Thread: Verify the Trig Identity

I've been working on this for a few hours... I have a mental block...

Verify the Identity:

[sin(2x)+sin(4x)] / [cos(2x) + cos(4x)] = tan(3x)


thanks so much, in advance

This should be easy if you can use Simpson's formulas (equations (1) and (3)) in the numerator and denominator...

..working it...

Very easy with Simpson's !!!!
Unfortunately, I'm in 11th grade Math and all we have to work with are sum/Difference/double angle identities... any more suggestions ???

It's indeed very easy with Simpson's.

If you can only use the multiple-angle formulas, I'm afraid it will be a bit more tricky. I don't see an elegant way just at first sight - you could reduce it all to angle x...

Looks like the function sum/diff (alias Simpson's .... is he taking credit for this these days ) was in my notes...

So I can use "Simpson's" ...
Thanks so much !!!!

(Happy)Hello, poolshark!

It's easy if we are allowed the Sum-to-Product identities:

. . $\displaystyle \begin{array}{ccc}\sin A + \sin B &=& 2\sin\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \\ \\ [-3mm]

\cos A + \cos B &=& 2\cos\left(\frac{A+B}{2}\right)\cos\left(\frac{A-B}{2}\right) \end{array}$

Verify the identity: .$\displaystyle \frac{\sin2x+ \sin4x} {\cos2x + \cos4x} \:=\: \tan3x\$

The numerator is: .$\displaystyle \sin4x + \sin 2x \:=\:2\sin3x\cos x$

The denominator is: .$\displaystyle \cos4x+\cos2x \:=\:2\cos3x \cos x$


The fraction becomes: .$\displaystyle \frac{{\color{blue}\rlap{/}}2\sin3x{\color{red}\rlap{/////}}\cos x}{{\color{blue}\rlap{/}}2\cos3x{\color{red}\rlap{/////}}\cos x} \;=\;\frac{\sin3x}{\cos3x} \;=\;\tan3x$

wow the teacher never taught us this in precal..

Originally Posted by poolshark

Looks like the function sum/diff (alias Simpson's .... is he taking credit for this these days ) was in my notes...

So I can use "Simpson's" ...
Thanks so much !!!!

I thought this would be the case, the problem "screams" for Simpson's