(cos(x)+cos(3x))/(sin(3x)-sin(x))=cot(x)

I like this question! Verify the following:

\Large \frac{\cos x + \cos {3x}}{\sin 3x - \sin {x}}=\cot x



It's helpful to know these two trig identities:

\cos{(3x)}=4\cos^3 x - 3 \cos x

\sin{(3x)}=-4\sin^3 x + 3 \sin x

If we apply those to the left side of the original equation we obtain:

\Large \frac{\cos x + \cos {3x}}{\sin 3x - \sin {x}}=\frac{4\cos^3 x - 2 \cos x}{-4\sin^3 x + 2 \sin x}=\frac{\cos x}{\sin x}\cdot \frac{2\cos^2 x - 1}{-2\sin^2 x + 1}=\cot x \cdot \frac{\cos{2x}}{\cos{2x}}=\cot x

jcaron2,
You mentioned, "It's helpful to know those two trig identities." Where can I find more of these identities short cuts and I was not those "two trig identities" in college.

I apologize,

Where can I find more of these identities short cuts and I was not taught** those "two trig identities" in college.

There's a pretty good list here (http://en.wikipedia.org/wiki/List_of_trigonometric_identities).

List of trigonometric identities - Wikipedia, the free encyclopedia (http://en.wikipedia.org/wiki/List_of_trigonometric_identities)

Wow, Thanks!

Well, you can use the basic ones:

\cos(3x) =\cos(2x + x) = \cos(2x)\cos(x) - \sin(x)\sin(2x)

From \cos(A+B) = \cos A\cos B - \sin A\sin B

Same for sine.

\sin(A+B) = \sin A \cos B - \cos A \sin B

And use those as long as you have double angles. This is longer, but makes use of the elementary identities.

Good point Jerry!

Either way, it's not often we get to see a trig verification with a triple angle. Thanks for posting, Comeondroogs!