Proving of trigonometric ID (solved)

Good Day,

I'm stuck on proving the following trigonometric identity:

**(sin 2x + sin 4x + sin 6x) / (cos 2x + cos 4x + cos 6x) = tan 4x**

I do hope that some one can give me some advice on how to prove this as I've been going at this for the past 2 days and I feel that I'm not getting anywhere near.

Thanks in advance.

Re: Proving of trigonometric ID

Quote:

Originally Posted by

**dd86** Good Day,

I'm stuck on proving the following trigonometric identity:

**(sin 2x + sin 4x + sin 6x) / (cos 2x + cos 4x + cos 6x) = tan 4x**

I do hope that some one can give me some advice on how to prove this as I've been going at this for the past 2 days and I feel that I'm not getting anywhere near.

Use the sum-to-product identities,

$\displaystyle \sin\theta+\sin\phi = 2\sin\left(\frac{\theta+\phi}2\right)\cos\left( \frac{\theta-\phi}2\right)$

$\displaystyle \cos\theta+\cos\phi = 2\cos\left(\frac{\theta+\phi}2\right)\cos\left( \frac{\theta-\phi}2\right)$

We have

$\displaystyle \frac{\sin2x+\sin4x+\sin6x}{\cos2x+\cos4x+\cos6x}$

$\displaystyle =\frac{\sin4x+\left(\sin2x + \sin6x\right)}{\cos4x + \left(\cos2x+\cos6x\right)}$

$\displaystyle =\frac{\sin4x+2\sin4x\cos2x}{\cos4x + 2\cos4x\cos2x}$

$\displaystyle =\frac{\sin4x\left(1+2\cos2x\right)}{\cos4x\left(1 + 2\cos2x\right)}$

$\displaystyle =\frac{\sin4x}{\cos4x} = \tan4x.$

Please note, however, that the cancellation step is not valid when $\displaystyle 1+2\cos2x = 0.$ The identity actually does not hold for these values of $\displaystyle x,$ as the original expression is undefined at such values. So the equation is only identically true for

$\displaystyle x\neq\pi\left(k\pm\frac13\right)\!,\ \ k\in\mathbb{Z}.$

Re: Proving of trigonometric ID

Hi Reckoner,

Thanks so much. Now it's clear where I went wrong.