Question

Question 5 Find the value of:

(i) sin 75°

(ii) tan 15°

Class X1 - Maths -Trigonometric Functions Page 73

Answer 1

(i) sin75°
= sin(45° + 30°)
use the formula,
Sin(A + B) = sinA.cosB+cosA.sinB
Here, A = 45°, and B = 30°
= sin45°.cos30° + cos45°.sin30°
= 1/√2 × √3/2 + 1/√2 × 1/2
= √3/2√2 + 1/2√2
= (√3 + 1)/2√2

(ii) tan15°
= tan(45°-30°)
use the formula,
Tan(A - B) = (tanA-tanB)/(1 + tanA.tanB)

Here, A = 45°, B = 30°

= (tan45° - tan30°)/(1+tan45°.tan30°)
= (1 - 1/√3)/(1+1/√3)
= (√3 - 1)/(√3 + 1)
Now, rationalize
= (√3 -1)(√3 -1)/(√3²-1²)
= (4-2√3)/2
= 2 - √3