Prove that sin2x-tanx=tanXcos2X

**mike789** · November 15th 2010, 12:18 PM

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?

**masters** · November 15th 2010, 12:30 PM

Originally Posted by mike789

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?

Hi mik789,

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

$\sin 2x - \tan x =$

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

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

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

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

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

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

**mike789** · November 15th 2010, 12:55 PM

Originally Posted by masters

Hi mik789,

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

$\sin 2x - \tan x =$

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

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

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

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

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

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

But how do you turn 2(cos^2)x-1 into cos2x

**mike789** · November 15th 2010, 01:09 PM

thanks. you cancelled the cos^2 to get cos2x

**arccos** · November 15th 2010, 03:45 PM

Originally Posted by mike789

thanks. you cancelled the cos^2 to get cos2x

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)

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