Question

(sin 3x + sin x) sin x + (cos 3x - cos x) cos x = 0
prove it​

Answer 1

Step-by-step explanation:

L.H.S. = (sin 3x + sin x)sin x + (cos 3x - cos x)cos x

= sin 3x sin x + sin²x + cos 3x cos x - cos²x

= (cos 3x cos x + sin 3x sin x) - (cos²x -

sin²px)

= cos (3x – x) - pcos 2x

= cos 2x - cos 2x

= 0

= R.H.S.

Answer 2

formula: sinC+ sinD= 2sinC+D/2 × CosC-D/2

cosC- cosD= 2sinC+D/2× Sin D-C/2

the put this formula in your question

2sin2x cosx sinx+ 2sin2x sin(-x) cosx

2sin2x cosx sinx -2sin2x cosx sinx

( sin(-x)= -sinx)

=0= RHS