Verify the Trig Identity

**poolshark** · December 21st 2008, 02:45 PM

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

**TD!** · December 21st 2008, 02:47 PM

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

**poolshark** · December 21st 2008, 02:55 PM

..working it...

**poolshark** · December 21st 2008, 03:09 PM

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 ???

**TD!** · December 21st 2008, 03:10 PM

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...

**poolshark** · December 21st 2008, 03:33 PM

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 !!!!

**Soroban** · December 21st 2008, 03:48 PM

(Happy)Hello, poolshark!

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

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

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

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

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

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

**Khaali91** · December 21st 2008, 07:12 PM

wow the teacher never taught us this in precal..

**TD!** · December 22nd 2008, 03:09 AM

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

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