Trigonometric Functions of Special Angles Worksheet

Trigonometric Functions of Special Angles Worksheet
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1.
Evaluate: sin 30° + cos 30° + sin 150° + cos 150°
 a. 1 + $\sqrt{3}$ b. $\sqrt{3}$ c. 1

Solution:

0° < 30° < 90°, so the terminal side of the angle 30° lies in quadrant I.

90° < 150° < 180°, so the terminal side of the angle 150° lies in quadrant II.

The sine is positive in both quadrants I, II and cosine is positive in quadrant I, and negative in quadrant II.

sin 30° + cos 30° + sin 150° + cos 150°

= 1 / 2 + 32 + 1 / 2 + (- 32) = 1

2.
Find the value of sin 120° + cos 120°.
 a. $\frac{\sqrt{2}}{2}$ b. $\frac{\sqrt{3}-1}{2}$ c. - $\frac{\sqrt{3}}{2}$

Solution:

90° < 120° < 180°, so the terminal side of the angle lies in quadrant II.

Sine function is positive and cosine function is negative in quadrant II.

sin 120° = 32 and cos 120° = - 12

sin 120° + cos 120° = 32 + - 12

= 3- 12

3.
Find the value of sec 225° - tan 225°.
 a. - $\sqrt{2}$ + 1 b. - $\sqrt{2}$ - 1 c. - $\frac{1}{\sqrt{2}}$ - 1 d. - $\frac{1}{\sqrt{2}}$ + 1

Solution:

180° < 225° < 270°, so the terminal side of the angle 225° lies in quadrant III.

Secant function is negative and tangent function is positive in quadrant III.

sec 225° = - 2 and tan 225° = 1

sec 225° - tan 225° = - 2 - 1

4.
Evaluate: (cot 300°)(cos 300°) + (tan 300°)(sin 300°).
 a. - $\frac{\left(9+\sqrt{3}\right)}{6}$ b. $\frac{9-\sqrt{3}}{6}$ c. - $\frac{\sqrt{3}}{4}$ d. $\frac{1-\sqrt{3}}{2}$

Solution:

270° < 300° < 360°, so the terminal side of the angle 300° lies in quadrant IV.

Cotangent function, tangent function, sine functions are negative and cosine function is positive in quadrant IV.

(cot 300°)(cos 300°) + (tan 300°)(sin 300°)

= (- 13)(12) + (- 3)(- 32)

= - 36 + 32

= 9-36

5.
Evaluate: $\frac{{\mathrm{tan 420}}_{}^{o}}{{\mathrm{cos 60}}_{}^{o}}$

 a. $\sqrt{6}$ b. 2 + $\sqrt{3}$ c. 2$\sqrt{3}$ d. 2 - $\sqrt{3}$

Solution:

420° - 360° = 60°, so a 420° angle is co-terminal with a 60° angle and its terminal side is in quadrant I.

The tangent function and cosine function are both positive in quadrant I.

tan 420ocos 60o = 312 = 23

6.
Evaluate: $\frac{{\mathrm{1 - sin \left(-30}}_{}^{o}\right)}{{\mathrm{1 + sin \left(-30}}_{}^{o}\right)}$

 a. $\frac{1}{4}$ b. $\frac{3}{4}$ c. $\frac{1}{3}$ d. 3

Solution:

The terminal side of an angle - 30° lies in quadrant IV.

The sine function is negative in quadrant IV.

1 - sin (-30o)1 + sin (-30o) = 1-(- 12)1+(- 12)

= 1+121-12 = 3212 = 3

7.
Find the value of sin 270°.
 a. - 1 b. $\frac{1}{2}$ c. 1

Solution:

270° is a quadrantal angle. Since the terminal side of a quadrantal angle lies on an axis, no reference triangle can be formed.

Choose a point, such as (0, - 1) on the terminal side of the angle in standard position.

Then x = 0, y = - 1, r = x2+y2 = 1

sin 270° = yr = - 11 = - 1

8.
Find the value of cos 390°.
 a. $\sqrt{3}$ b. 1 c. $\frac{\sqrt{3}}{2}$ d. $\frac{1}{2}$

Solution:

390° - 360° = 30°, so a 390° angle is co-terminal with a 30° angle and its terminal side is in quadrant I.

The cosine function is positive in quadrant I.

cos 390° = cos 30° = 32

9.
Find the value of csc 90°.
 a. - 1 b. not defined c. 1

Solution:

90° is a quadrantal angle. Since the terminal side of a quadrantal angle lies on an axis, no reference triangle can be formed.

Choose a point, such as (0, 1) on the terminal side of the angle in standard position.

Then x = 0, y = 1, r = x2+y2 = 1

csc 90° = ry = 1 / 1 = 1.

10.
Find the value of cos 270°.
 a. $\frac{1}{2}$ b. 1 c. - 1

Solution:

270° is a quadrantal angle. Since the terminal side of a quadrantal angle lies on an axis, no reference triangle can be formed.

Choose a point, such as (0, - 1) on the terminal side of the angle in standard position.

Then x = 0, y = - 1, r = x2+y2 = 1

cos 270° = xr = 0 / 1 = 0