sup guys how do u prove this other than expanding everything, i tried contracting it but it didnt work out well for me :/
sin(5x) + sin(3x) - 2sin(2x)cos(x)=2sin(2x)cos(3x)
sup guys how do u prove this other than expanding everything, i tried contracting it but it didnt work out well for me :/
sin(5x) + sin(3x) - 2sin(2x)cos(x)=2sin(2x)cos(3x)
LHS
= sin 5x + sin 3x - 2sin 2x cos x
= sin 5x + sin 3x - {sin 2x cos x + cos 2x sin x + sin 2x cos x - cos 2x sin x}
= sin 5x + sin 3x - {sin (2x + x) + sin (2x - x)}
= sin 5x + sin x
= sin (3x + 2x) - sin (3x - 2x)
= sin 3x cos 2x + cos 3x sin 2x - sin 3x cos 2x + cos 3x sin 2x
= 2sin 2x cos 3x
= RHS
LHS
= sin 5x + sin 3x - 2sin 2x cos x
= sin 5x + sin 3x - {sin 2x cos x + cos 2x sin x + sin 2x cos x - cos 2x sin x}
= sin 5x + sin 3x - {sin (2x + x) + sin (2x - x)}
= sin 5x + sin x
= sin (3x + 2x) - sin (3x - 2x)
= sin 3x cos 2x + cos 3x sin 2x - sin 3x cos 2x + cos 3x sin 2x
= 2sin 2x cos 3x
= RHS