# Prove identities, can anyone help?

• Feb 14th 2009, 08:53 AM
joemc22
Prove identities, can anyone help?
Hi, I am working through proving some trigonomic identities and need some help on the following, can someone please help?

cos3x - cos7x = tan2x
sin7x + sin3x

I know I need to end up with: sin2x = tan2x
cos2x

Need some help getting there?

Thanks for any help.
• Feb 14th 2009, 08:56 AM
Moo
Hello,
Quote:

Originally Posted by joemc22
Hi, I am working through proving some trigonomic identities and need some help on the following, can someone please help?

cos3x - cos7x = tan2x
sin7x + sin3x

I know I need to end up with: sin2x = tan2x
cos2x

Need some help getting there?

Thanks for any help.

Use these formulae :

$\displaystyle \sin a+\sin b=2 \sin \tfrac{a+b}{2} \cos \tfrac{a-b}{2}$
$\displaystyle \sin a-\sin b=2 \cos \tfrac{a+b}{2} \sin \tfrac{a-b}{2}$
$\displaystyle \cos a+\cos b=2 \cos \tfrac{a+b}{2} \cos \tfrac{a-b}{2}$
$\displaystyle \cos a-\cos b=-2 \sin \tfrac{a+b}{2} \sin \tfrac{a-b}{2}$

(They directly come from the addition & subtraction formulae of cos and sin)
• Feb 14th 2009, 09:17 AM
joemc22
Prove identities
Thanks, but(Worried)
Not sure what to do with that, I'm working from a list of basic identities and need to know the next steps, haven't seen that formulae before?
• Feb 14th 2009, 01:04 PM
Trig Identity
Hello joemc22

As Moo said, use the formulae :

$\displaystyle \sin a+\sin b=2 \sin \tfrac{a+b}{2} \cos \tfrac{a-b}{2}$

$\displaystyle \cos a-\cos b=-2 \sin \tfrac{a+b}{2} \sin \tfrac{a-b}{2}$

$\displaystyle \frac{\cos 3x - \cos 7x}{\sin 7x + \sin 3x} = \frac{-2 \sin \tfrac{3x+7x}{2} \sin \tfrac{3x - 7x}{2}}{2 \sin \tfrac{7x+3x}{2}\cos\tfrac{7x-3x}{2}}$

$\displaystyle = \frac{-\sin 5x\sin (-2x)}{\sin 5x \cos 2x}$

$\displaystyle = \frac{\sin 2x}{\cos 2x}$

$\displaystyle =\tan 2x$