Question

If sin (A − B) = sin A cos B − cos A sin B and cos (A − B) = cos A cos B + sin A sin B, find the values of sin 15° and cos 15°.

Answer 1

SOLUTION :

Given :  

sin(A – B) = sin A cos B – cos A sin B

and, cos (A – B) = cos A cos B + sin A sin B

We need to find, sin 15° and cos 15°.

Let A = 45° and B = 30°

sin(A – B) = sin A cos B – cos A sin B

sin (45° - 30°) = sin 45° cos 30° – cos 45° sin 30°

sin 15° = sin 45° cos 30° – cos 45° sin 30°

=(1/√2 × √3/2) - (1/√2 ×1/2)

[sin 45°=1/√2 , cos 30°= √3/2,  cos 45°= 1/√2,  sin 30°= ½]

= √3/2√2 −1/2√2

sin 15° = (√3 -1) /2√2

cos (A – B) = cos A cos B + sin A sin B

cos (45° - 30°) = cos 45° cos 30° – sin 45° sin 30°

cos 15° = cos 45° cos 30° – sin 45° sin 30°

= (1/√2 × √3/2)  + (1/√2 ×1/2)

[sin 45°=1/√2 , cos 30°= √3/2,  cos 45°= 1/√2,  sin 30°= ½]

= √3/2√2 + 1/2√2

cos 15° = (√3 +1) /2√2

Hence, the values of sin 15° = (√3 -1) /2√2 and cos 15° = (√3 +1) /2√2.

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