# A little push on this trig identity

1. Feb 26, 2005

### aisha

A little push on this trig identity plz

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

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

2. Feb 26, 2005

### vincentchan

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

3. Feb 26, 2005

### aisha

I have to make the left hand side equal the right hand side I dont think it matters if you use cos or tan which ever one is easier.

4. Feb 26, 2005

### 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

$$\tan^2x$$

5. Feb 26, 2005

### aisha

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