Honors Pre-Calculus Unit 5 Review: Trigonometric Functions

An angle is in standard position when its vertex is at the origin and its initial side lies on the positive x-axis. The terminal side is where the rotation stops. Counterclockwise rotation gives a positive angle, and clockwise rotation gives a negative one.

You can measure angles in degrees or radians. One radian is the angle that cuts off an arc equal in length to the radius. A full turn is $360^\circ = 2\pi$ radians, so $\pi$ radians $= 180^\circ$. To convert degrees to radians, multiply by $\frac{\pi}{180}$. To convert back, multiply by $\frac{180}{\pi}$.

Coterminal angles share a terminal side, so they differ by full turns of $360^\circ$ or $2\pi$. For example, $45^\circ$, $405^\circ$, and $-315^\circ$ all point the same direction and have the same trig values.

Radians make circle formulas clean. For a nonnegative angle swept $\theta$ in radians, arc length is $s = r\theta$. Over an elapsed time $t$, average angular speed is $\omega = \frac{\theta}{t}$, and average linear speed is $v = r\omega$.

Worked example. A wheel with radius 30 cm turns through $150^\circ$. How far does a point on the rim travel? First convert the angle: $150 \cdot \frac{\pi}{180} = \frac{5\pi}{6}$. Then $s = r\theta = 30 \cdot \frac{5\pi}{6} = 25\pi \approx 78.5$ cm.

The unit circle has radius 1 and is centered at $(0, 0)$. For any angle $\theta$ in standard position, the terminal side meets the circle at the point $(\cos\theta, \sin\theta)$. Cosine is the x-coordinate and sine is the y-coordinate.

Learn the first-quadrant points. At $\frac{\pi}{6}$ the point is $\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$, at $\frac{\pi}{4}$ it is $\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$, and at $\frac{\pi}{3}$ it is $\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$. The quadrantal points are $(1, 0)$, $(0, 1)$, $(-1, 0)$, and $(0, -1)$.

Angles with one of these first-quadrant reference angles reuse the corresponding coordinate values. For a nonquadrantal angle, the reference angle is the acute angle between the terminal side and the x-axis. It gives the size of each value, and the quadrant gives the sign. The phrase "All Students Take Calculus" tracks which functions are positive in Quadrants I through IV.

You can rotate forever in either direction, so sine and cosine have a domain of all real numbers. Every point on the circle satisfies $x^2 + y^2 = 1$, so their range is $[-1, 1]$. Both repeat every $2\pi$, so $\sin(\theta + 2\pi) = \sin\theta$.

Worked example. Find $\sin\frac{5\pi}{3}$ and $\cos\frac{5\pi}{3}$. The angle is between $\frac{3\pi}{2}$ and $2\pi$, so it lies in Quadrant IV. The reference angle is $2\pi - \frac{5\pi}{3} = \frac{\pi}{3}$. In Quadrant IV cosine is positive and sine is negative, so $\cos\frac{5\pi}{3} = \frac{1}{2}$ and $\sin\frac{5\pi}{3} = -\frac{\sqrt{3}}{2}$.

The other four trig functions all come from sine and cosine. Tangent is $\tan\theta = \frac{\sin\theta}{\cos\theta}$, which on the unit circle is $\frac{y}{x}$. The other definitions are $\cot\theta = \frac{\cos\theta}{\sin\theta}$, $\sec\theta = \frac{1}{\cos\theta}$, and $\csc\theta = \frac{1}{\sin\theta}$. Cotangent also equals $\frac{1}{\tan\theta}$ wherever tangent is defined and nonzero.

When a denominator is zero, the function is undefined. Tangent and secant are undefined where $\cos\theta = 0$, such as $\frac{\pi}{2}$. Cotangent and cosecant are undefined where $\sin\theta = 0$, such as $0$ and $\pi$.

Each reciprocal keeps the sign of its partner, so secant follows cosine and cosecant follows sine. Tangent and cotangent are positive in Quadrants I and III. Cosine and secant are even, so $\cos(-x) = \cos x$. The other four are odd, so $\sin(-x) = -\sin x$.

Dividing $\sin^2\theta + \cos^2\theta = 1$ by $\cos^2\theta$ gives $1 + \tan^2\theta = \sec^2\theta$. Dividing by $\sin^2\theta$ gives $1 + \cot^2\theta = \csc^2\theta$.

Worked example. Suppose $\sin\theta = \frac{3}{5}$ and $\theta$ is in Quadrant II. Then $\cos^2\theta = 1 - \frac{9}{25} = \frac{16}{25}$. Cosine is negative in Quadrant II, so $\cos\theta = -\frac{4}{5}$. That gives $\tan\theta = -\frac{3}{4}$, $\cot\theta = -\frac{4}{3}$, $\sec\theta = -\frac{5}{4}$, and $\csc\theta = \frac{5}{3}$.

In a right triangle, pick one acute angle $\theta$ and label the sides relative to it. The hypotenuse is across from the right angle, the opposite side is across from $\theta$, and the adjacent leg touches $\theta$ and is not the hypotenuse. SOH-CAH-TOA gives $\sin\theta = \frac{\text{opp}}{\text{hyp}}$, $\cos\theta = \frac{\text{adj}}{\text{hyp}}$, and $\tan\theta = \frac{\text{opp}}{\text{adj}}$. Flip each ratio to get csc, sec, and cot.

These ratios match the unit circle. Put $\theta$ at the origin with a hypotenuse of 1, and the legs become $\cos\theta$ and $\sin\theta$.

Two special triangles produce the exact values. A 30-60-90 triangle has sides in the ratio $1 : \sqrt{3} : 2$, and a 45-45-90 triangle has sides in the ratio $1 : 1 : \sqrt{2}$.

To find a missing side, set up a ratio and solve. To find a missing angle, use an inverse function such as $\theta = \tan^{-1}\left(\frac{\text{opp}}{\text{adj}}\right)$.

The two acute angles add to $90^\circ$, so the side opposite one angle is adjacent to the other. That gives cofunction identities like $\sin\theta = \cos(90^\circ - \theta)$.

Worked example. A 12-foot ladder leans against a wall and makes a $70^\circ$ angle with the ground. The wall height is opposite the angle and the ladder is the hypotenuse, so $h = 12\sin 70^\circ \approx 12(0.940) \approx 11.3$ feet. The base sits $12\cos 70^\circ \approx 4.1$ feet from the wall. As a check, $11.3^2 + 4.1^2 \approx 144 = 12^2$.