Anonymous
Anonymous asked in Science & MathematicsMathematics · 8 years ago

# Prove that sin a + 2sin 3a +sin 5a/sin 3a + 2 sin5a +sin7a = sin3a/sin5a?

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• EM
Lv 7
8 years ago

Let's construct an identity that will be useful later.

sin(x + y) cos(x - y) =

By angle sum identity for sine,

[sinx cosy + siny cosx][cosx cosy + sinx siny] =

(sinx cosx cos²y + sinx cosx sin²y) + (siny cosy cos²x + sin²x siny cosy) =

(sinx cosx)(cos²y + sin²y) + (siny cosy)(cos²x + sin²x) =

By one of the Pythagorean identities,

sinx cosx + siny cosy =

By the double angle identity for sine,

(1/2)sin2x + (1/2)sin2y =

(1/2)(sin2x + sin2y) = sin(x + y) cos(x - y) [This will be useful here.]

Now on to the original problem:

(sina + 2sin3a + sin5a) / (sin3a + 2sin5a + sin7a) =

[(sina + sin3a) + (sin3a + sin5a)] / [(sin3a + sin5a) + (sin5a + sin7a)] =

Employing our "new" identity,

[2sin{(3a + a)/2} cos{(3a - a)/2} + 2sin{(5a + 3a)/2} cos{(5a - 3a)/2}] / [2sin{(5a + 3a)/2} cos{(5a - 3a)/2} + 2sin{(7a + 5a)/2} cos{(7a - 5a)/2}] =

[2sin2a cosa + 2sin4a cosa] / [2sin4a cosa + 2sin6a cosa] =

2cosa [sin2a + sin4a] / [{2cosa}{sin4a + sin6a}] =

(sin2a + sin4a) / (sin4a + sin6a) =

Employing our "new" identity again,

2sin{(4a + 2a)/2} cos{(4a - 2a)/2} / [2sin{(6a + 4a)/2} cos{(6a - 4a)/2}] =

2sin3a cosa / [2sin5a cosa] =

sin3a / sin5a