Question

( sin3x + sinx )sinx + (cos3x - cosx)cosx=0 prove it

Answer 1

sinx(sin3x + sinx) = sinx(2 sin2x cosx)     ...(1)
cosx(cos 3x - cos x) = cosx(-2sin2x sinx)        .....(2)
(1) +(2)  =2sinx sin2x cosx - 2sinx sin2x cosx = 0  

Answer 2

LHS=

( sin3x + sinx )sinx + (cos3x - cosx)cosx

$\displaystyle \sf \: { \left[ 2sin \left( \frac{3x + x}{2} \right)cos \left( \frac{3x - x}{2} \right)\right] sinx + \left[ 2sin \left( \frac{3x + x}{2} \right)sin \left( \frac{x - 3x}{2} \right)\right]\: \: cosx } \\ \\ \displaystyle \sf \: \left[2sin2x \: \times cosx \right]sinx + \left[ 2sin2x \times sin( - x)\right]cosx \\ \\ \displaystyle \sf \: 2sin2x \: sinx \: cosx + \left[ - 2sin2x \: sinx\right]cosx \\ \\ \displaystyle \sf \: 2sin2x \: sinx \: cosx \: - 2sin2x \: sinx \: cosx \: \\ \\ \sf \: = 0$