Evaluating Each Trigonometric Function worksheets - Free Printable

Let's solve each trigonometric function step by step and find the exact values. We'll use unit circle values, reference angles, periodicity, and trigonometric identities.

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1) sin 330°

- 330° is in the fourth quadrant, where sine is negative.
- Reference angle: $ 360^\circ - 330^\circ = 30^\circ $
- $ \sin 330^\circ = -\sin 30^\circ = -\frac{1}{2} $

✔ Answer: $-\frac{1}{2}$

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2) cos 330°

- 330° is in the fourth quadrant, cosine is positive.
- Reference angle: $30^\circ$
- $ \cos 330^\circ = \cos 30^\circ = \frac{\sqrt{3}}{2} $

✔ Answer: $\frac{\sqrt{3}}{2}$

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3) tan 120°

- 120° is in the second quadrant, tangent is negative.
- Reference angle: $180^\circ - 120^\circ = 60^\circ$
- $ \tan 120^\circ = -\tan 60^\circ = -\sqrt{3} $

✔ Answer: $-\sqrt{3}$

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4) cos $\frac{7\pi}{4}$

- Convert to degrees: $ \frac{7\pi}{4} = 315^\circ $
- 315° is in the fourth quadrant, cosine is positive
- Reference angle: $360^\circ - 315^\circ = 45^\circ$
- $ \cos \frac{7\pi}{4} = \cos 45^\circ = \frac{\sqrt{2}}{2} $

✔ Answer: $\frac{\sqrt{2}}{2}$

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5) cot 225°

- 225° is in the third quadrant, cotangent is positive (since both sin and cos are negative)
- Reference angle: $225^\circ - 180^\circ = 45^\circ$
- $ \cot 225^\circ = \cot 45^\circ = 1 $

✔ Answer: $1$

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6) sin $\frac{5\pi}{3}$

- $ \frac{5\pi}{3} = 300^\circ $
- Third quadrant? No: 300° is in fourth quadrant, sine is negative
- Reference angle: $360^\circ - 300^\circ = 60^\circ$
- $ \sin \frac{5\pi}{3} = -\sin 60^\circ = -\frac{\sqrt{3}}{2} $

✔ Answer: $-\frac{\sqrt{3}}{2}$

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7) sin 150°

- 150° is in the second quadrant, sine is positive
- Reference angle: $180^\circ - 150^\circ = 30^\circ$
- $ \sin 150^\circ = \sin 30^\circ = \frac{1}{2} $

✔ Answer: $\frac{1}{2}$

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8) cos 480°

- 480° > 360° → subtract 360°: $480^\circ - 360^\circ = 120^\circ$
- So $ \cos 480^\circ = \cos 120^\circ $
- 120° is in second quadrant, cosine is negative
- Reference angle: $60^\circ$
- $ \cos 120^\circ = -\cos 60^\circ = -\frac{1}{2} $

✔ Answer: $-\frac{1}{2}$

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9) tan 225°

- 225° is in third quadrant, tangent is positive
- Reference angle: $225^\circ - 180^\circ = 45^\circ$
- $ \tan 225^\circ = \tan 45^\circ = 1 $

✔ Answer: $1$

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10) tan $\frac{9\pi}{4}$

- $ \frac{9\pi}{4} = \frac{9\pi}{4} - 2\pi = \frac{9\pi}{4} - \frac{8\pi}{4} = \frac{\pi}{4} $
- So $ \tan \frac{9\pi}{4} = \tan \frac{\pi}{4} = 1 $

✔ Answer: $1$

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11) cos $\frac{17\pi}{6}$

- Simplify: $ \frac{17\pi}{6} = 2\pi + \frac{5\pi}{6} $ → subtract $2\pi$: $ \frac{5\pi}{6} $
- $ \frac{5\pi}{6} = 150^\circ $
- Second quadrant, cosine is negative
- Reference angle: $30^\circ$
- $ \cos \frac{5\pi}{6} = -\cos 30^\circ = -\frac{\sqrt{3}}{2} $

✔ Answer: $-\frac{\sqrt{3}}{2}$

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12) sin $\frac{10\pi}{3}$

- $ \frac{10\pi}{3} = \frac{10\pi}{3} - 2\pi = \frac{10\pi}{3} - \frac{6\pi}{3} = \frac{4\pi}{3} $
- $ \frac{4\pi}{3} = 240^\circ $, third quadrant, sine is negative
- Reference angle: $60^\circ$
- $ \sin \frac{4\pi}{3} = -\sin 60^\circ = -\frac{\sqrt{3}}{2} $

✔ Answer: $-\frac{\sqrt{3}}{2}$

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13) cos 675°

- 675° > 360° → subtract 360°: $675 - 360 = 315^\circ$
- $ \cos 675^\circ = \cos 315^\circ $
- 315° is in fourth quadrant, cosine is positive
- Reference angle: $45^\circ$
- $ \cos 315^\circ = \cos 45^\circ = \frac{\sqrt{2}}{2} $

✔ Answer: $\frac{\sqrt{2}}{2}$

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14) tan $\frac{8\pi}{3}$

- $ \frac{8\pi}{3} = \frac{8\pi}{3} - 2\pi = \frac{8\pi}{3} - \frac{6\pi}{3} = \frac{2\pi}{3} $
- $ \frac{2\pi}{3} = 120^\circ $, second quadrant, tangent is negative
- Reference angle: $60^\circ$
- $ \tan \frac{2\pi}{3} = -\tan 60^\circ = -\sqrt{3} $

✔ Answer: $-\sqrt{3}$

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15) sin 225°

- 225° is in third quadrant, sine is negative
- Reference angle: $45^\circ$
- $ \sin 225^\circ = -\sin 45^\circ = -\frac{\sqrt{2}}{2} $

✔ Answer: $-\frac{\sqrt{2}}{2}$

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16) sin $(-\frac{5\pi}{4})$

- Use identity: $ \sin(-\theta) = -\sin(\theta) $
- So $ \sin\left(-\frac{5\pi}{4}\right) = -\sin\left(\frac{5\pi}{4}\right) $
- $ \frac{5\pi}{4} = 225^\circ $, third quadrant, sine is negative
- $ \sin\left(\frac{5\pi}{4}\right) = -\frac{\sqrt{2}}{2} $
- Therefore: $ -(-\frac{\sqrt{2}}{2}) = \frac{\sqrt{2}}{2} $

✔ Answer: $\frac{\sqrt{2}}{2}$

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17) cos $(-\frac{13\pi}{6})$

- Use identity: $ \cos(-\theta) = \cos(\theta) $
- So $ \cos\left(-\frac{13\pi}{6}\right) = \cos\left(\frac{13\pi}{6}\right) $
- $ \frac{13\pi}{6} = 2\pi + \frac{\pi}{6} $ → subtract $2\pi$: $ \frac{\pi}{6} $
- $ \cos\left(\frac{13\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2} $

✔ Answer: $\frac{\sqrt{3}}{2}$

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18) sin 390°

- $390^\circ - 360^\circ = 30^\circ$
- $ \sin 390^\circ = \sin 30^\circ = \frac{1}{2} $

✔ Answer: $\frac{1}{2}$

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✔ Final Answers:

| Problem | Answer |
|--------|--------|
| 1) sin 330° | $-\frac{1}{2}$ |
| 2) cos 330° | $\frac{\sqrt{3}}{2}$ |
| 3) tan 120° | $-\sqrt{3}$ |
| 4) cos $\frac{7\pi}{4}$ | $\frac{\sqrt{2}}{2}$ |
| 5) cot 225° | $1$ |
| 6) sin $\frac{5\pi}{3}$ | $-\frac{\sqrt{3}}{2}$ |
| 7) sin 150° | $\frac{1}{2}$ |
| 8) cos 480° | $-\frac{1}{2}$ |
| 9) tan 225° | $1$ |
| 10) tan $\frac{9\pi}{4}$ | $1$ |
| 11) cos $\frac{17\pi}{6}$ | $-\frac{\sqrt{3}}{2}$ |
| 12) sin $\frac{10\pi}{3}$ | $-\frac{\sqrt{3}}{2}$ |
| 13) cos 675° | $\frac{\sqrt{2}}{2}$ |
| 14) tan $\frac{8\pi}{3}$ | $-\sqrt{3}$ |
| 15) sin 225° | $-\frac{\sqrt{2}}{2}$ |
| 16) sin $(-\frac{5\pi}{4})$ | $\frac{\sqrt{2}}{2}$ |
| 17) cos $(-\frac{13\pi}{6})$ | $\frac{\sqrt{3}}{2}$ |
| 18) sin 390° | $\frac{1}{2}$ |

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