Prove that sinx+sin3x+sin5x=(1+2cos2x)*sin3x .

Since the added trigonometric functions are matching,
we'll apply the identity:

sin a + sin b = 2 sin
[(a+b)/2]*cos[(a-b)/2]

We'll add the first and the 3rd term
from the left side:

sin x + sin 5x = 2 sin
[(x+5x)/2]*cos[(x-5x)/2]

sin x + sin 5x = 2 sin
3x*cos(-2x)

Since the cosine function is even,we'll
get:

sin x + sin 5x = 2 sin 3x*cos
2x

We'll re-write the
expression:

sin 3x + 2 sin 3x*cos 2x =
(1+2cos2x)*sin3x

We'll factorize by sin 3x to the left
side:

sin 3x*(1 + 2cos 2x) =
(1+2cos2x)*sin3x

We notice that managing the
left side, we'll get LHS = RHS , therefore the given identity sin x + sin 3x + sin 5x =
(1+2cos2x)*sin3x is verified.