# Trig Proof

1. Oct 11, 2007

### danago

If $$z=cis\theta$$, verify that $$\tan \theta = \frac{{z - z^{ - 1} }}{{i(z + z^{ - 1} )}}$$. Use this result to prove that $$\cos (2\theta ) = \frac{{1 - \tan ^2 \theta }}{{1 + \tan ^2 \theta }}$$

Ok, ive managed to verify the first equation given, but im not really sure how to use it to prove the second identity. Im really not sure where to start. If somebody could give me a hint about where to start id be very appreciative.

Thanks,
Dan.

2. Oct 11, 2007

### mjsd

Does your $$z^{-1}$$ actually mean conjugate of z?

3. Oct 11, 2007

### danago

Nah its the reciprocal of z.

4. Oct 11, 2007

### HallsofIvy

Staff Emeritus
Haven't seen "cis" in a long time! It's engineering shorthand for $cos(\theta)+ i sin(\theta)= e^{i\theta}$. From the last form, or comparing $\theta$ for z and z-1, it should be clear that if $z= cis(\theta)$ then $z^{-1}= cis(-\theta)= cos(\theta)- i sin(\theta)$. Putting those in for z and z-1 in
$$\frac{z- z^{-1}}{i(z+ z^{-1}}[/itex] By the way, when |z|= 1, as is the case here, z-1 is the complex conjugate: [tex]\frac{1}{x+ iy}= \frac{1}{x+iy}\frac{x-iy}{x-iy}= \frac{x-iy}{x^2+ y^2}= x- iy$$

5. Oct 11, 2007

### mjsd

basically, you have $$\tan \theta$$ equal something, so all you need to do to prove your 2nd identity is just sub in this where $$\tan \theta$$'s appear and simplify then you shall see that it actually equals to $$\cos (2\theta)$$
hint:
$$\cos (x)=\frac{e^{ix}+e^{-ix}}{2}$$

6. Oct 11, 2007

### mjsd

Nah, my question was a rhetorical one

also note that $$z+z^*=2\text{Re}(z)$$ , $$z-z^* = 2i\text{Im}(z)$$ and $$z z^* = |z|$$

7. Oct 11, 2007

### danago

Got it thanks for the help guys :D