Proving an identity

**arze** · July 1st 2009, 07:38 PM

Hi all, this is my first post here, i need some help proving this identity
(sinAsin2A +sin3Asin6A)/(sinAcis2A+sin3Acos6A) =tan5A
Thanks

**red_dog** · July 1st 2009, 09:31 PM

Use these formulas:

$\sin a\sin b=\frac{1}{2}[\cos(a-b)-\cos(a+b)]$

$\sin a\cos b=\frac{1}{2}[\sin(a+b)+\sin(a-b)]$

And the right hand member must be $-\tan 5A$

**yeongil** · July 1st 2009, 09:56 PM

You'll have to use the product-to-sum formulas:

$\sin x \sin y = \frac{\cos(x - y) - \cos(x + y)}{2}$ AND
$\sin x \cos y = \frac{\sin(x + y) + \sin(x - y)}{2}$

So,
$\frac{\sin A \sin 2A + \sin 3A \sin 6A}{\sin A \cos 2A + \sin 3A \cos 6A}$
$\begin{aligned} &= \frac{\frac{\cos(A - 2A) - \cos(A + 2A)}{2} + \frac{\cos(3A - 6A) - \cos(3A + 6A)}{2}}{\frac{\sin(A + 2A) + \sin(A - 2A)}{2} + \frac{\sin(3A + 6A) + \sin(3A - 6A)}{2}} \\ &= \frac{\cos(-A) - \cos 3A + \cos (-3A) - \cos 9A}{\sin 3A + \sin(-A) + \sin 9A + \sin(-3A)} \\ &= \frac{\cos A - \cos 3A + \cos 3A - \cos 9A}{\sin 3A - \sin A + \sin 9A - \sin 3A} \\ &= \frac{\cos A - \cos 9A}{\sin 9A - \sin A} \end{aligned}$

Then, use the sum-to-product formulas:
$\sin x \pm \sin y = 2\sin \left(\frac{x \pm y}{2}\right)\cos\left(\frac{x \mp y}{2}\right)$ AND
$\cos x - \cos y = -2\sin \left(\frac{x + y}{2}\right)\sin\left(\frac{x - y}{2}\right)$

$\frac{\cos A - \cos 9A}{\sin 9A - \sin A}$
$\begin{aligned} &= \frac{-2\sin \left(\frac{A + 9A}{2}\right)\sin\left(\frac{A - 9A}{2}\right)}{2\sin \left(\frac{A - 9A}{2}\right)\cos\left(\frac{A + 9A}{2}\right)} \\ &= \frac{-2\sin 5A \sin (-4A)}{2\sin (-4A) \cos 5A} \\ &= \frac{-\sin 5A}{\cos 5A} \\ &= -\tan 5A \end{aligned}$

Hmm, I'm getting -tan 5A...

01

EDIT: posted too late. red dog also got -tan 5A.

**arze** · July 2nd 2009, 09:08 PM

In my book it DOES say that tan5A so i guess it would be a printing error because there is at least one other problem where the "-" sign is missing.
but thanks anyway!

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