Please help me out with this question

Prove that sin2x-tanx=tanXcos2X

This is what i did:

Sin2x-tanx=0

2sinXcosX-tanx=0

2sinXcosX-(sinX/cosX)=0

Don’t know how to take it any further to solve it?

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- Nov 15th 2010, 12:18 PMmike789Prove that sin2x-tanx=tanXcos2X
Please help me out with this question

Prove that sin2x-tanx=tanXcos2X

This is what i did:

Sin2x-tanx=0

2sinXcosX-tanx=0

2sinXcosX-(sinX/cosX)=0

Don’t know how to take it any further to solve it? - Nov 15th 2010, 12:30 PMmasters

Hi mik789,

Here's what I did. I used the left side only.

$\displaystyle \sin 2x - \tan x =$

$\displaystyle 2\sin x \cos x - \tan x =$

$\displaystyle 2 \sin x \cos x - \dfrac{\sin x}{\cos x}=$

$\displaystyle \dfrac{2 \sin x \cos^2 x-\sin x}{\cos x}=$

$\displaystyle \dfrac{\sin x(2 \cos ^2 x-1)}{\cos x}=$

$\displaystyle \dfrac{\sin x}{\cos x} (2\cos^2x-1)=$

$\displaystyle \boxed{\tan x \cos 2x}$

- Nov 15th 2010, 12:55 PMmike789
- Nov 15th 2010, 01:09 PMmike789
thanks. you cancelled the cos^2 to get cos2x

- Nov 15th 2010, 03:45 PMarccos
not exactly.the double angle formula for cos states that cos2x = 2cos^2x - 1 . You cannot "cancel" away cos on it's own. It is always accompanied by a something(in this case,x)

and also for proving, you are only allowed to algebraically manipulate one side such that it looks like the other. cross multiplication and things like that are not allowed because if you do that it means that it is an equation and not an identity(correct me if I am wrong here)