# EVALUATING TRIGONOMETRIC RATIOS FOR SPECIAL ANGLES

Evaluating Trigonometric Ratios for Special Angles :

In this section, you will learn how to find values of trigonometric ratios for special angles.

## Evaluating Trigonometric Ratios for Special Angles - Practice Questions

Question 1 :

Verify the following equalities:

(i)  sin260°+ cos260°  =  1

Solution :

sin260° + cos260°  =  (√3/2)2 + (1/2)2

sin260° + cos260°  =  (3/4) + (1/4)

sin260° + cos260°  =  (3 + 1)/4

sin260° + cos260°  =  4/4

sin260° + cos260°  =  1

(ii)  1 + tan230°  =  sec230°

Solution :

1 + tan230°  =  1 + (1/√3)2

1 + tan230°  =  1 + (1/3)

1 + tan230°  =  4/3 -----(1)

sec230°  =  (2/√3)2

sec230°  =  4/3 -----(2)

From (1) and (2), we get

1 + tan230°  =  sec230°

(iii) cos90°  =  1 - 2 sin245°  =  2 cos245° - 1

cos 90°  =  0 -----(1)

1 - 2 sin245°  =  1 - 2(1/√2)2

1 - 2 sin245°  =  1 - 2(1/2)

1 - 2 sin245°  =  0 -----(2)

2 cos2 45° - 1  =  2(1/√2)2 - 1

2 cos2 45° - 1  =  2(1/2) - 1

2 cos2 45° - 1  =  0 -------(3)

From (1), (2) and (3), we get

cos90°  =  1 - 2 sin245°  =  2 cos245° - 1

(iv)  sin 30° cos 60° + cos 30° sin 60°  =  sin 90°

Solution :

sin 30° cos 60° + cos 30° sin 60° :

=  (1/2) (1/2) + (√3/2) (√3/2)

=  (1/4) + (3/4)

=  (1 + 3)/4

=  4/4

=  1 -----(1)

sin 90°  =  1 -----(2)

From (1) and (2), we get

sin 30° cos 60° + cos 30° sin 60°  =  sin 90°

Question 2 :

Find the value of the following :

(tan45°/cosec30°) + (sec60°/cot45°) - (5sin90°/2cos0°)

Solution :

(tan45°/cosec30°) + (sec60°/cot45°) - (5sin90°/2cos0°) :

=  (1/2) + (2/1) - (5(1)/2(1))

=  (1/2) + (2/1) - (5/2)

=  (1 + 2 - 5)/2

=  -2/2

=  -1 -----(1)

(ii) (sin 90°+cos 60°+cos 45°) × (sin 30°+cos 0°-cos 45°)

Solution :

(sin 90°+cos 60°+cos 45°) × (sin 30°+cos 0°-cos 45°) :

=  [1 + (1/2) + (1/√2)] × [(1/2) + 1 - (1/√2)]

=  [(3/2) + (1/√2)] × [(3/2) - (1/√2)]

=  [(3/2)2 - (1/√2)2]

=  [(9/4) - (1/2)]

=  (9 - 2)/4

=  7/4

(iii)  sin2 30° - 2 cos3 60° + 3 tan4 45°

Solution :

sin2 30° - 2 cos3 60° + 3 tan4 45° :

=  (1/2)2 - 2(1/2)3 + 3(1)4

=  (1/4) - (1/4) + 3

=  3

Question 3 :

Verify cos 3A = 4 cos3A - 3 cos A , when A  = 30°

Solution :

cos 3A  =  cos 3(30°)

cos 3A  =  cos 90°

cos 3A  =  0  ---(1)

4 cos3A - 3 cosA  =  4 cos330° - 3 cos 30°

4 cos3A - 3 cosA  =  4 (√3/2)3 - 3 (√3/2)

4 cos3A - 3 cosA  =  (12√3/8) - (3√3/2)

4 cos3A - 3 cosA  =  (3√3/2) - (3√3/2)

4 cos3A - 3 cosA  =  0 -----(2)

From (1) and (2), we get

cos 3A = 4 cos3A - 3 cos A

when A  = 30°.

Question 4 :

Find the value of 8 sin 2x cos 4x sin6x, when x = 15°

Solution :

8 sin 2x cos 4x sin6x :

=  8 sin 2(15°) cos 4(15°) sin 6(15°)

=  8 sin 30° cos 60° sin 90°

=  8(1/2) (1/2) (1)

=  2 After having gone through the stuff given above, we hope that the students would have understood to evaluate trigonometric ratios for special angles.

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