Special Angles and Trigonometric Evaluation

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What are the exact coordinates of the terminal point for $\theta = \frac{\pi}{3}$ ($60^\circ$)?

A unit circle diagram showing a 60-degree angle in the first quadrant with a reference triangle whose terminal point has coordinates (1/2, square root of 3 over 2).

$$\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right)$$ Thus, $\cos\frac{\pi}{3} = \frac{1}{2}$, $\sin\frac{\pi}{3} = \frac{\sqrt{3}}{2}$, and $\tan\frac{\pi}{3} = \sqrt{3}$.

How do you evaluate $\sin\left(\frac{2\pi}{3}\right)$ and $\cos\left(\frac{2\pi}{3}\right)$?

$\frac{2\pi}{3}$ lies in Quadrant II with reference angle $\theta' = \frac{\pi}{3}$. In Quadrant II, sine is positive and cosine is negative: $$\sin\left(\frac{2\pi}{3}\right) = \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}$$ $$\cos\left(\frac{2\pi}{3}\right) = -\cos\left(\frac{\pi}{3}\right) = -\frac{1}{2}$$

How do you evaluate $\sin\left(\frac{3\pi}{4}\right)$ and $\cos\left(\frac{3\pi}{4}\right)$?

$\frac{3\pi}{4}$ lies in Quadrant II with reference angle $\theta' = \frac{\pi}{4}$: $$\sin\left(\frac{3\pi}{4}\right) = \frac{\sqrt{2}}{2}$$ $$\cos\left(\frac{3\pi}{4}\right) = -\frac{\sqrt{2}}{2}$$

How do you evaluate $\tan\left(\frac{5\pi}{6}\right)$?

$\frac{5\pi}{6}$ lies in Quadrant II with reference angle $\theta' = \frac{\pi}{6}$. In Quadrant II, tangent is negative: $$\tan\left(\frac{5\pi}{6}\right) = -\tan\left(\frac{\pi}{6}\right) = -\frac{\sqrt{3}}{3}$$

How do you evaluate $\sin\left(\frac{7\pi}{6}\right)$ and $\cos\left(\frac{7\pi}{6}\right)$?

$\frac{7\pi}{6}$ lies in Quadrant III with reference angle $\theta' = \frac{\pi}{6}$. Both coordinates are negative: $$\cos\left(\frac{7\pi}{6}\right) = -\frac{\sqrt{3}}{2}$$ $$\sin\left(\frac{7\pi}{6}\right) = -\frac{1}{2}$$

How do you evaluate $\tan\left(\frac{5\pi}{4}\right)$?

$\frac{5\pi}{4}$ lies in Quadrant III with reference angle $\theta' = \frac{\pi}{4}$. In Quadrant III, tangent is positive: $$\tan\left(\frac{5\pi}{4}\right) = +\tan\left(\frac{\pi}{4}\right) = 1$$

How do you evaluate $\sin\left(\frac{5\pi}{3}\right)$ and $\cos\left(\frac{5\pi}{3}\right)$?

$\frac{5\pi}{3}$ lies in Quadrant IV with reference angle $\theta' = \frac{\pi}{3}$. In Quadrant IV, cosine is positive and sine is negative: $$\cos\left(\frac{5\pi}{3}\right) = \frac{1}{2}$$ $$\sin\left(\frac{5\pi}{3}\right) = -\frac{\sqrt{3}}{2}$$

How do you evaluate $\cos\left(\frac{7\pi}{4}\right)$ and $\sin\left(\frac{7\pi}{4}\right)$?

$\frac{7\pi}{4}$ lies in Quadrant IV with reference angle $\theta' = \frac{\pi}{4}$: $$\cos\left(\frac{7\pi}{4}\right) = \frac{\sqrt{2}}{2}$$ $$\sin\left(\frac{7\pi}{4}\right) = -\frac{\sqrt{2}}{2}$$

How are the three reciprocal trigonometric functions defined on the unit circle?

$$\sec\theta = \frac{1}{x} = \frac{1}{\cos\theta}$$ $$\csc\theta = \frac{1}{y} = \frac{1}{\sin\theta}$$ $$\cot\theta = \frac{x}{y} = \frac{1}{\tan\theta}$$

Evaluate $\sec\left(\frac{2\pi}{3}\right)$.

$$\sec\left(\frac{2\pi}{3}\right) = \frac{1}{\cos\left(\frac{2\pi}{3}\right)} = \frac{1}{-\frac{1}{2}} = -2$$

Evaluate $\csc\left(\frac{7\pi}{6}\right)$.

$$\csc\left(\frac{7\pi}{6}\right) = \frac{1}{\sin\left(\frac{7\pi}{6}\right)} = \frac{1}{-\frac{1}{2}} = -2$$

Evaluate $\cot\left(\frac{3\pi}{2}\right)$.

$$\cot\left(\frac{3\pi}{2}\right) = \frac{\cos\left(\frac{3\pi}{2}\right)}{\sin\left(\frac{3\pi}{2}\right)} = \frac{0}{-1} = 0$$ Note: While $\tan\left(\frac{3\pi}{2}\right)$ is undefined, $\cot\left(\frac{3\pi}{2}\right)$ is defined and equals $0$.

What are the even-odd identities for $\cos(-\theta)$ and $\sin(-\theta)$?

Cosine is an even function: $$\cos(-\theta) = \cos\theta$$ Sine is an odd function: $$\sin(-\theta) = -\sin\theta$$

What are the exact coordinates of the terminal point for $\theta = \frac{\pi}{6}$ ($30^\circ$)?

A unit circle showing an angle of pi over 6 (30 degrees) formed by a radius of length 1, terminating at coordinates (square root of 3 over 2, 1/2) with perpendicular legs representing cosine and sine.

$$\left(\frac{\sqrt{3}}{2}, \frac{1}{2}\right)$$ Thus, $\cos\frac{\pi}{6} = \frac{\sqrt{3}}{2}$, $\sin\frac{\pi}{6} = \frac{1}{2}$, and $\tan\frac{\pi}{6} = \frac{1}{\sqrt{3}} = \frac{\sqrt{3}}{3}$.

What are the exact coordinates of the terminal point for $\theta = \frac{\pi}{4}$ ($45^\circ$)?

A unit circle diagram in the first quadrant for a 45-degree angle (pi over 4), showing a right triangle with hypotenuse 1, legs square root of 2 over 2, and terminal point coordinates (square root of 2 over 2, square root of 2 over 2).

$$\left(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\right)$$ Thus, $\cos\frac{\pi}{4} = \frac{\sqrt{2}}{2}$, $\sin\frac{\pi}{4} = \frac{\sqrt{2}}{2}$, and $\tan\frac{\pi}{4} = 1$.

Special Angles and Trigonometric Evaluation

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