UPSC NDA Math Model Paper 4

Given: sin A sin (60° - A) sin (60° + A) = k sin 3A
As we know that, sin (A + B) = sin A cos B + sin B cos A and sin (A - B) = sin A cos B – sin B cos A
L.H.S: sin A sin (60° - A) sin (60° + A) = sin A (sin 60° cos A – sin A cos 60°) (sin 60° cos A + sin A cos 60°)
⇒ sin A sin (60° - A) sin (60° + A) = sin A (3/ 4 – sin‌2A)
⇒sin‌A[

3

4

−sin2A]=k‌sin‌3‌A
⇒

3‌sin‌A−4sin3A

4

=k‌sin‌3‌A
⇒3‌sin‌A−4sin3A=4k‌sin‌3‌A
As we know that, sin‌3‌A=3‌sin‌A−4sin3A
⇒ sin 3A = 4k sin 3A
⇒ k = 1/4