Trigonometric Functions of Special Angles Worksheet

Trigonometric Functions of Special Angles Worksheet
  • Page 1
 1.  
Evaluate: sin 30° + cos 30° + sin 150° + cos 150°
a.
1 + 3
b.
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


Correct answer : (4)
 2.  
Find the value of sin 120° + cos 120°.
a.
22
b.
3-12
c.
- 32


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


Correct answer : (3)
 3.  
Find the value of sec 225° - tan 225°.
a.
- 2 + 1
b.
- 2 - 1
c.
- 12 - 1
d.
- 12 + 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


Correct answer : (2)
 4.  
Evaluate: (cot 300°)(cos 300°) + (tan 300°)(sin 300°).
a.
- (9+3)6
b.
9-36
c.
- 34
d.
1-32


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


Correct answer : (2)
 5.  
Evaluate: tan 420ocos 60o

a.
6
b.
2 + 3
c.
23
d.
2 - 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


Correct answer : (3)
 6.  
Evaluate: 1 - sin (-30o)1 + sin (-30o)

a.
1 4
b.
3 4
c.
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


Correct answer : (4)
 7.  
Find the value of sin 270°.
a.
- 1
b.
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


Correct answer : (1)
 8.  
Find the value of cos 390°.
a.
3
b.
1
c.
32
d.
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


Correct answer : (3)
 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.


Correct answer : (4)
 10.  
Find the value of cos 270°.
a.
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


Correct answer : (1)

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