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