Unit 6 & 7: Trigonometric Functions

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Angles definitions

An angle is formed by rotating a ray around a fixed endpoint. The starting ray is called the initial side, and the ending ray is called the terminal side.

An angle is in standard position if:

• its vertex is at the origin,
• its initial side lies on the positive $x$-axis,
• its terminal side is determined by rotating from the positive $x$-axis.

By definition, counterclockwise rotations are positive, and clockwise rotations are negative.

For example, $135^\circ$ is a positive angle, while $-45^\circ$ is a negative angle.

Degrees and radians

Degrees measure angles by splitting a full circle into $360$ equal parts. Radians measure angles by comparing arc length to radius, where one radian is the angle an arc length that measures out $r$ units.

One full circle is

$$360^\circ=2\pi\text{ radians}.$$

so therefore

$$180^\circ=\pi\text{ radians}.$$

Extension. The radian is a unitless measure, meaning that it is not arbitrarily set. Why is there always $2\pi$ radians in a circle?

Example. Convert $225^\circ$ to radians.

Multiply by $\frac{\pi}{180}$:

$$225^\circ\cdot\frac{\pi}{180} =\frac{225\pi}{180} =\frac{5\pi}{4}.$$

Thus

$$225^\circ=\frac{5\pi}{4}.$$

Example. Convert $-\frac{4\pi}{3}$ radians to degrees.

Multiply by $\frac{180}{\pi}$:

$$-\frac{4\pi}{3}\cdot\frac{180}{\pi} =-240^\circ.$$

Thus

$$-\frac{4\pi}{3}=-240^\circ.$$

Common angles

These are the most important degree-radian conversions, but many other conversions can be done through trig rules which will be discussed later:

$$\begin{array}{c|c} \text{Degrees} & \text{Radians}\\ \hline 0^\circ & 0\\ 30^\circ & \frac{\pi}{6}\\ 45^\circ & \frac{\pi}{4}\\ 60^\circ & \frac{\pi}{3}\\ 90^\circ & \frac{\pi}{2}\\ 120^\circ & \frac{2\pi}{3}\\ 135^\circ & \frac{3\pi}{4}\\ 150^\circ & \frac{5\pi}{6}\\ 180^\circ & \pi\\ 210^\circ & \frac{7\pi}{6}\\ 225^\circ & \frac{5\pi}{4}\\ 240^\circ & \frac{4\pi}{3}\\ 270^\circ & \frac{3\pi}{2}\\ 300^\circ & \frac{5\pi}{3}\\ 315^\circ & \frac{7\pi}{4}\\ 330^\circ & \frac{11\pi}{6}\\ 360^\circ & 2\pi \end{array}$$

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Arc length and sector area

Radians are useful because they connect angles directly to lengths on a circle.

If a central angle $\theta$ cuts off an arc of length $s$ on a circle of radius $r$, then

$$\theta=\frac{s}{r}.$$

Solving for arc length gives

$$s=r\theta.$$

The angle $\theta$ must be measured in radians.

Warning

The arc length and sector formulas require $\theta$ in radians, not degrees. On your calculator, if you go to MODE, you can switch from degrees to radians.

Example. Find the arc length of a circle of radius $3$ meters subtended by a central angle of $120^\circ$.

First convert the angle to radians:

$$120^\circ\cdot\frac{\pi}{180}=\frac{2\pi}{3}.$$

Then use $s=r\theta$:

$$s=3\cdot\frac{2\pi}{3}=2\pi.$$

So the arc length is

$$2\pi\text{ meters}.$$

Example. Find the radius of a circle whose arc length is $6$ meters and whose central angle is $\frac14$ radian.

Use

$$s=r\theta.$$

Then

$$6=r\left(\frac14\right).$$

Thus

$$r=24\text{ meters}.$$

Sector area

A sector is a region cut out by two radii and the arc between them. If $\theta$ is measured in radians, the area of a sector is

$$A=\frac12r^2\theta.$$

This comes from taking the fraction $\frac{\theta}{2\pi}$ of the full circle area $\pi r^2$:

$$A=\frac{\theta}{2\pi}\cdot \pi r^2=\frac12r^2\theta.$$

Example. Find the area of a sector with radius $6$ and central angle $\frac{5\pi}{6}$.

Use the sector area formula:

$$A=\frac12r^2\theta.$$

Then

$$A=\frac12(6)^2\left(\frac{5\pi}{6}\right) =18\cdot\frac{5\pi}{6} =15\pi.$$

Thus

$$15\pi.$$

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Circular motion

When an object moves around a circle, there are two related speeds:

• Angular speed measures how quickly the angle changes.
• Linear speed measures how quickly the object moves along the circle.

If an angle $\theta$ is swept out in time $t$, then the angular speed is

$$\omega=\frac{\theta}{t}.$$

If a distance $d$ is traveled along the circle in time $t$, then the linear speed is

$$v=\frac{d}{t}.$$

Since arc length is $s=r\theta$, linear speed and angular speed are connected by

$$v=r\omega.$$

Again, angular speed must be measured in radians per unit time.

Example. A wheel rotates at $15$ revolutions per second. The radius of the wheel is $20$ cm. Find the angular speed in radians per second and the linear speed of a point on the edge of the wheel.

One revolution is $2\pi$ radians, so

$$\omega=15\cdot 2\pi=30\pi.$$

Thus the angular speed is

$$30\pi\text{ rad/sec}.$$

Now use $v=r\omega$:

$$v=20(30\pi)=600\pi.$$

So the linear speed is

$$600\pi\text{ cm/sec}.$$

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Right-triangle trigonometry

For an acute angle $\theta$ in a right triangle, the three main trigonometric ratios are:

$$\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}},$$ $$\cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}},$$

and

$$\tan\theta=\frac{\text{opposite}}{\text{adjacent}}.$$

A common memory device is SOH-CAH-TOA:

• Sine = Opposite over Hypotenuse.
• Cosine = Adjacent over Hypotenuse.
• Tangent = Opposite over Adjacent.

The reciprocal trigonometric functions are also defined as well:

$$\csc\theta=\frac{1}{\sin\theta} =\frac{\text{hypotenuse}}{\text{opposite}},$$ $$\sec\theta=\frac{1}{\cos\theta} =\frac{\text{hypotenuse}}{\text{adjacent}},$$

and

$$\cot\theta=\frac{1}{\tan\theta} =\frac{\text{adjacent}}{\text{opposite}}$$

The three functions are called cosecant, secant, and cotangent, respectively. To memorize, remember that you have to add co- to any trig function without a co- and do not add one otherwise.

Example. A right triangle has legs $6$ and $2$. Find the six trigonometric functions for the acute angle opposite the side of length $6$.

First find the hypotenuse:

$$c=\sqrt{6^2+2^2}=\sqrt{40}=2\sqrt{10}.$$

For the angle opposite the side of length $6$:

$$\sin\theta=\frac{6}{2\sqrt{10}}=\frac{3}{\sqrt{10}},$$ $$\cos\theta=\frac{2}{2\sqrt{10}}=\frac{1}{\sqrt{10}},$$

and

$$\tan\theta=\frac{6}{2}=3.$$

The reciprocal functions are

$$\csc\theta=\frac{\sqrt{10}}{3},$$ $$\sec\theta=\sqrt{10},$$

and

$$\cot\theta=\frac13.$$

Cofunction identities

The acute angles in a right triangle are complementary. This means their measures add to $90^\circ$, or $\frac{\pi}{2}$ radians.

Cofunctions of complementary angles are equal:

$$\sin\theta=\cos\left(\frac{\pi}{2}-\theta\right),$$ $$\cos\theta=\sin\left(\frac{\pi}{2}-\theta\right),$$ $$\tan\theta=\cot\left(\frac{\pi}{2}-\theta\right),$$

and

$$\sec\theta=\csc\left(\frac{\pi}{2}-\theta\right).$$

For example,

$$\sin 37^\circ=\cos 53^\circ.$$

Extension. Prove the six theorems above.

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Special right triangles

Two special triangles produce many exact trigonometric values.

The 30-60-90 triangle

A $30^\circ$-$60^\circ$-$90^\circ$ triangle has side ratio

$$x:x\sqrt3:2x.$$

The side opposite $30^\circ$ is $x$, the side opposite $60^\circ$ is $x\sqrt3$, and the hypotenuse is $2x$.

Therefore:

$$\sin 30^\circ=\frac12, \qquad \cos 30^\circ=\frac{\sqrt3}{2}, \qquad \tan 30^\circ=\frac{1}{\sqrt3}.$$

Also:

$$\sin 60^\circ=\frac{\sqrt3}{2}, \qquad \cos 60^\circ=\frac12, \qquad \tan 60^\circ=\sqrt3.$$

The 45-45-90 triangle

A $45^\circ$-$45^\circ$-$90^\circ$ triangle has side ratio

$$x:x:x\sqrt2.$$

Therefore:

$$\sin 45^\circ=\frac{1}{\sqrt2}=\frac{\sqrt2}{2},$$ $$\cos 45^\circ=\frac{1}{\sqrt2}=\frac{\sqrt2}{2},$$

and

$$\tan 45^\circ=1.$$

All of the values seen on a traditional unit circle come from these two special right triangles.

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Trig functions of any angle

Right-triangle definitions only directly handle acute angles. To define trig functions for any angle, use the unit circle.

The unit circle is the circle

$$x^2+y^2=1.$$

If an angle $\theta$ is in standard position and its terminal side intersects the unit circle at $P(x,y)$, then

$$x=\cos\theta$$

and

$$y=\sin\theta.$$

So the point on the unit circle is

$$P=(\cos\theta,\sin\theta).$$

This also gives

$$\tan\theta=\frac{y}{x}=\frac{\sin\theta}{\cos\theta},$$

as long as $x\ne0$.

The reciprocal functions are:

$$\csc\theta=\frac1{\sin\theta}, \qquad \sec\theta=\frac1{\cos\theta}, \qquad \cot\theta=\frac1{\tan\theta}=\frac{\cos\theta}{\sin\theta}.$$

Quadrant signs

The signs of trig functions depend on the quadrant:

$$\begin{array}{c|c|c|c} \text{Quadrant} & \sin\theta & \cos\theta & \tan\theta\\ \hline \text{I} & + & + & +\\ \text{II} & + & - & -\\ \text{III} & - & - & +\\ \text{IV} & - & + & - \end{array}$$

Since sine is the $y$-coordinate, it is positive above the $x$-axis and negative below it. Since cosine is the $x$-coordinate, it is positive to the right of the $y$-axis and negative to the left.

Reference angles

The reference angle is the acute angle formed by the terminal side of $\theta$ and the $x$-axis.

Problem-solving strategy

• If $\theta$ is in Quadrant I, the reference angle is $\theta$.
• If $\theta$ is in Quadrant II, the reference angle is $\pi-\theta$.
• If $\theta$ is in Quadrant III, the reference angle is $\theta-\pi$.
• If $\theta$ is in Quadrant IV, the reference angle is $2\pi-\theta$.

The reference angle gives the magnitude of the trig value. The quadrant gives the sign.

Example. Evaluate $\cos 315^\circ$.

The angle $315^\circ$ is in Quadrant IV. Its reference angle is

$$360^\circ-315^\circ=45^\circ.$$

Cosine is positive in Quadrant IV, so

$$\cos 315^\circ=\cos 45^\circ=\frac{\sqrt2}{2}.$$

Thus

$$\cos 315^\circ=\frac{\sqrt2}{2}.$$

Example. Evaluate $\sin\left(\frac{4\pi}{3}\right)$.

The angle $\frac{4\pi}{3}$ is in Quadrant III. Its reference angle is

$$\frac{4\pi}{3}-\pi=\frac{\pi}{3}.$$

Sine is negative in Quadrant III, so

$$\sin\left(\frac{4\pi}{3}\right) =-\sin\left(\frac{\pi}{3}\right) =-\frac{\sqrt3}{2}.$$

Thus

$$\sin\left(\frac{4\pi}{3}\right)=-\frac{\sqrt3}{2}.$$

An image of the unit circle is shown below (with filled in values as described later):

Coterminal angles

Two angles are coterminal if they share the same terminal side.

In degrees, coterminal angles differ by a multiple of $360^\circ$:

$$\theta+360^\circ k,\qquad k\in\mathbb{Z}.$$

In radians, coterminal angles differ by a multiple of $2\pi$:

$$\theta+2\pi k,\qquad k\in\mathbb{Z}.$$

For example, $120^\circ$ and $-240^\circ$ are coterminal because

$$120^\circ-360^\circ=-240^\circ.$$

All coterminal angles will have the same trig values, so usually we define the angles from $0^\circ$ to $360^\circ$ or from $0$ to $2\pi$ radians.

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Unit circle values

The most common unit circle coordinates are split by half-circle so the values stay readable:

$$\begin{array}{c|c|c|c} \theta & \text{Degrees} & \cos\theta & \sin\theta\\ \hline 0 & 0^\circ & 1 & 0\\ \frac{\pi}{6} & 30^\circ & \frac{\sqrt3}{2} & \frac12\\ \frac{\pi}{4} & 45^\circ & \frac{\sqrt2}{2} & \frac{\sqrt2}{2}\\ \frac{\pi}{3} & 60^\circ & \frac12 & \frac{\sqrt3}{2}\\ \frac{\pi}{2} & 90^\circ & 0 & 1\\ \frac{2\pi}{3} & 120^\circ & -\frac12 & \frac{\sqrt3}{2}\\ \frac{3\pi}{4} & 135^\circ & -\frac{\sqrt2}{2} & \frac{\sqrt2}{2}\\ \frac{5\pi}{6} & 150^\circ & -\frac{\sqrt3}{2} & \frac12\\ \pi & 180^\circ & -1 & 0 \end{array}$$ $$\begin{array}{c|c|c|c} \theta & \text{Degrees} & \cos\theta & \sin\theta\\ \hline \frac{7\pi}{6} & 210^\circ & -\frac{\sqrt3}{2} & -\frac12\\ \frac{5\pi}{4} & 225^\circ & -\frac{\sqrt2}{2} & -\frac{\sqrt2}{2}\\ \frac{4\pi}{3} & 240^\circ & -\frac12 & -\frac{\sqrt3}{2}\\ \frac{3\pi}{2} & 270^\circ & 0 & -1\\ \frac{5\pi}{3} & 300^\circ & \frac12 & -\frac{\sqrt3}{2}\\ \frac{7\pi}{4} & 315^\circ & \frac{\sqrt2}{2} & -\frac{\sqrt2}{2}\\ \frac{11\pi}{6} & 330^\circ & \frac{\sqrt3}{2} & -\frac12\\ 2\pi & 360^\circ & 1 & 0 \end{array}$$

The first-quadrant values are repeated around the circle with signs determined by the quadrant.

For example:

$$\cos\left(\frac{2\pi}{3}\right)=-\frac12, \qquad \sin\left(\frac{2\pi}{3}\right)=\frac{\sqrt3}{2}.$$

and

$$\cos\left(\frac{7\pi}{4}\right)=\frac{\sqrt2}{2}, \qquad \sin\left(\frac{7\pi}{4}\right)=-\frac{\sqrt2}{2}.$$

Some reciprocal or quotient trig functions are undefined when their denominator is $0$:

• $\tan\theta=\frac{\sin\theta}{\cos\theta}$ is undefined when $\cos\theta=0$.
• $\sec\theta=\frac1{\cos\theta}$ is undefined when $\cos\theta=0$.
• $\csc\theta=\frac1{\sin\theta}$ is undefined when $\sin\theta=0$.
• $\cot\theta=\frac{\cos\theta}{\sin\theta}$ is undefined when $\sin\theta=0$.

For example,

$$\tan\left(\frac{3\pi}{2}\right) =\frac{\sin\left(\frac{3\pi}{2}\right)}{\cos\left(\frac{3\pi}{2}\right)} =\frac{-1}{0},$$

so

$$\tan\left(\frac{3\pi}{2}\right)\text{ is undefined}.$$

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Trig identities

The unit circle equation

$$x^2+y^2=1$$

becomes the most important trigonometric identity because $x=\cos\theta$ and $y=\sin\theta$:

Key infoPythagorean identity

$$\sin^2\theta+\cos^2\theta=1.$$

Dividing both sides by $\cos^2\theta$ gives

$$\frac{\sin^2\theta}{\cos^2\theta} +\frac{\cos^2\theta}{\cos^2\theta} =\frac1{\cos^2\theta}.$$

Thus

$$1+\tan^2\theta=\sec^2\theta.$$

Dividing both sides by $\sin^2\theta$ gives

$$1+\cot^2\theta=\csc^2\theta.$$

Finding trig values from one trig value

If one trig value is known, the Pythagorean identity and the quadrant can determine the others.

Example. Suppose $90^\circ<\beta<180^\circ$ and $\sin\beta=\frac14$. Find $\cos\beta$ and $\tan\beta$.

Since $\beta$ is in Quadrant II, cosine is negative and tangent is negative.

Use

$$\sin^2\beta+\cos^2\beta=1.$$

Substitute $\sin\beta=\frac14$:

$$\left(\frac14\right)^2+\cos^2\beta=1.$$

Then

$$\cos^2\beta=1-\frac1{16}=\frac{15}{16}.$$

So

$$\cos\beta=\pm\frac{\sqrt{15}}{4}.$$

Because $\beta$ is in Quadrant II,

$$\cos\beta=-\frac{\sqrt{15}}4.$$

Now

$$\tan\beta=\frac{\sin\beta}{\cos\beta} =\frac{\frac14}{-\frac{\sqrt{15}}4} =-\frac1{\sqrt{15}}.$$

Thus

$$\tan\beta=-\frac1{\sqrt{15}}.$$

Proving identities

To prove a trigonometric identity, work on one side of the equation and transform it into the other side.

Problem-solving strategy

Common strategies include:

• rewrite everything in terms of sine and cosine,
• use the Pythagorean identities,
• combine fractions,
• substitute if necessary,
• factor,
• multiply by a conjugate when useful.

Example. Prove that

$$\tan\theta\sin\theta=\sec\theta-\cos\theta.$$

Start with the left-hand side:

$$\tan\theta\sin\theta =\frac{\sin\theta}{\cos\theta}\cdot\sin\theta.$$

So

$$\tan\theta\sin\theta =\frac{\sin^2\theta}{\cos\theta}.$$

Use $\sin^2\theta=1-\cos^2\theta$:

$$\frac{\sin^2\theta}{\cos\theta} =\frac{1-\cos^2\theta}{\cos\theta}.$$

Split the fraction:

$$\frac{1-\cos^2\theta}{\cos\theta} =\frac1{\cos\theta}-\frac{\cos^2\theta}{\cos\theta}.$$

Simplify:

$$\frac1{\cos\theta}-\cos\theta =\sec\theta-\cos\theta.$$

Thus

$$\tan\theta\sin\theta=\sec\theta-\cos\theta.$$

Example. Prove that

$$\frac{\sin A}{1+\cos A}+\frac{1+\cos A}{\sin A}=2\csc A.$$

Start with the left-hand side:

$$\frac{\sin A}{1+\cos A}+\frac{1+\cos A}{\sin A}.$$

Use the common denominator $\sin A(1+\cos A)$:

$$\frac{\sin^2 A+(1+\cos A)^2}{\sin A(1+\cos A)}.$$

Expand the numerator:

$$\sin^2 A+1+2\cos A+\cos^2 A.$$

Use $\sin^2 A+\cos^2 A=1$:

$$\sin^2 A+1+2\cos A+\cos^2 A =2+2\cos A.$$

So the expression becomes

$$\frac{2+2\cos A}{\sin A(1+\cos A)}.$$

Factor the numerator:

$$\frac{2(1+\cos A)}{\sin A(1+\cos A)}.$$

Cancel:

$$\frac2{\sin A}=2\csc A.$$

Thus

$$\frac{\sin A}{1+\cos A}+\frac{1+\cos A}{\sin A}=2\csc A.$$

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Even, odd, and periodic behavior

The unit circle also explains the symmetry between trig functions.

Cosine is an even function:

$$\cos(-\theta)=\cos\theta.$$

Sine and tangent are odd functions:

$$\sin(-\theta)=-\sin\theta$$

and

$$\tan(-\theta)=-\tan\theta.$$

Sine and cosine are periodic with period $2\pi$:

$$\sin(\theta+2\pi k)=\sin\theta$$

and

$$\cos(\theta+2\pi k)=\cos\theta,$$

where $k$ is any integer.

Tangent has period $\pi$:

$$\tan(\theta+\pi k)=\tan\theta.$$

Example. If $\sin t=\frac23$, find $\sin(-t)$.

Since sine is odd,

$$\sin(-t)=-\sin t.$$

Therefore

$$\sin(-t)=-\frac23.$$

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Graphs of sine and cosine

The parent sine function is

$$y=\sin x.$$

It has:

• Domain: $(-\infty,\infty)$.
• Range: $[-1,1]$.
• Period: $2\pi$.
• Amplitude: $1$.
• Midline: $y=0$.

One full cycle of $y=\sin x$ goes through these key points:

$$\left(0,0\right), \left(\frac{\pi}{2},1\right), \left(\pi,0\right), \left(\frac{3\pi}{2},-1\right), \left(2\pi,0\right).$$

The parent cosine function is

$$y=\cos x.$$

It has:

• Domain: $(-\infty,\infty)$.
• Range: $[-1,1]$.
• Period: $2\pi$.
• Amplitude: $1$.
• Midline: $y=0$.

One full cycle of $y=\cos x$ goes through these key points:

$$\left(0,1\right), \left(\frac{\pi}{2},0\right), \left(\pi,-1\right), \left(\frac{3\pi}{2},0\right), \left(2\pi,1\right).$$

Sine and cosine are phase shifts of each other. For example,

$$\cos x=\sin\left(x+\frac{\pi}{2}\right).$$

Solving basic trig equations with graphs

Graphs and the unit circle both help solve equations like $\sin x=0.75$ or $\cos x=-0.35$ on an interval.

Example. Solve $\sin x=0.75$ on $[0,2\pi)$.

The calculator gives the first solution

$$x\approx \sin^{-1}(0.75)\approx 0.848.$$

Since sine is also positive in Quadrant II, the second solution is

$$x=\pi-0.848\approx 2.294.$$

Thus

$$x\approx0.848\quad\text{or}\quad x\approx2.294.$$

Example. Solve $\cos x=-0.35$ on $[0,2\pi)$.

Cosine is negative in Quadrants II and III. The calculator gives

$$x\approx \cos^{-1}(-0.35)\approx 1.928.$$

The second solution is

$$x=2\pi-1.928\approx 4.355.$$

Thus

$$x\approx1.928\quad\text{or}\quad x\approx4.355.$$

Warning

REMEMBER to always check for multiple solutions using the trig rules!

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Practice

1. Let $\theta=-\frac{29\pi}{6}$.

   $(A)$ Find the least positive coterminal angle with $\theta$.

   $(B)$ Convert both angles to degrees.

   $(C)$ Find the reference angle and quadrant of $\theta$.

   $(D)$ Evaluate all six trigonometric functions of $\theta$ exactly.

2. A sector of a circle has perimeter $40$ cm and central angle $\frac{5\pi}{6}$. Find the radius, arc length, and area of the sector exactly.

3. A wheel of radius $18$ cm rotates counterclockwise at $45$ revolutions per minute. A bug starts at the point on the wheel closest to the ground. After $7$ seconds, find the bug’s angle in standard position, its coordinates relative to the center of the wheel, and its linear speed in cm/sec.

4. A pulley system has two wheels connected by a belt without slipping. Wheel A has radius $4$ inches and rotates at $150$ revolutions per minute. Wheel B rotates at $60$ revolutions per minute. Find the radius of Wheel B. Then find the linear belt speed in inches per second.

5. Let $\theta$ be in Quadrant II and suppose $\tan\theta=-\frac{8}{15}$. Find exact values of $\sin\theta$, $\cos\theta$, $\sec\theta$, $\csc\theta$, and $\cot\theta$. Then evaluate $\sin(\pi-\theta)$ and $\cos(\theta+\pi)$.

6. Let $P=(x,y)$ be a point on the unit circle in Quadrant III. If $x-y=\frac{\sqrt2}{2}$, find $P$ and the angle $\theta\in[0,2\pi)$ whose terminal side passes through $P$.

7. Evaluate exactly: $6\sin\left(-\frac{7\pi}{6}\right)-4\cos\left(\frac{11\pi}{3}\right)+3\tan\left(-\frac{13\pi}{4}\right)-2\sec\left(\frac{17\pi}{6}\right).$

8. Solve exactly on $[0,4\pi)$: $2\sin^2x-\sin x-1=0.$

9. Solve exactly on $[0,2\pi)$:$2\cos^2x+\sqrt3\cos x-1=0.$

10. Solve exactly on $[0,3\pi)$:$\tan^2x-3=0.$

11. The radius of the circle in the figure is 2 units. Express the length of $DC$ in terms of $\alpha$.

12. Prove the identity: $\frac{1-\cos\theta}{\sin\theta}+\frac{\sin\theta}{1-\cos\theta}=2\csc\theta.$ Then state all values of $\theta$ in $[0,2\pi)$ for which the original identity is undefined.

13. Prove the identity: $\frac{1}{\cos\theta\sin^2\theta}=\sec\theta+\csc\theta\cot\theta.$

14. For each of the following trigonometric expressions, find a segment in the diagram that has length equal to the trigonometric expression: $\sin\theta, \cos\theta, \sec\theta, \csc\theta, \tan\theta, \cot\theta$. Note that you are not asked to express each trigonometric function in terms of multiple segments in the diagram. You must find a segment whose whole length equals the corresponding trig function. The graph is given below:

15. On $[0,2\pi)$, solve the equation numerically to three decimal places: $3\sin x-2\cos x=1.$ (Hint: Try the substitution $t=\tan(x/2)$, and solve for $x$ using the $\tan^{-1}$ button on the calculator.)

16. (Bonus, rational points on the unit circle)

The unit circle is

$$x^2+y^2=1.$$

One obvious rational point on the unit circle is $(-1,0)$. Now draw a line with rational slope $m$ through $(-1,0)$:

$$y=m(x+1).$$

$(A)$ Substitute $y=m(x+1)$ into $x^2+y^2=1$ and show that the line intersects the unit circle at $(-1,0)$ and one other point.

$(B)$ Find the coordinates of the second intersection point in terms of $m$.

$(C)$ Explain why every rational value of $m$ gives a rational point on the unit circle.

$(D)$ Use your formula to find a rational point on the unit circle when $m=\frac23$, then interpret that point as $(\cos\theta,\sin\theta)$ for some angle $\theta$.

$(E)$ Why does this method not produce the point $(-1,0)$ as the second intersection point? What slope would be needed to reach the point $(1,0)$?