Question

Can anyone show me how to prove the following trig identity

(sin(3x)/(sinx)-(cos(3x)/cosx)=2

Answer 1

Is this complete??

I mean where is LHS and where is RHS..

You want to prove this equal to what thing??

Answer 2

equals 2 I am sorry I forgot to post that.....

Answer 3

See:

Take the LCM first and combine the terms:

\[\frac{\sin(3x)}{\sin(x)} - \frac{\cos(3x)}{\cos(x)} \implies \frac{\sin(3x) \cdot \cos(x) - \cos(3x) \cdot \sin(x)}{\sin(x)\cos(x)}\]

\[\implies \frac{\sin(3x - x)}{\sin(x) \cdot \cos(x)} \implies \frac{\sin(2x)}{\sin(x) \cdot \cos(x)} \implies \frac{2\sin(x) \cdot \cos(x)}{\sin(x) \cdot \cos(x)} \implies 2\]

Answer 4

Oh I see now thanks you for your response.

Answer 5

Identities used are:

\[\sin(A) \cdot \cos(B) - \cos(A) \cdot \sin(B) = \sin(A-B)\]

\[\sin(2A) = 2\sin(A) \cdot \cos(A)\]

Answer 6

Welcome dear..