A little push on this trig identity

[aisha]

A little push on this trig identity please

$$\tan^2(x)= \frac {1-\cos(2x)} {1+cos(2x)}$$

I need a little push I know from my other post that $$\cos(2x)=\cos^2 (x) - \sin^2 (x)$$ (can someone explain why?)

 

[vincentchan]

what is your questioin? proving the tan identity or the cos one?

 

[aisha]

I have to make the left hand side equal the right hand side I don't think it matters if you use cos or tan which ever one is easier.

 

[blue_soda025]

It would be easier to work with the right side.
So it would be like:

$$\frac {1-\cos(2x)} {1+cos(2x)}$$

$$\frac {1 - (1 - 2\sin^2x)}{1 + 2\cos^2x - 1}$$ Double angle identities

$$\frac{2\sin^2x}{2\cos^2x}$$ 2's cancel out

$$\tan^2x$$

 

[aisha]

Thanks sooo much BLUE SODA I am not good with the double angle identity thanks again