΞΈ β = 10 5 β 105 β 105^\circ 10 5 β
sin β‘ 105 β = 1 β cos β‘ 210 β 2 . \sin105^\circ
=\sqrt{\frac{1-\cos210^\circ}{2}}. sin 10 5 β = 2 1 β cos 21 0 β β β . Because
cos β‘ 210 β = β 3 2 , \cos210^\circ=-\frac{\sqrt3}{2}, cos 21 0 β = β 2 3 β β , we get
sin β‘ 105 β = 1 + 3 2 2 = 2 + 3 4 = 2 + 3 2 . \sin105^\circ
=\sqrt{\frac{1+\frac{\sqrt3}{2}}{2}}
=\sqrt{\frac{2+\sqrt3}{4}}
=\frac{\sqrt{2+\sqrt3}}{2}. sin 10 5 β = 2 1 + 2 3 β β β β = 4 2 + 3 β β β = 2 2 + 3 β β β .
Product-to-sum formulas rewrite products of trig functions as sums or differences.
sin β‘ A sin β‘ B = 1 2 [ cos β‘ ( A β B ) β cos β‘ ( A + B ) ] \sin A\sin B=\frac12[\cos(A-B)-\cos(A+B)] sin A sin B = 2 1 β [ cos ( A β B ) β cos ( A + B )] cos β‘ A cos β‘ B = 1 2 [ cos β‘ ( A β B ) + cos β‘ ( A + B ) ] \cos A\cos B=\frac12[\cos(A-B)+\cos(A+B)] cos A cos B = 2 1 β [ cos ( A β B ) + cos ( A + B )] sin β‘ A cos β‘ B = 1 2 [ sin β‘ ( A + B ) + sin β‘ ( A β B ) ] \sin A\cos B=\frac12[\sin(A+B)+\sin(A-B)] sin A cos B = 2 1 β [ sin ( A + B ) + sin ( A β B )] cos β‘ A sin β‘ B = 1 2 [ sin β‘ ( A + B ) β sin β‘ ( A β B ) ] \cos A\sin B=\frac12[\sin(A+B)-\sin(A-B)] cos A sin B = 2 1 β [ sin ( A + B ) β sin ( A β B )] Sum-to-product formulas reverse the idea.
sin β‘ A + sin β‘ B = 2 sin β‘ ( A + B 2 ) cos β‘ ( A β B 2 ) \sin A+\sin B=2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) sin A + sin B = 2 sin ( 2 A + B β ) cos ( 2 A β B β ) sin β‘ A β sin β‘ B = 2 cos β‘ ( A + B 2 ) sin β‘ ( A β B 2 ) \sin A-\sin B=2\cos\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) sin A β sin B = 2 cos ( 2 A + B β ) sin ( 2 A β B β ) cos β‘ A + cos β‘ B = 2 cos β‘ ( A + B 2 ) cos β‘ ( A β B 2 ) \cos A+\cos B=2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) cos A + cos B = 2 cos ( 2 A + B β ) cos ( 2 A β B β ) cos β‘ A β cos β‘ B = β 2 sin β‘ ( A + B 2 ) sin β‘ ( A β B 2 ) \cos A-\cos B=-2\sin\left(\frac{A+B}{2}\right)\sin\left(\frac{A-B}{2}\right) cos A β cos B = β 2 sin ( 2 A + B β ) sin ( 2 A β B β ) These formulas are useful for simplifying expressions and for rewriting functions in a graphable form.
Proof (Sum-to-product/Product-to-sum formulas). Prove the formulas above.
Add the cosine addition and subtraction formulas:
cos β‘ ( A β B ) = cos β‘ A cos β‘ B + sin β‘ A sin β‘ B \cos(A-B)=\cos A\cos B+\sin A\sin B cos ( A β B ) = cos A cos B + sin A sin B and
cos β‘ ( A + B ) = cos β‘ A cos β‘ B β sin β‘ A sin β‘ B . \cos(A+B)=\cos A\cos B-\sin A\sin B. cos ( A + B ) = cos A cos B β sin A sin B . Adding them gives
cos β‘ ( A β B ) + cos β‘ ( A + B ) = 2 cos β‘ A cos β‘ B . \cos(A-B)+\cos(A+B)=2\cos A\cos B. cos ( A β B ) + cos ( A + B ) = 2 cos A cos B . Therefore
cos β‘ A cos β‘ B = 1 2 [ cos β‘ ( A β B ) + cos β‘ ( A + B ) ] . \cos A\cos B=\frac12[\cos(A-B)+\cos(A+B)]. cos A cos B = 2 1 β [ cos ( A β B ) + cos ( A + B )] . Subtracting instead gives
cos β‘ ( A β B ) β cos β‘ ( A + B ) = 2 sin β‘ A sin β‘ B , \cos(A-B)-\cos(A+B)=2\sin A\sin B, cos ( A β B ) β cos ( A + B ) = 2 sin A sin B , so
sin β‘ A sin β‘ B = 1 2 [ cos β‘ ( A β B ) β cos β‘ ( A + B ) ] . \sin A\sin B=\frac12[\cos(A-B)-\cos(A+B)]. sin A sin B = 2 1 β [ cos ( A β B ) β cos ( A + B )] . The other product-to-sum formulas come from adding or subtracting the sine addition and subtraction formulas and is left as an exercise to the reader.
Example. Rewrite
cos β‘ ( 5 x ) + cos β‘ ( 3 x ) \cos(5x)+\cos(3x) cos ( 5 x ) + cos ( 3 x ) as a product.
Use
cos β‘ A + cos β‘ B = 2 cos β‘ ( A + B 2 ) cos β‘ ( A β B 2 ) . \cos A+\cos B=2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right). cos A + cos B = 2 cos ( 2 A + B β ) cos ( 2 A β B β ) . Let A = 5 x A=5x A = 5 x B = 3 x B=3x B = 3 x
cos β‘ ( 5 x ) + cos β‘ ( 3 x ) = 2 cos β‘ ( 5 x + 3 x 2 ) cos β‘ ( 5 x β 3 x 2 ) . \cos(5x)+\cos(3x)
=2\cos\left(\frac{5x+3x}{2}\right)\cos\left(\frac{5x-3x}{2}\right). cos ( 5 x ) + cos ( 3 x ) = 2 cos ( 2 5 x + 3 x β ) cos ( 2 5 x β 3 x β ) . So
cos β‘ ( 5 x ) + cos β‘ ( 3 x ) = 2 cos β‘ ( 4 x ) cos β‘ x . \cos(5x)+\cos(3x)=2\cos(4x)\cos x. cos ( 5 x ) + cos ( 3 x ) = 2 cos ( 4 x ) cos x .
When solving trigonometric equations, the main goal is to reduce the equation to a familiar trig statement such as
sin β‘ x = a , cos β‘ x = a , tan β‘ x = a . \sin x=a,\qquad \cos x=a,\qquad \tan x=a. sin x = a , cos x = a , tan x = a . A general strategy is:
Problem-solving strategy
Use identities to rewrite the equation using one trig function when possible.
Factor if the equation is quadratic in a trig expression.
Find the reference angle using the unit circle.
Use the interval and quadrant signs to list every solution. Always reject values outside the allowed domain and range!
Check for extraneous solutions if the work involved squaring, dividing by a variable expression, or using reciprocal functions.
Example. Solve exactly on [ 0 , 2 Ο ) [0,2\pi) [ 0 , 2 Ο )
2 sin β‘ 2 x + sin β‘ x β 1 = 0. 2\sin^2x+\sin x-1=0. 2 sin 2 x + sin x β 1 = 0. Factor:
( 2 sin β‘ x β 1 ) ( sin β‘ x + 1 ) = 0. (2\sin x-1)(\sin x+1)=0. ( 2 sin x β 1 ) ( sin x + 1 ) = 0. So
sin β‘ x = 1 2 or sin β‘ x = β 1. \sin x=\frac12
\qquad\text{or}\qquad
\sin x=-1. sin x = 2 1 β or sin x = β 1. On [ 0 , 2 Ο ) [0,2\pi) [ 0 , 2 Ο )
sin β‘ x = 1 2 \sin x=\frac12 sin x = 2 1 β at
x = Ο 6 , 5 Ο 6 . x=\frac{\pi}{6},\frac{5\pi}{6}. x = 6 Ο β , 6 5 Ο β . Also,
sin β‘ x = β 1 \sin x=-1 sin x = β 1 at
x = 3 Ο 2 . x=\frac{3\pi}{2}. x = 2 3 Ο β . Thus
x = Ο 6 , 5 Ο 6 , 3 Ο 2 . x=\frac{\pi}{6},\frac{5\pi}{6},\frac{3\pi}{2}. x = 6 Ο β , 6 5 Ο β , 2 3 Ο β .
If the problem asks for all real solutions, use periodicity.
For sine and cosine, add 2 Ο k 2\pi k 2 Ο k
x = solution + 2 Ο k , k β Z . x=\text{solution}+2\pi k,\qquad k\in\mathbb Z. x = solution + 2 Ο k , k β Z . For tangent and cotangent, add Ο k \pi k Ο k
x = solution + Ο k , k β Z . x=\text{solution}+\pi k,\qquad k\in\mathbb Z. x = solution + Ο k , k β Z . If the equation has a coefficient inside the trig function, solve the inside angle first, then isolate x x x
A function needs to be one-to-one in order to have an inverse function. The original sine, cosine, and tangent graphs are not one-to-one on their full domains, so their domains are restricted before defining inverse trig functions.
For inverse sine (arcsine),
y = sin β‘ β 1 x y=\sin^{-1}x y = sin β 1 x means
sin β‘ y = x . \sin y=x. sin y = x . The restricted sine function for the standard arcsine function uses the interval
[ β Ο 2 , Ο 2 ] . \left[-\frac{\pi}{2},\frac{\pi}{2}\right]. [ β 2 Ο β , 2 Ο β ] . So
Domain of sin β‘ β 1 x \sin^{-1}x sin β 1 x [ β 1 , 1 ] [-1,1] [ β 1 , 1 ]
Range of sin β‘ β 1 x \sin^{-1}x sin β 1 x [ β Ο 2 , Ο 2 ] \left[-\frac{\pi}{2},\frac{\pi}{2}\right] [ β 2 Ο β , 2 Ο β ]
For inverse cosine (arccosine),
y = cos β‘ β 1 x y=\cos^{-1}x y = cos β 1 x means
cos β‘ y = x . \cos y=x. cos y = x . The restricted cosine function for the standard arccosine function uses the interval
[ 0 , Ο ] . [0,\pi]. [ 0 , Ο ] . So
Domain of cos β‘ β 1 x \cos^{-1}x cos β 1 x [ β 1 , 1 ] [-1,1] [ β 1 , 1 ]
Range of cos β‘ β 1 x \cos^{-1}x cos β 1 x [ 0 , Ο ] [0,\pi] [ 0 , Ο ]
For inverse tangent (arctangent),
y = tan β‘ β 1 x y=\tan^{-1}x y = tan β 1 x means
tan β‘ y = x . \tan y=x. tan y = x . The restricted tangent function for the standard arctangent function uses the interval
( β Ο 2 , Ο 2 ) . \left(-\frac{\pi}{2},\frac{\pi}{2}\right). ( β 2 Ο β , 2 Ο β ) . So
Domain of tan β‘ β 1 x \tan^{-1}x tan β 1 x ( β β , β ) (-\infty,\infty) ( β β , β )
Range of tan β‘ β 1 x \tan^{-1}x tan β 1 x ( β Ο 2 , Ο 2 ) \left(-\frac{\pi}{2},\frac{\pi}{2}\right) ( β 2 Ο β , 2 Ο β )
Inverse trig compositions work cleanly only when the angle lies in the chosen restricted range.
For all x x x
sin β‘ ( sin β‘ β 1 x ) = x , \sin(\sin^{-1}x)=x, sin ( sin β 1 x ) = x , cos β‘ ( cos β‘ β 1 x ) = x , \cos(\cos^{-1}x)=x, cos ( cos β 1 x ) = x , and
tan β‘ ( tan β‘ β 1 x ) = x . \tan(\tan^{-1}x)=x. tan ( tan β 1 x ) = x . Warning
Expressions like sin β‘ β 1 ( sin β‘ x ) \sin^{-1}(\sin x) sin β 1 ( sin x ) x x x [ β Ο 2 , Ο 2 ] \left[-\frac{\pi}{2},\frac{\pi}{2}\right] [ β 2 Ο β , 2 Ο β ]
When evaluating a composition such as
sin β‘ ( cos β‘ β 1 x ) , \sin(\cos^{-1}x), sin ( cos β 1 x ) , draw a triangle or use an identity and always inverse trig functions represent angles. If ΞΈ = cos β‘ β 1 x \theta=\cos^{-1}x ΞΈ = cos β 1 x cos β‘ ΞΈ = x \cos\theta=x cos ΞΈ = x ΞΈ \theta ΞΈ [ 0 , Ο ] [0,\pi] [ 0 , Ο ]
Example. Evaluate exactly:
sin β‘ ( cos β‘ β 1 3 5 ) . \sin\left(\cos^{-1}\frac35\right). sin ( cos β 1 5 3 β ) . Let
ΞΈ = cos β‘ β 1 3 5 . \theta=\cos^{-1}\frac35. ΞΈ = cos β 1 5 3 β . Then
cos β‘ ΞΈ = 3 5 , \cos\theta=\frac35, cos ΞΈ = 5 3 β , and ΞΈ \theta ΞΈ [ 0 , Ο ] [0,\pi] [ 0 , Ο ] ΞΈ \theta ΞΈ
Draw a right triangle with adjacent side 3 3 3 5 5 5
5 2 β 3 2 = 16 = 4. \sqrt{5^2-3^2}=\sqrt{16}=4. 5 2 β 3 2 β = 16 β = 4. Therefore
sin β‘ ΞΈ = 4 5 . \sin\theta=\frac45. sin ΞΈ = 5 4 β . So
sin β‘ ( cos β‘ β 1 3 5 ) = 4 5 . \sin\left(\cos^{-1}\frac35\right)=\frac45. sin ( cos β 1 5 3 β ) = 5 4 β .
For f ( x ) = β 3 sin β‘ ( 2 ( x β Ο 6 ) ) + 1 f(x)=-3\sin\left(2\left(x-\frac{\pi}{6}\right)\right)+1 f ( x ) = β 3 sin ( 2 ( x β 6 Ο β ) ) + 1
The function is
f ( x ) = β 3 sin β‘ ( 2 ( x β Ο 6 ) ) + 1. f(x)=-3\sin\left(2\left(x-\frac{\pi}{6}\right)\right)+1. f ( x ) = β 3 sin ( 2 ( x β 6 Ο β ) ) + 1. The amplitude is
3 . \boxed{3}. 3 β . The period is
2 Ο β£ 2 β£ = Ο . \frac{2\pi}{\lvert 2 \rvert}=\pi. β£ 2 β£ 2 Ο β = Ο . So
period = Ο . \boxed{\text{period}=\pi}. period = Ο β . The phase shift is
Ο 6 Β right . \boxed{\frac{\pi}{6}\text{ right}}. 6 Ο β Β right β . The midline is
y = 1 . \boxed{y=1}. y = 1 β . The range is
1 β 3 β€ y β€ 1 + 3 , 1-3\le y\le 1+3, 1 β 3 β€ y β€ 1 + 3 , so
[ β 2 , 4 ] . \boxed{[-2,4]}. [ β 2 , 4 ] β . The key-point increment is one fourth of the period:
Ο 4 . \frac{\pi}{4}. 4 Ο β . Starting at x = Ο 6 x=\frac{\pi}{6} x = 6 Ο β x x x
Ο 6 , 5 Ο 12 , 2 Ο 3 , 11 Ο 12 , 7 Ο 6 . \frac{\pi}{6},\quad
\frac{5\pi}{12},\quad
\frac{2\pi}{3},\quad
\frac{11\pi}{12},\quad
\frac{7\pi}{6}. 6 Ο β , 12 5 Ο β , 3 2 Ο β , 12 11 Ο β , 6 7 Ο β . Because the sine graph is reflected over the x x x y y y
1 , β 2 , 1 , 4 , 1. 1,\quad -2,\quad 1,\quad 4,\quad 1. 1 , β 2 , 1 , 4 , 1. Thus the five key points are
( Ο 6 , 1 ) , ( 5 Ο 12 , β 2 ) , ( 2 Ο 3 , 1 ) , ( 11 Ο 12 , 4 ) , ( 7 Ο 6 , 1 ) . \boxed{\left(\frac{\pi}{6},1\right),
\left(\frac{5\pi}{12},-2\right),
\left(\frac{2\pi}{3},1\right),
\left(\frac{11\pi}{12},4\right),
\left(\frac{7\pi}{6},1\right)}. ( 6 Ο β , 1 ) , ( 12 5 Ο β , β 2 ) , ( 3 2 Ο β , 1 ) , ( 12 11 Ο β , 4 ) , ( 6 7 Ο β , 1 ) β . A graph with many key points is shown below:
ΒΌ 6 ΒΌ 2 5 ΒΌ 6 7 ΒΌ 6 Β‘ 2 1 4 x y
A sinusoidal function has maximum value 7 7 7 x = Ο 3 x=\frac{\pi}{3} x = 3 Ο β β 1 -1 β 1 x = 7 Ο 6 x=\frac{7\pi}{6} x = 6 7 Ο β
The maximum is 7 7 7 β 1 -1 β 1
D = 7 + ( β 1 ) 2 = 3. D=\frac{7+(-1)}{2}=3. D = 2 7 + ( β 1 ) β = 3. The amplitude is half the distance between max and min:
A = 7 β ( β 1 ) 2 = 4. A=\frac{7-(-1)}{2}=4. A = 2 7 β ( β 1 ) β = 4. The distance from a maximum to the next minimum is half a period:
7 Ο 6 β Ο 3 = 7 Ο 6 β 2 Ο 6 = 5 Ο 6 . \frac{7\pi}{6}-\frac{\pi}{3}
=\frac{7\pi}{6}-\frac{2\pi}{6}
=\frac{5\pi}{6}. 6 7 Ο β β 3 Ο β = 6 7 Ο β β 6 2 Ο β = 6 5 Ο β . So the period is
2 β
5 Ο 6 = 5 Ο 3 . 2\cdot\frac{5\pi}{6}=\frac{5\pi}{3}. 2 β
6 5 Ο β = 3 5 Ο β . Thus
B = 2 Ο 5 Ο 3 = 6 5 . B=\frac{2\pi}{\frac{5\pi}{3}}=\frac65. B = 3 5 Ο β 2 Ο β = 5 6 β . Since the function has a maximum at x = Ο 3 x=\frac{\pi}{3} x = 3 Ο β
f ( x ) = 4 cos β‘ ( 6 5 ( x β Ο 3 ) ) + 3 . \boxed{f(x)=4\cos\left(\frac65\left(x-\frac{\pi}{3}\right)\right)+3}. f ( x ) = 4 cos ( 5 6 β ( x β 3 Ο β ) ) + 3 β . The key-point increment is
1 4 β
5 Ο 3 = 5 Ο 12 . \frac14\cdot\frac{5\pi}{3}=\frac{5\pi}{12}. 4 1 β β
3 5 Ο β = 12 5 Ο β . Starting at the maximum x = Ο 3 x=\frac{\pi}{3} x = 3 Ο β
( Ο 3 , 7 ) , ( 3 Ο 4 , 3 ) , ( 7 Ο 6 , β 1 ) , ( 19 Ο 12 , 3 ) , ( 2 Ο , 7 ) . \boxed{\left(\frac{\pi}{3},7\right),
\left(\frac{3\pi}{4},3\right),
\left(\frac{7\pi}{6},-1\right),
\left(\frac{19\pi}{12},3\right),
\left(2\pi,7\right)}. ( 3 Ο β , 7 ) , ( 4 3 Ο β , 3 ) , ( 6 7 Ο β , β 1 ) , ( 12 19 Ο β , 3 ) , ( 2 Ο , 7 ) β . A graph with many key points is shown below:
ΒΌ 3 ΒΌ 2 ΒΌ Β‘ 1 3 7 x y
For g ( x ) = 2 tan β‘ ( 3 ( x + Ο 12 ) ) β 1 g(x)=2\tan\left(3\left(x+\frac{\pi}{12}\right)\right)-1 g ( x ) = 2 tan ( 3 ( x + 12 Ο β ) ) β 1 x x x
The function is
g ( x ) = 2 tan β‘ ( 3 ( x + Ο 12 ) ) β 1. g(x)=2\tan\left(3\left(x+\frac{\pi}{12}\right)\right)-1. g ( x ) = 2 tan ( 3 ( x + 12 Ο β ) ) β 1. The period of tangent is
Ο β£ 3 β£ = Ο 3 . \frac{\pi}{\lvert 3 \rvert}=\frac{\pi}{3}. β£ 3 β£ Ο β = 3 Ο β . So
period = Ο 3 . \boxed{\text{period}=\frac{\pi}{3}}. period = 3 Ο β β . The center point occurs where the tangent input is 0 0 0
3 ( x + Ο 12 ) = 0. 3\left(x+\frac{\pi}{12}\right)=0. 3 ( x + 12 Ο β ) = 0. Thus
x = β Ο 12 . x=-\frac{\pi}{12}. x = β 12 Ο β . At the center,
g ( β Ο 12 ) = 2 tan β‘ ( 0 ) β 1 = β 1. g\left(-\frac{\pi}{12}\right)=2\tan(0)-1=-1. g ( β 12 Ο β ) = 2 tan ( 0 ) β 1 = β 1. So the center point is
( β Ο 12 , β 1 ) . \boxed{\left(-\frac{\pi}{12},-1\right)}. ( β 12 Ο β , β 1 ) β . The distance from the center to each vertical asymptote is half the period:
1 2 β
Ο 3 = Ο 6 . \frac12\cdot\frac{\pi}{3}=\frac{\pi}{6}. 2 1 β β
3 Ο β = 6 Ο β . So the surrounding vertical asymptotes are
β Ο 12 β Ο 6 = β Ο 4 -\frac{\pi}{12}-\frac{\pi}{6}=-\frac{\pi}{4} β 12 Ο β β 6 Ο β = β 4 Ο β and
β Ο 12 + Ο 6 = Ο 12 . -\frac{\pi}{12}+\frac{\pi}{6}=\frac{\pi}{12}. β 12 Ο β + 6 Ο β = 12 Ο β . Thus
x = β Ο 4 and x = Ο 12 . \boxed{x=-\frac{\pi}{4}\quad\text{and}\quad x=\frac{\pi}{12}}. x = β 4 Ο β and x = 12 Ο β β . For the x x x g ( x ) = 0 g(x)=0 g ( x ) = 0
2 tan β‘ ( 3 ( x + Ο 12 ) ) β 1 = 0. 2\tan\left(3\left(x+\frac{\pi}{12}\right)\right)-1=0. 2 tan ( 3 ( x + 12 Ο β ) ) β 1 = 0. Then
tan β‘ ( 3 ( x + Ο 12 ) ) = 1 2 . \tan\left(3\left(x+\frac{\pi}{12}\right)\right)=\frac12. tan ( 3 ( x + 12 Ο β ) ) = 2 1 β . On the center branch,
3 ( x + Ο 12 ) = tan β‘ β 1 ( 1 2 ) . 3\left(x+\frac{\pi}{12}\right)=\tan^{-1}\left(\frac12\right). 3 ( x + 12 Ο β ) = tan β 1 ( 2 1 β ) . Therefore
x = 1 3 tan β‘ β 1 ( 1 2 ) β Ο 12 . \boxed{x=\frac13\tan^{-1}\left(\frac12\right)-\frac{\pi}{12}}. x = 3 1 β tan β 1 ( 2 1 β ) β 12 Ο β β . The branch should increase from the vertical asymptote x = β Ο 4 x=-\frac{\pi}{4} x = β 4 Ο β x = Ο 12 x=\frac{\pi}{12} x = 12 Ο β ( β Ο 12 , β 1 ) \left(-\frac{\pi}{12},-1\right) ( β 12 Ο β , β 1 ) x x x
A graph with many key points is shown below:
Β‘ ΒΌ 12 Β‘ 1 x y
For h ( x ) = β 2 sec β‘ ( 1 2 ( x β Ο ) ) + 3 h(x)=-2\sec\left(\frac12(x-\pi)\right)+3 h ( x ) = β 2 sec ( 2 1 β ( x β Ο ) ) + 3 x = Ο x=\pi x = Ο
The function is
h ( x ) = β 2 sec β‘ ( 1 2 ( x β Ο ) ) + 3. h(x)=-2\sec\left(\frac12(x-\pi)\right)+3. h ( x ) = β 2 sec ( 2 1 β ( x β Ο ) ) + 3. The period of secant is
2 Ο β£ 1 2 β£ = 4 Ο . \frac{2\pi}{\left\lvert \frac12\right\rvert}=4\pi. β 2 1 β β 2 Ο β = 4 Ο . So
period = 4 Ο . \boxed{\text{period}=4\pi}. period = 4 Ο β . The midline is
y = 3 . \boxed{y=3}. y = 3 β . One period starting at x = Ο x=\pi x = Ο
x = Ο x=\pi x = Ο to
x = Ο + 4 Ο = 5 Ο . x=\pi+4\pi=5\pi. x = Ο + 4 Ο = 5 Ο . Secant has vertical asymptotes when the cosine inside is 0 0 0
1 2 ( x β Ο ) = Ο 2 + k Ο . \frac12(x-\pi)=\frac{\pi}{2}+k\pi. 2 1 β ( x β Ο ) = 2 Ο β + k Ο . Multiply by 2 2 2
x β Ο = Ο + 2 k Ο . x-\pi=\pi+2k\pi. x β Ο = Ο + 2 k Ο . Thus
x = 2 Ο + 2 k Ο . x=2\pi+2k\pi. x = 2 Ο + 2 k Ο . In the period [ Ο , 5 Ο ] [\pi,5\pi] [ Ο , 5 Ο ]
x = 2 Ο and x = 4 Ο . \boxed{x=2\pi\quad\text{and}\quad x=4\pi}. x = 2 Ο and x = 4 Ο β . The parent secant has range ( β β , β 1 ] βͺ [ 1 , β ) (-\infty,-1]\cup[1,\infty) ( β β , β 1 ] βͺ [ 1 , β ) β 2 -2 β 2 ( β β , β 2 ] βͺ [ 2 , β ) (-\infty,-2]\cup[2,\infty) ( β β , β 2 ] βͺ [ 2 , β ) 3 3 3
( β β , 1 ] βͺ [ 5 , β ) . \boxed{(-\infty,1]\cup[5,\infty)}. ( β β , 1 ] βͺ [ 5 , β ) β . The guiding cosine curve is
y = β 2 cos β‘ ( 1 2 ( x β Ο ) ) + 3. y=-2\cos\left(\frac12(x-\pi)\right)+3. y = β 2 cos ( 2 1 β ( x β Ο ) ) + 3. On [ Ο , 5 Ο ] [\pi,5\pi] [ Ο , 5 Ο ]
( Ο , 1 ) , ( 2 Ο , 3 ) , ( 3 Ο , 5 ) , ( 4 Ο , 3 ) , ( 5 Ο , 1 ) . (\pi,1),\quad (2\pi,3),\quad (3\pi,5),\quad (4\pi,3),\quad (5\pi,1). ( Ο , 1 ) , ( 2 Ο , 3 ) , ( 3 Ο , 5 ) , ( 4 Ο , 3 ) , ( 5 Ο , 1 ) . The secant graph has vertices at ( Ο , 1 ) (\pi,1) ( Ο , 1 ) ( 3 Ο , 5 ) (3\pi,5) ( 3 Ο , 5 ) ( 5 Ο , 1 ) (5\pi,1) ( 5 Ο , 1 ) x = 2 Ο x=2\pi x = 2 Ο x = 4 Ο x=4\pi x = 4 Ο
A graph with many key points is shown below:
ΒΌ 2 ΒΌ 3 ΒΌ 4 ΒΌ 5 ΒΌ 1 3 5 7 x y
Solve exactly on [ 0 , 4 Ο ) [0,4\pi) [ 0 , 4 Ο ) y = 2 sin β‘ 2 x β sin β‘ x β 1 y=2\sin^2x-\sin x-1 y = 2 sin 2 x β sin x β 1 [ 0 , 4 Ο ) [0,4\pi) [ 0 , 4 Ο ) x x x 2 sin β‘ 2 x β sin β‘ x β 1 = 0. 2\sin^2x-\sin x-1=0. 2 sin 2 x β sin x β 1 = 0.
Let
u = sin β‘ x . u=\sin x. u = sin x . Then
2 u 2 β u β 1 = 0. 2u^2-u-1=0. 2 u 2 β u β 1 = 0. Factor:
( 2 u + 1 ) ( u β 1 ) = 0. (2u+1)(u-1)=0. ( 2 u + 1 ) ( u β 1 ) = 0. Thus
sin β‘ x = β 1 2 or sin β‘ x = 1. \sin x=-\frac12
\qquad\text{or}\qquad
\sin x=1. sin x = β 2 1 β or sin x = 1. On [ 0 , 4 Ο ) [0,4\pi) [ 0 , 4 Ο )
sin β‘ x = 1 \sin x=1 sin x = 1 at
x = Ο 2 , 5 Ο 2 . x=\frac{\pi}{2},\frac{5\pi}{2}. x = 2 Ο β , 2 5 Ο β . Also,
sin β‘ x = β 1 2 \sin x=-\frac12 sin x = β 2 1 β at
x = 7 Ο 6 , 11 Ο 6 , 19 Ο 6 , 23 Ο 6 . x=\frac{7\pi}{6},\frac{11\pi}{6},\frac{19\pi}{6},\frac{23\pi}{6}. x = 6 7 Ο β , 6 11 Ο β , 6 19 Ο β , 6 23 Ο β . Therefore
x = Ο 2 , 7 Ο 6 , 11 Ο 6 , 5 Ο 2 , 19 Ο 6 , 23 Ο 6 . \boxed{x=\frac{\pi}{2},\frac{7\pi}{6},\frac{11\pi}{6},\frac{5\pi}{2},\frac{19\pi}{6},\frac{23\pi}{6}}. x = 2 Ο β , 6 7 Ο β , 6 11 Ο β , 2 5 Ο β , 6 19 Ο β , 6 23 Ο β β . The graph of y = 2 sin β‘ 2 x β sin β‘ x β 1 y=2\sin^2x-\sin x-1 y = 2 sin 2 x β sin x β 1 x x x [ 0 , 4 Ο ) [0,4\pi) [ 0 , 4 Ο )
A graph with many key points is shown below:
ΒΌ 2 ΒΌ 3 ΒΌ 4 ΒΌ Β‘ 1 1 2 x y
Solve on [ 0 , 4 Ο ) [0, 4\pi) [ 0 , 4 Ο ) 4 sin β‘ ( 4 x ) cos β‘ ( 6 x ) = 2 sin β‘ ( 10 x ) + 1 4\sin(4x)\cos(6x)= 2\sin(10x)+1 4 sin ( 4 x ) cos ( 6 x ) = 2 sin ( 10 x ) + 1
Start with
4 sin β‘ ( 4 x ) cos β‘ ( 6 x ) = 2 sin β‘ ( 10 x ) + 1. 4\sin(4x)\cos(6x)=2\sin(10x)+1. 4 sin ( 4 x ) cos ( 6 x ) = 2 sin ( 10 x ) + 1. Use the product-to-sum identity
sin β‘ A cos β‘ B = 1 2 [ sin β‘ ( A + B ) + sin β‘ ( A β B ) ] . \sin A\cos B=\frac12[\sin(A+B)+\sin(A-B)]. sin A cos B = 2 1 β [ sin ( A + B ) + sin ( A β B )] . Then
4 sin β‘ ( 4 x ) cos β‘ ( 6 x ) = 2 [ sin β‘ ( 10 x ) + sin β‘ ( β 2 x ) ] . 4\sin(4x)\cos(6x)
=2[\sin(10x)+\sin(-2x)]. 4 sin ( 4 x ) cos ( 6 x ) = 2 [ sin ( 10 x ) + sin ( β 2 x )] . Since sin β‘ ( β 2 x ) = β sin β‘ ( 2 x ) \sin(-2x)=-\sin(2x) sin ( β 2 x ) = β sin ( 2 x )
4 sin β‘ ( 4 x ) cos β‘ ( 6 x ) = 2 sin β‘ ( 10 x ) β 2 sin β‘ ( 2 x ) . 4\sin(4x)\cos(6x)=2\sin(10x)-2\sin(2x). 4 sin ( 4 x ) cos ( 6 x ) = 2 sin ( 10 x ) β 2 sin ( 2 x ) . So the equation becomes
2 sin β‘ ( 10 x ) β 2 sin β‘ ( 2 x ) = 2 sin β‘ ( 10 x ) + 1. 2\sin(10x)-2\sin(2x)=2\sin(10x)+1. 2 sin ( 10 x ) β 2 sin ( 2 x ) = 2 sin ( 10 x ) + 1. Subtract 2 sin β‘ ( 10 x ) 2\sin(10x) 2 sin ( 10 x )
β 2 sin β‘ ( 2 x ) = 1. -2\sin(2x)=1. β 2 sin ( 2 x ) = 1. Thus
sin β‘ ( 2 x ) = β 1 2 . \sin(2x)=-\frac12. sin ( 2 x ) = β 2 1 β . Let u = 2 x u=2x u = 2 x x β [ 0 , 4 Ο ) x\in[0,4\pi) x β [ 0 , 4 Ο )
u β [ 0 , 8 Ο ) . u\in[0,8\pi). u β [ 0 , 8 Ο ) . On [ 0 , 8 Ο ) [0,8\pi) [ 0 , 8 Ο ) sin β‘ u = β 1 2 \sin u=-\frac12 sin u = β 2 1 β
u = 7 Ο 6 , 11 Ο 6 , 19 Ο 6 , 23 Ο 6 , 31 Ο 6 , 35 Ο 6 , 43 Ο 6 , 47 Ο 6 . u=\frac{7\pi}{6},\frac{11\pi}{6},\frac{19\pi}{6},\frac{23\pi}{6},
\frac{31\pi}{6},\frac{35\pi}{6},\frac{43\pi}{6},\frac{47\pi}{6}. u = 6 7 Ο β , 6 11 Ο β , 6 19 Ο β , 6 23 Ο β , 6 31 Ο β , 6 35 Ο β , 6 43 Ο β , 6 47 Ο β . Divide by 2 2 2
x = 7 Ο 12 , 11 Ο 12 , 19 Ο 12 , 23 Ο 12 , 31 Ο 12 , 35 Ο 12 , 43 Ο 12 , 47 Ο 12 . \boxed{x=\frac{7\pi}{12},\frac{11\pi}{12},\frac{19\pi}{12},\frac{23\pi}{12},
\frac{31\pi}{12},\frac{35\pi}{12},\frac{43\pi}{12},\frac{47\pi}{12}}. x = 12 7 Ο β , 12 11 Ο β , 12 19 Ο β , 12 23 Ο β , 12 31 Ο β , 12 35 Ο β , 12 43 Ο β , 12 47 Ο β β .
Solve exactly on [ 0 , 2 Ο ) [0,2\pi) [ 0 , 2 Ο ) tan β‘ x + cot β‘ x = 4. \tan x+\cot x=4. tan x + cot x = 4.
The equation is
tan β‘ x + cot β‘ x = 4. \tan x+\cot x=4. tan x + cot x = 4. Let
t = tan β‘ x . t=\tan x. t = tan x . Then cot β‘ x = 1 t \cot x=\frac1t cot x = t 1 β
t + 1 t = 4. t+\frac1t=4. t + t 1 β = 4. Multiply by t t t
t 2 + 1 = 4 t . t^2+1=4t. t 2 + 1 = 4 t . So
t 2 β 4 t + 1 = 0. t^2-4t+1=0. t 2 β 4 t + 1 = 0. Use the quadratic formula:
t = 4 Β± 16 β 4 2 = 2 Β± 3 . t=\frac{4\pm\sqrt{16-4}}{2}
=2\pm\sqrt3. t = 2 4 Β± 16 β 4 β β = 2 Β± 3 β . Since
tan β‘ ( Ο 12 ) = 2 β 3 \tan\left(\frac{\pi}{12}\right)=2-\sqrt3 tan ( 12 Ο β ) = 2 β 3 β and
tan β‘ ( 5 Ο 12 ) = 2 + 3 , \tan\left(\frac{5\pi}{12}\right)=2+\sqrt3, tan ( 12 5 Ο β ) = 2 + 3 β , the solutions on [ 0 , 2 Ο ) [0,2\pi) [ 0 , 2 Ο )
x = Ο 12 , 5 Ο 12 , 13 Ο 12 , 17 Ο 12 . \boxed{x=\frac{\pi}{12},\frac{5\pi}{12},\frac{13\pi}{12},\frac{17\pi}{12}}. x = 12 Ο β , 12 5 Ο β , 12 13 Ο β , 12 17 Ο β β .
Evaluate each of these exactly: sin β‘ 75 β , cos β‘ 15 β , tan β‘ 105 β . \sin 75^\circ, \cos 15^\circ,\tan 105^\circ. sin 7 5 β , cos 1 5 β , tan 10 5 β . 15 β 15^\circ 1 5 β 75 β 75^\circ 7 5 β
Use angle addition and subtraction formulas.
First,
sin β‘ 75 β = sin β‘ ( 45 β + 30 β ) . \sin75^\circ=\sin(45^\circ+30^\circ). sin 7 5 β = sin ( 4 5 β + 3 0 β ) . So
sin β‘ 75 β = sin β‘ 45 β cos β‘ 30 β + cos β‘ 45 β sin β‘ 30 β . \sin75^\circ
=\sin45^\circ\cos30^\circ+\cos45^\circ\sin30^\circ. sin 7 5 β = sin 4 5 β cos 3 0 β + cos 4 5 β sin 3 0 β . Thus
sin β‘ 75 β = 2 2 β
3 2 + 2 2 β
1 2 = 6 + 2 4 . \sin75^\circ
=\frac{\sqrt2}{2}\cdot\frac{\sqrt3}{2}
+\frac{\sqrt2}{2}\cdot\frac12
=\frac{\sqrt6+\sqrt2}{4}. sin 7 5 β = 2 2 β β β
2 3 β β + 2 2 β β β
2 1 β = 4 6 β + 2 β β . Next,
cos β‘ 15 β = cos β‘ ( 45 β β 30 β ) . \cos15^\circ=\cos(45^\circ-30^\circ). cos 1 5 β = cos ( 4 5 β β 3 0 β ) . So
cos β‘ 15 β = cos β‘ 45 β cos β‘ 30 β + sin β‘ 45 β sin β‘ 30 β = 6 + 2 4 . \cos15^\circ
=\cos45^\circ\cos30^\circ+\sin45^\circ\sin30^\circ
=\frac{\sqrt6+\sqrt2}{4}. cos 1 5 β = cos 4 5 β cos 3 0 β + sin 4 5 β sin 3 0 β = 4 6 β + 2 β β . Finally,
tan β‘ 105 β = tan β‘ ( 60 β + 45 β ) . \tan105^\circ=\tan(60^\circ+45^\circ). tan 10 5 β = tan ( 6 0 β + 4 5 β ) . Then
tan β‘ 105 β = 3 + 1 1 β 3 . \tan105^\circ
=\frac{\sqrt3+1}{1-\sqrt3}. tan 10 5 β = 1 β 3 β 3 β + 1 β . Rationalize:
3 + 1 1 β 3 β
1 + 3 1 + 3 = 4 + 2 3 β 2 = β 2 β 3 . \frac{\sqrt3+1}{1-\sqrt3}\cdot\frac{1+\sqrt3}{1+\sqrt3}
=\frac{4+2\sqrt3}{-2}
=-2-\sqrt3. 1 β 3 β 3 β + 1 β β
1 + 3 β 1 + 3 β β = β 2 4 + 2 3 β β = β 2 β 3 β . Therefore
sin β‘ 75 β = 6 + 2 4 , \boxed{\sin75^\circ=\frac{\sqrt6+\sqrt2}{4}}, sin 7 5 β = 4 6 β + 2 β β β , cos β‘ 15 β = 6 + 2 4 , \boxed{\cos15^\circ=\frac{\sqrt6+\sqrt2}{4}}, cos 1 5 β = 4 6 β + 2 β β β , and
tan β‘ 105 β = β 2 β 3 . \boxed{\tan105^\circ=-2-\sqrt3}. tan 10 5 β = β 2 β 3 β β .
Prove the identity: sin β‘ ( x + y ) + sin β‘ ( x β y ) cos β‘ ( x + y ) + cos β‘ ( x β y ) = tan β‘ x . \frac{\sin(x+y)+\sin(x-y)}{\cos(x+y)+\cos(x-y)}=\tan x. c o s ( x + y ) + c o s ( x β y ) s i n ( x + y ) + s i n ( x β y ) β = tan x .
Start with the left-hand side:
sin β‘ ( x + y ) + sin β‘ ( x β y ) cos β‘ ( x + y ) + cos β‘ ( x β y ) . \frac{\sin(x+y)+\sin(x-y)}{\cos(x+y)+\cos(x-y)}. cos ( x + y ) + cos ( x β y ) sin ( x + y ) + sin ( x β y ) β . Use sum-to-product formulas:
sin β‘ ( x + y ) + sin β‘ ( x β y ) = 2 sin β‘ x cos β‘ y . \sin(x+y)+\sin(x-y)
=2\sin x\cos y. sin ( x + y ) + sin ( x β y ) = 2 sin x cos y . Also,
cos β‘ ( x + y ) + cos β‘ ( x β y ) = 2 cos β‘ x cos β‘ y . \cos(x+y)+\cos(x-y)
=2\cos x\cos y. cos ( x + y ) + cos ( x β y ) = 2 cos x cos y . Therefore
sin β‘ ( x + y ) + sin β‘ ( x β y ) cos β‘ ( x + y ) + cos β‘ ( x β y ) = 2 sin β‘ x cos β‘ y 2 cos β‘ x cos β‘ y . \frac{\sin(x+y)+\sin(x-y)}{\cos(x+y)+\cos(x-y)}
=\frac{2\sin x\cos y}{2\cos x\cos y}. cos ( x + y ) + cos ( x β y ) sin ( x + y ) + sin ( x β y ) β = 2 cos x cos y 2 sin x cos y β . Cancel:
2 sin β‘ x cos β‘ y 2 cos β‘ x cos β‘ y = sin β‘ x cos β‘ x = tan β‘ x . \frac{2\sin x\cos y}{2\cos x\cos y}
=\frac{\sin x}{\cos x}
=\tan x. 2 cos x cos y 2 sin x cos y β = cos x sin x β = tan x . Thus
sin β‘ ( x + y ) + sin β‘ ( x β y ) cos β‘ ( x + y ) + cos β‘ ( x β y ) = tan β‘ x . \boxed{\frac{\sin(x+y)+\sin(x-y)}{\cos(x+y)+\cos(x-y)}=\tan x}. cos ( x + y ) + cos ( x β y ) sin ( x + y ) + sin ( x β y ) β = tan x β .
Solve exactly on [ 0 , 2 Ο ) [0,2\pi) [ 0 , 2 Ο ) cos β‘ ( 5 x ) + cos β‘ ( 3 x ) = 0. \cos(5x)+\cos(3x)=0. cos ( 5 x ) + cos ( 3 x ) = 0.
Use the sum-to-product formula:
cos β‘ A + cos β‘ B = 2 cos β‘ ( A + B 2 ) cos β‘ ( A β B 2 ) . \cos A+\cos B
=2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right). cos A + cos B = 2 cos ( 2 A + B β ) cos ( 2 A β B β ) . With A = 5 x A=5x A = 5 x B = 3 x B=3x B = 3 x
cos β‘ ( 5 x ) + cos β‘ ( 3 x ) = 2 cos β‘ ( 4 x ) cos β‘ x . \cos(5x)+\cos(3x)
=2\cos(4x)\cos x. cos ( 5 x ) + cos ( 3 x ) = 2 cos ( 4 x ) cos x . So the equation becomes
2 cos β‘ ( 4 x ) cos β‘ x = 0. 2\cos(4x)\cos x=0. 2 cos ( 4 x ) cos x = 0. Thus
cos β‘ ( 4 x ) = 0 \cos(4x)=0 cos ( 4 x ) = 0 or
cos β‘ x = 0. \cos x=0. cos x = 0. For cos β‘ x = 0 \cos x=0 cos x = 0 [ 0 , 2 Ο ) [0,2\pi) [ 0 , 2 Ο )
x = Ο 2 , 3 Ο 2 . x=\frac{\pi}{2},\frac{3\pi}{2}. x = 2 Ο β , 2 3 Ο β . For cos β‘ ( 4 x ) = 0 \cos(4x)=0 cos ( 4 x ) = 0
4 x = Ο 2 + k Ο . 4x=\frac{\pi}{2}+k\pi. 4 x = 2 Ο β + k Ο . So
x = Ο 8 + k Ο 4 . x=\frac{\pi}{8}+\frac{k\pi}{4}. x = 8 Ο β + 4 k Ο β . On [ 0 , 2 Ο ) [0,2\pi) [ 0 , 2 Ο )
x = Ο 8 , 3 Ο 8 , 5 Ο 8 , 7 Ο 8 , 9 Ο 8 , 11 Ο 8 , 13 Ο 8 , 15 Ο 8 . x=\frac{\pi}{8},\frac{3\pi}{8},\frac{5\pi}{8},\frac{7\pi}{8},
\frac{9\pi}{8},\frac{11\pi}{8},\frac{13\pi}{8},\frac{15\pi}{8}. x = 8 Ο β , 8 3 Ο β , 8 5 Ο β , 8 7 Ο β , 8 9 Ο β , 8 11 Ο β , 8 13 Ο β , 8 15 Ο β . Therefore
x = Ο 8 , 3 Ο 8 , Ο 2 , 5 Ο 8 , 7 Ο 8 , 9 Ο 8 , 11 Ο 8 , 3 Ο 2 , 13 Ο 8 , 15 Ο 8 . \boxed{x=\frac{\pi}{8},\frac{3\pi}{8},\frac{\pi}{2},\frac{5\pi}{8},\frac{7\pi}{8},\frac{9\pi}{8},\frac{11\pi}{8},\frac{3\pi}{2},\frac{13\pi}{8},\frac{15\pi}{8}}. x = 8 Ο β , 8 3 Ο β , 2 Ο β , 8 5 Ο β , 8 7 Ο β , 8 9 Ο β , 8 11 Ο β , 2 3 Ο β , 8 13 Ο β , 8 15 Ο β β . The product-to-sum form
cos β‘ ( 5 x ) + cos β‘ ( 3 x ) = 2 cos β‘ ( 4 x ) cos β‘ x \cos(5x)+\cos(3x)=2\cos(4x)\cos x cos ( 5 x ) + cos ( 3 x ) = 2 cos ( 4 x ) cos x shows that zeros occur whenever either factor is zero. A sketch should mark the eight zeros from cos β‘ ( 4 x ) = 0 \cos(4x)=0 cos ( 4 x ) = 0 cos β‘ x = 0 \cos x=0 cos x = 0
ΒΌ 2 ΒΌ 3 ΒΌ 2 2 ΒΌ Β‘ 2 Β‘ 1 1 2 x y
Evaluate exactly: sin β‘ ( cos β‘ β 1 3 5 ) + cos β‘ ( sin β‘ β 1 ( β 5 13 ) ) + tan β‘ ( cos β‘ β 1 ( β 4 5 ) ) . \sin\left(\cos^{-1}\frac35\right)+\cos\left(\sin^{-1}\left(-\frac5{13}\right)\right)+\tan\left(\cos^{-1}\left(-\frac45\right)\right). sin ( cos β 1 5 3 β ) + cos ( sin β 1 ( β 13 5 β ) ) + tan ( cos β 1 ( β 5 4 β ) ) .
Let
Ξ± = cos β‘ β 1 3 5 . \alpha=\cos^{-1}\frac35. Ξ± = cos β 1 5 3 β . Then cos β‘ Ξ± = 3 5 \cos\alpha=\frac35 cos Ξ± = 5 3 β Ξ± β [ 0 , Ο ] \alpha\in[0,\pi] Ξ± β [ 0 , Ο ] Ξ± \alpha Ξ±
sin β‘ Ξ± = 4 5 . \sin\alpha=\frac45. sin Ξ± = 5 4 β . So
sin β‘ ( cos β‘ β 1 3 5 ) = 4 5 . \sin\left(\cos^{-1}\frac35\right)=\frac45. sin ( cos β 1 5 3 β ) = 5 4 β . Next, let
Ξ² = sin β‘ β 1 ( β 5 13 ) . \beta=\sin^{-1}\left(-\frac5{13}\right). Ξ² = sin β 1 ( β 13 5 β ) . Since Ξ² β [ β Ο 2 , Ο 2 ] \beta\in\left[-\frac{\pi}{2},\frac{\pi}{2}\right] Ξ² β [ β 2 Ο β , 2 Ο β ] Ξ² \beta Ξ²
cos β‘ Ξ² = 12 13 . \cos\beta=\frac{12}{13}. cos Ξ² = 13 12 β . So
cos β‘ ( sin β‘ β 1 ( β 5 13 ) ) = 12 13 . \cos\left(\sin^{-1}\left(-\frac5{13}\right)\right)=\frac{12}{13}. cos ( sin β 1 ( β 13 5 β ) ) = 13 12 β . Finally, let
Ξ³ = cos β‘ β 1 ( β 4 5 ) . \gamma=\cos^{-1}\left(-\frac45\right). Ξ³ = cos β 1 ( β 5 4 β ) . Then Ξ³ β [ 0 , Ο ] \gamma\in[0,\pi] Ξ³ β [ 0 , Ο ] Ξ³ \gamma Ξ³
sin β‘ Ξ³ = 3 5 , \sin\gamma=\frac35, sin Ξ³ = 5 3 β , and
tan β‘ Ξ³ = 3 5 β 4 5 = β 3 4 . \tan\gamma=\frac{\frac35}{-\frac45}=-\frac34. tan Ξ³ = β 5 4 β 5 3 β β = β 4 3 β . Now add:
4 5 + 12 13 β 3 4 . \frac45+\frac{12}{13}-\frac34. 5 4 β + 13 12 β β 4 3 β . Using common denominator 260 260 260
208 260 + 240 260 β 195 260 = 253 260 . \frac{208}{260}+\frac{240}{260}-\frac{195}{260}
=\frac{253}{260}. 260 208 β + 260 240 β β 260 195 β = 260 253 β . Therefore
253 260 . \boxed{\frac{253}{260}}. 260 253 β β .
How many solutions does the equation tan β‘ ( 2 x ) = cos β‘ ( x 2 ) \tan(2x)=\cos(\tfrac{x}{2}) tan ( 2 x ) = cos ( 2 x β ) [ 0 , 2 Ο ] [0,2\pi] [ 0 , 2 Ο ]
We need count solutions to
tan β‘ ( 2 x ) = cos β‘ ( x 2 ) \tan(2x)=\cos\left(\frac{x}{2}\right) tan ( 2 x ) = cos ( 2 x β ) on [ 0 , 2 Ο ] [0,2\pi] [ 0 , 2 Ο ]
The tangent side is undefined when
2 x = Ο 2 + k Ο . 2x=\frac{\pi}{2}+k\pi. 2 x = 2 Ο β + k Ο . Thus
x = Ο 4 + k Ο 2 . x=\frac{\pi}{4}+\frac{k\pi}{2}. x = 4 Ο β + 2 k Ο β . On [ 0 , 2 Ο ] [0,2\pi] [ 0 , 2 Ο ]
Ο 4 , 3 Ο 4 , 5 Ο 4 , 7 Ο 4 . \frac{\pi}{4},\frac{3\pi}{4},\frac{5\pi}{4},\frac{7\pi}{4}. 4 Ο β , 4 3 Ο β , 4 5 Ο β , 4 7 Ο β . These split the interval into five pieces:
[ 0 , Ο 4 ) , ( Ο 4 , 3 Ο 4 ) , ( 3 Ο 4 , 5 Ο 4 ) , ( 5 Ο 4 , 7 Ο 4 ) , ( 7 Ο 4 , 2 Ο ] . \left[0,\frac{\pi}{4}\right),\quad
\left(\frac{\pi}{4},\frac{3\pi}{4}\right),\quad
\left(\frac{3\pi}{4},\frac{5\pi}{4}\right),\quad
\left(\frac{5\pi}{4},\frac{7\pi}{4}\right),\quad
\left(\frac{7\pi}{4},2\pi\right]. [ 0 , 4 Ο β ) , ( 4 Ο β , 4 3 Ο β ) , ( 4 3 Ο β , 4 5 Ο β ) , ( 4 5 Ο β , 4 7 Ο β ) , ( 4 7 Ο β , 2 Ο ] . Let
F ( x ) = tan β‘ ( 2 x ) β cos β‘ ( x 2 ) . F(x)=\tan(2x)-\cos\left(\frac{x}{2}\right). F ( x ) = tan ( 2 x ) β cos ( 2 x β ) . On each of the five pieces, tan β‘ ( 2 x ) \tan(2x) tan ( 2 x ) β β -\infty β β β \infty β cos β‘ ( x 2 ) \cos\left(\frac{x}{2}\right) cos ( 2 x β ) 1 1 1 β 1 -1 β 1 [ 0 , 2 Ο ] [0,2\pi] [ 0 , 2 Ο ]
More formally,
F β² ( x ) = 2 sec β‘ 2 ( 2 x ) + 1 2 sin β‘ ( x 2 ) > 0 F'(x)=2\sec^2(2x)+\frac12\sin\left(\frac{x}{2}\right)>0 F β² ( x ) = 2 sec 2 ( 2 x ) + 2 1 β sin ( 2 x β ) > 0 wherever F F F [ 0 , 2 Ο ] [0,2\pi] [ 0 , 2 Ο ] F F F
On each piece, the function changes from negative to positive:
On the first interval, F ( 0 ) = β 1 F(0)=-1 F ( 0 ) = β 1 F ( x ) β β F(x)\to\infty F ( x ) β β x β Ο 4 β x\to\frac{\pi}{4}^{-} x β 4 Ο β β
On each middle interval, F ( x ) β β β F(x)\to-\infty F ( x ) β β β F ( x ) β β F(x)\to\infty F ( x ) β β
On the last interval, F ( x ) β β β F(x)\to-\infty F ( x ) β β β x β 7 Ο 4 + x\to\frac{7\pi}{4}^{+} x β 4 7 Ο β + F ( 2 Ο ) = 1 F(2\pi)=1 F ( 2 Ο ) = 1
Therefore there is exactly one solution in each of the five pieces.
5 \boxed{5} 5 β
Suppose Ο 2 < ΞΈ < Ο \frac{\pi}{2}<\theta<\pi 2 Ο β < ΞΈ < Ο cos β‘ ΞΈ = β 3 5 \cos\theta=-\frac35 cos ΞΈ = β 5 3 β sin β‘ ( ΞΈ 2 ) , cos β‘ ( ΞΈ 2 ) , \sin\left(\frac{\theta}{2}\right), \cos\left(\frac{\theta}{2}\right), sin ( 2 ΞΈ β ) , cos ( 2 ΞΈ β ) , tan β‘ ( ΞΈ 2 ) . \tan\left(\frac{\theta}{2}\right). tan ( 2 ΞΈ β ) .
We are given
Ο 2 < ΞΈ < Ο , \frac{\pi}{2}<\theta<\pi, 2 Ο β < ΞΈ < Ο , so ΞΈ \theta ΞΈ
ΞΈ 2 \frac{\theta}{2} 2 ΞΈ β is in Quadrant I.
Since
cos β‘ ΞΈ = β 3 5 , \cos\theta=-\frac35, cos ΞΈ = β 5 3 β , the half-angle formulas give
sin β‘ ( ΞΈ 2 ) = 1 β cos β‘ ΞΈ 2 . \sin\left(\frac{\theta}{2}\right)
=\sqrt{\frac{1-\cos\theta}{2}}. sin ( 2 ΞΈ β ) = 2 1 β cos ΞΈ β β . Substitute:
sin β‘ ( ΞΈ 2 ) = 1 β ( β 3 5 ) 2 = 8 5 2 = 4 5 = 2 5 5 . \sin\left(\frac{\theta}{2}\right)
=\sqrt{\frac{1-\left(-\frac35\right)}{2}}
=\sqrt{\frac{\frac85}{2}}
=\sqrt{\frac45}
=\frac{2\sqrt5}{5}. sin ( 2 ΞΈ β ) = 2 1 β ( β 5 3 β ) β β = 2 5 8 β β β = 5 4 β β = 5 2 5 β β . Also,
cos β‘ ( ΞΈ 2 ) = 1 + cos β‘ ΞΈ 2 = 1 β 3 5 2 = 2 5 2 = 1 5 = 5 5 . \cos\left(\frac{\theta}{2}\right)
=\sqrt{\frac{1+\cos\theta}{2}}
=\sqrt{\frac{1-\frac35}{2}}
=\sqrt{\frac{\frac25}{2}}
=\sqrt{\frac15}
=\frac{\sqrt5}{5}. cos ( 2 ΞΈ β ) = 2 1 + cos ΞΈ β β = 2 1 β 5 3 β β β = 2 5 2 β β β = 5 1 β β = 5 5 β β . Thus
tan β‘ ( ΞΈ 2 ) = sin β‘ ( ΞΈ 2 ) cos β‘ ( ΞΈ 2 ) = 2. \tan\left(\frac{\theta}{2}\right)
=\frac{\sin\left(\frac{\theta}{2}\right)}{\cos\left(\frac{\theta}{2}\right)}
=2. tan ( 2 ΞΈ β ) = cos ( 2 ΞΈ β ) sin ( 2 ΞΈ β ) β = 2. Therefore
sin β‘ ( ΞΈ 2 ) = 2 5 5 , cos β‘ ( ΞΈ 2 ) = 5 5 , tan β‘ ( ΞΈ 2 ) = 2 . \boxed{\sin\left(\frac{\theta}{2}\right)=\frac{2\sqrt5}{5}},
\qquad
\boxed{\cos\left(\frac{\theta}{2}\right)=\frac{\sqrt5}{5}},
\qquad
\boxed{\tan\left(\frac{\theta}{2}\right)=2}. sin ( 2 ΞΈ β ) = 5 2 5 β β β , cos ( 2 ΞΈ β ) = 5 5 β β β , tan ( 2 ΞΈ β ) = 2 β .
Let x x x y y y sin β‘ x sin β‘ y = 3 \frac{\sin x}{\sin y} = 3 s i n y s i n x β = 3 cos β‘ x cos β‘ y = 1 2 \frac{\cos x}{\cos y} = \frac{1}{2} c o s y c o s x β = 2 1 β sin β‘ 2 x sin β‘ 2 y + cos β‘ 2 x cos β‘ 2 y \frac{\sin 2x}{\sin 2y} + \frac{\cos 2x}{\cos 2y} s i n 2 y s i n 2 x β + c o s 2 y c o s 2 x β p q \frac{p}{q} q p β p p p q q q p + q p+q p + q
We are given
sin β‘ x sin β‘ y = 3 \frac{\sin x}{\sin y}=3 sin y sin x β = 3 and
cos β‘ x cos β‘ y = 1 2 . \frac{\cos x}{\cos y}=\frac12. cos y cos x β = 2 1 β . So
sin β‘ x = 3 sin β‘ y \sin x=3\sin y sin x = 3 sin y and
cos β‘ x = 1 2 cos β‘ y . \cos x=\frac12\cos y. cos x = 2 1 β cos y . Let
S = sin β‘ 2 y and C = cos β‘ 2 y . S=\sin^2 y
\qquad\text{and}\qquad
C=\cos^2 y. S = sin 2 y and C = cos 2 y . Since S + C = 1 S+C=1 S + C = 1
sin β‘ 2 x + cos β‘ 2 x = 1. \sin^2x+\cos^2x=1. sin 2 x + cos 2 x = 1. Substitute the given ratios:
( 3 sin β‘ y ) 2 + ( 1 2 cos β‘ y ) 2 = 1. (3\sin y)^2+\left(\frac12\cos y\right)^2=1. ( 3 sin y ) 2 + ( 2 1 β cos y ) 2 = 1. Thus
9 S + 1 4 C = 1. 9S+\frac14C=1. 9 S + 4 1 β C = 1. Since C = 1 β S C=1-S C = 1 β S
9 S + 1 4 ( 1 β S ) = 1. 9S+\frac14(1-S)=1. 9 S + 4 1 β ( 1 β S ) = 1. Multiply by 4 4 4
36 S + 1 β S = 4. 36S+1-S=4. 36 S + 1 β S = 4. So
35 S = 3 , 35S=3, 35 S = 3 , and
S = 3 35 . S=\frac3{35}. S = 35 3 β . Then
C = 1 β 3 35 = 32 35 . C=1-\frac3{35}=\frac{32}{35}. C = 1 β 35 3 β = 35 32 β . Now,
sin β‘ 2 x sin β‘ 2 y = 2 sin β‘ x cos β‘ x 2 sin β‘ y cos β‘ y = sin β‘ x sin β‘ y β
cos β‘ x cos β‘ y = 3 β
1 2 = 3 2 . \frac{\sin 2x}{\sin 2y}
=\frac{2\sin x\cos x}{2\sin y\cos y}
=\frac{\sin x}{\sin y}\cdot\frac{\cos x}{\cos y}
=3\cdot\frac12
=\frac32. sin 2 y sin 2 x β = 2 sin y cos y 2 sin x cos x β = sin y sin x β β
cos y cos x β = 3 β
2 1 β = 2 3 β . Also,
cos β‘ 2 x cos β‘ 2 y = cos β‘ 2 x β sin β‘ 2 x cos β‘ 2 y β sin β‘ 2 y . \frac{\cos 2x}{\cos 2y}
=\frac{\cos^2x-\sin^2x}{\cos^2y-\sin^2y}. cos 2 y cos 2 x β = cos 2 y β sin 2 y cos 2 x β sin 2 x β . Compute the numerator:
cos β‘ 2 x β sin β‘ 2 x = 1 4 C β 9 S . \cos^2x-\sin^2x
=\frac14C-9S. cos 2 x β sin 2 x = 4 1 β C β 9 S . Using C = 32 35 C=\frac{32}{35} C = 35 32 β S = 3 35 S=\frac3{35} S = 35 3 β
1 4 C β 9 S = 1 4 β
32 35 β 9 β
3 35 = 8 35 β 27 35 = β 19 35 . \frac14C-9S
=\frac14\cdot\frac{32}{35}-9\cdot\frac3{35}
=\frac8{35}-\frac{27}{35}
=-\frac{19}{35}. 4 1 β C β 9 S = 4 1 β β
35 32 β β 9 β
35 3 β = 35 8 β β 35 27 β = β 35 19 β . The denominator is
C β S = 32 35 β 3 35 = 29 35 . C-S=\frac{32}{35}-\frac3{35}
=\frac{29}{35}. C β S = 35 32 β β 35 3 β = 35 29 β . Therefore
cos β‘ 2 x cos β‘ 2 y = β 19 35 29 35 = β 19 29 . \frac{\cos 2x}{\cos 2y}
=\frac{-\frac{19}{35}}{\frac{29}{35}}
=-\frac{19}{29}. cos 2 y cos 2 x β = 35 29 β β 35 19 β β = β 29 19 β . Now add:
sin β‘ 2 x sin β‘ 2 y + cos β‘ 2 x cos β‘ 2 y = 3 2 β 19 29 . \frac{\sin 2x}{\sin 2y}+\frac{\cos 2x}{\cos 2y}
=\frac32-\frac{19}{29}. sin 2 y sin 2 x β + cos 2 y cos 2 x β = 2 3 β β 29 19 β . Use denominator 58 58 58
3 2 β 19 29 = 87 58 β 38 58 = 49 58 . \frac32-\frac{19}{29}
=\frac{87}{58}-\frac{38}{58}
=\frac{49}{58}. 2 3 β β 29 19 β = 58 87 β β 58 38 β = 58 49 β . Thus
p = 49 and q = 58. p=49
\qquad\text{and}\qquad
q=58. p = 49 and q = 58. So
p + q = 107 . \boxed{p+q=107}. p + q = 107 β .
A tide height is modeled by a sinusoidal function of time. At t = 2 t=2 t = 2 11 11 11 t = 8 t=8 t = 8 3 3 3
( A ) (A) ( A ) H ( t ) H(t) H ( t )
( B ) (B) ( B )
( C ) (C) ( C ) t = 2 t=2 t = 2 9 9 9
( D ) (D) ( D ) H ( t ) = 9 H(t)=9 H ( t ) = 9 t = 2 t=2 t = 2
The high tide is 11 11 11 3 3 3
D = 11 + 3 2 = 7. D=\frac{11+3}{2}=7. D = 2 11 + 3 β = 7. The amplitude is
A = 11 β 3 2 = 4. A=\frac{11-3}{2}=4. A = 2 11 β 3 β = 4. The time from high to the next low is half a period:
8 β 2 = 6. 8-2=6. 8 β 2 = 6. So the period is
12. 12. 12. Thus
B = 2 Ο 12 = Ο 6 . B=\frac{2\pi}{12}=\frac{\pi}{6}. B = 12 2 Ο β = 6 Ο β . Since the tide is at a high when t = 2 t=2 t = 2
H ( t ) = 4 cos β‘ ( Ο 6 ( t β 2 ) ) + 7 . \boxed{H(t)=4\cos\left(\frac{\pi}{6}(t-2)\right)+7}. H ( t ) = 4 cos ( 6 Ο β ( t β 2 ) ) + 7 β . The period is
12 Β hours , \boxed{12\text{ hours}}, 12 Β hours β , and the midline is
H = 7 . \boxed{H=7}. H = 7 β . Now solve for when H ( t ) = 9 H(t)=9 H ( t ) = 9
9 = 4 cos β‘ ( Ο 6 ( t β 2 ) ) + 7. 9=4\cos\left(\frac{\pi}{6}(t-2)\right)+7. 9 = 4 cos ( 6 Ο β ( t β 2 ) ) + 7. Then
2 = 4 cos β‘ ( Ο 6 ( t β 2 ) ) , 2=4\cos\left(\frac{\pi}{6}(t-2)\right), 2 = 4 cos ( 6 Ο β ( t β 2 ) ) , so
cos β‘ ( Ο 6 ( t β 2 ) ) = 1 2 . \cos\left(\frac{\pi}{6}(t-2)\right)=\frac12. cos ( 6 Ο β ( t β 2 ) ) = 2 1 β . The first time after the high occurs when the angle is Ο 3 \frac{\pi}{3} 3 Ο β
Ο 6 ( t β 2 ) = Ο 3 . \frac{\pi}{6}(t-2)=\frac{\pi}{3}. 6 Ο β ( t β 2 ) = 3 Ο β . Multiply by 6 Ο \frac6\pi Ο 6 β
t β 2 = 2. t-2=2. t β 2 = 2. So
t = 4 Β hours . \boxed{t=4\text{ hours}}. t = 4 Β hours β . One full period runs from t = 2 t=2 t = 2 t = 14 t=14 t = 14
( 2 , 11 ) , ( 5 , 7 ) , ( 8 , 3 ) , ( 11 , 7 ) , ( 14 , 11 ) . (2,11),\quad (5,7),\quad (8,3),\quad (11,7),\quad (14,11). ( 2 , 11 ) , ( 5 , 7 ) , ( 8 , 3 ) , ( 11 , 7 ) , ( 14 , 11 ) . The point where H ( t ) = 9 H(t)=9 H ( t ) = 9 t = 2 t=2 t = 2
( 4 , 9 ) . \boxed{(4,9)}. ( 4 , 9 ) β . 2 4 5 8 11 14 16 3 7 9 11 t H ( t )
(Bonus, ViΓ¨teβs formula for Ο \pi Ο
Let
a 1 = 2 a_1=\sqrt2 a 1 β = 2 β and define
a n + 1 = 2 + a n . a_{n+1}=\sqrt{2+a_n}. a n + 1 β = 2 + a n β β . The nested radicals
2 , 2 + 2 , 2 + 2 + 2 , β¦ \sqrt2,\quad \sqrt{2+\sqrt2},\quad \sqrt{2+\sqrt{2+\sqrt2}},\quad \ldots 2 β , 2 + 2 β β , 2 + 2 + 2 β β β , β¦ are connected to repeated half-angle identities.
( A ) (A) ( A )
a n = 2 cos β‘ ( Ο 2 n + 1 ) . a_n=2\cos\left(\frac{\pi}{2^{n+1}}\right). a n β = 2 cos ( 2 n + 1 Ο β ) . ( B ) (B) ( B ) ( A ) (A) ( A ) sin β‘ ( 2 x ) = 2 sin β‘ x cos β‘ x \sin(2x)=2\sin x\cos x sin ( 2 x ) = 2 sin x cos x a n 2 \frac{a_n}{2} 2 a n β β 2 Ο \frac{2}{\pi} Ο 2 β
( C ) (C) ( C )
2 Ο = 2 2 β
2 + 2 2 β
2 + 2 + 2 2 β― β . \frac{2}{\pi}
=\frac{\sqrt2}{2}\cdot
\frac{\sqrt{2+\sqrt2}}{2}\cdot
\frac{\sqrt{2+\sqrt{2+\sqrt2}}}{2}\cdots. Ο 2 β = 2 2 β β β
2 2 + 2 β β β β
2 2 + 2 + 2 β β β β β― . For part ( A ) (A) ( A )
a 1 = 2 = 2 β
2 2 = 2 cos β‘ ( Ο 4 ) . a_1=\sqrt2=2\cdot\frac{\sqrt2}{2}=2\cos\left(\frac{\pi}{4}\right). a 1 β = 2 β = 2 β
2 2 β β = 2 cos ( 4 Ο β ) . Since
Ο 4 = Ο 2 2 , \frac{\pi}{4}=\frac{\pi}{2^{2}}, 4 Ο β = 2 2 Ο β , this matches
a 1 = 2 cos β‘ ( Ο 2 1 + 1 ) . a_1=2\cos\left(\frac{\pi}{2^{1+1}}\right). a 1 β = 2 cos ( 2 1 + 1 Ο β ) . Now assume
a n = 2 cos β‘ ( Ο 2 n + 1 ) . a_n=2\cos\left(\frac{\pi}{2^{n+1}}\right). a n β = 2 cos ( 2 n + 1 Ο β ) . Then
a n + 1 = 2 + a n = 2 + 2 cos β‘ ( Ο 2 n + 1 ) . a_{n+1}
=\sqrt{2+a_n}
=\sqrt{2+2\cos\left(\frac{\pi}{2^{n+1}}\right)}. a n + 1 β = 2 + a n β β = 2 + 2 cos ( 2 n + 1 Ο β ) β . Use the half-angle identity
cos β‘ ( u 2 ) = 1 + cos β‘ u 2 \cos\left(\frac{u}{2}\right)=\sqrt{\frac{1+\cos u}{2}} cos ( 2 u β ) = 2 1 + cos u β β for angles in Quadrant I. Rearranging gives
2 cos β‘ ( u 2 ) = 2 + 2 cos β‘ u . 2\cos\left(\frac{u}{2}\right)=\sqrt{2+2\cos u}. 2 cos ( 2 u β ) = 2 + 2 cos u β . With
u = Ο 2 n + 1 , u=\frac{\pi}{2^{n+1}}, u = 2 n + 1 Ο β , we get
a n + 1 = 2 cos β‘ ( Ο 2 n + 2 ) . a_{n+1}
=2\cos\left(\frac{\pi}{2^{n+2}}\right). a n + 1 β = 2 cos ( 2 n + 2 Ο β ) . Thus
a n = 2 cos β‘ ( Ο 2 n + 1 ) . \boxed{a_n=2\cos\left(\frac{\pi}{2^{n+1}}\right)}. a n β = 2 cos ( 2 n + 1 Ο β ) β . For part ( B ) (B) ( B ) ( A ) (A) ( A )
a n 2 = cos β‘ ( Ο 2 n + 1 ) . \frac{a_n}{2}=\cos\left(\frac{\pi}{2^{n+1}}\right). 2 a n β β = cos ( 2 n + 1 Ο β ) . Consider the finite product
P N = β n = 1 N a n 2 = β n = 1 N cos β‘ ( Ο 2 n + 1 ) . P_N=\prod_{n=1}^{N}\frac{a_n}{2}
=\prod_{n=1}^{N}\cos\left(\frac{\pi}{2^{n+1}}\right). P N β = n = 1 β N β 2 a n β β = n = 1 β N β cos ( 2 n + 1 Ο β ) . Use the repeated double-angle identity
sin β‘ ( 2 x ) = 2 sin β‘ x cos β‘ x . \sin(2x)=2\sin x\cos x. sin ( 2 x ) = 2 sin x cos x . Starting with x = Ο 2 N + 1 x=\frac{\pi}{2^{N+1}} x = 2 N + 1 Ο β
sin β‘ ( Ο 2 ) = 2 N sin β‘ ( Ο 2 N + 1 ) β n = 1 N cos β‘ ( Ο 2 n + 1 ) . \sin\left(\frac{\pi}{2}\right)
=2^N\sin\left(\frac{\pi}{2^{N+1}}\right)
\prod_{n=1}^{N}\cos\left(\frac{\pi}{2^{n+1}}\right). sin ( 2 Ο β ) = 2 N sin ( 2 N + 1 Ο β ) n = 1 β N β cos ( 2 n + 1 Ο β ) . Since sin β‘ ( Ο 2 ) = 1 \sin\left(\frac{\pi}{2}\right)=1 sin ( 2 Ο β ) = 1
1 = 2 N sin β‘ ( Ο 2 N + 1 ) P N . 1=2^N\sin\left(\frac{\pi}{2^{N+1}}\right)P_N. 1 = 2 N sin ( 2 N + 1 Ο β ) P N β . Therefore
P N = 1 2 N sin β‘ ( Ο 2 N + 1 ) . P_N=\frac{1}{2^N\sin\left(\frac{\pi}{2^{N+1}}\right)}. P N β = 2 N sin ( 2 N + 1 Ο β ) 1 β . For part ( C ) (C) ( C )
u N = Ο 2 N + 1 . u_N=\frac{\pi}{2^{N+1}}. u N β = 2 N + 1 Ο β . Then
2 N = Ο 2 u N . 2^N=\frac{\pi}{2u_N}. 2 N = 2 u N β Ο β . So
P N = 1 2 N sin β‘ u N = 1 Ο 2 u N sin β‘ u N = 2 Ο β
u N sin β‘ u N . P_N=\frac{1}{2^N\sin u_N}
=\frac{1}{\frac{\pi}{2u_N}\sin u_N}
=\frac{2}{\pi}\cdot\frac{u_N}{\sin u_N}. P N β = 2 N sin u N β 1 β = 2 u N β Ο β sin u N β 1 β = Ο 2 β β
sin u N β u N β β . As N β β N\to\infty N β β u N β 0 u_N\to0 u N β β 0
u N sin β‘ u N β 1. \frac{u_N}{\sin u_N}\to1. sin u N β u N β β β 1. Thus
lim β‘ N β β P N = 2 Ο . \lim_{N\to\infty}P_N=\frac2\pi. N β β lim β P N β = Ο 2 β . Since
a 1 2 = 2 2 , \frac{a_1}{2}=\frac{\sqrt2}{2}, 2 a 1 β β = 2 2 β β , a 2 2 = 2 + 2 2 , \frac{a_2}{2}=\frac{\sqrt{2+\sqrt2}}{2}, 2 a 2 β β = 2 2 + 2 β β β , and so on, we get ViΓ¨teβs product:
2 Ο = 2 2 β
2 + 2 2 β
2 + 2 + 2 2 β― . \boxed{
\frac{2}{\pi}
=\frac{\sqrt2}{2}\cdot
\frac{\sqrt{2+\sqrt2}}{2}\cdot
\frac{\sqrt{2+\sqrt{2+\sqrt2}}}{2}\cdots
}. Ο 2 β = 2 2 β β β
2 2 + 2 β β β β
2 2 + 2 + 2 β β β β β― β .