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comeondroogs · Mar 1, 2011, 07:14 PM

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

jcaron2 · Mar 1, 2011, 08:23 PM

I like this question! Verify the following:

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

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:

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

comeondroogs · Mar 1, 2011, 08:59 PM

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.

comeondroogs · Mar 1, 2011, 09:02 PM

I apologize,

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

jcaron2 · Mar 1, 2011, 09:18 PM

There's a pretty good list here.

List of trigonometric identities - Wikipedia, the free encyclopedia

comeondroogs · Mar 1, 2011, 09:59 PM

Wow, Thanks!

Unknown008 · Mar 2, 2011, 12:31 AM

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.

jcaron2 · Mar 2, 2011, 05:59 AM

Good point Jerry!

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