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

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### 2 Answers

We have to prove that sin x+ sin 3x + sin 5x =(1+ 2*cos 2x)*sin 3x

Use the relation sin x + sin y = 2*sin[(x + y)/2]*cos[(x - y)/2]

sin x + sin 3x + sin 5x

as sin 3x is there on the right hand side use apply the relation to sin x + sin 5x

=> 2*sin[(x + 5x)/2]*cos[(x - 5x)/2] + sin 3x

=> 2*sin[6x/2]*cos[-4x/2] + sin 3x

=> 2*sin 3x*cos[-4x/2] + sin 3x

cos (-x) = cos x

=> sin 3x*[2*cos 2x + 1]

which is the right hand side

**This proves: sin x+ sin 3x + sin 5x =(1+ 2*cos 2x)*sin 3x**

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.**