Thread: identity another

[1-tan^2(x/2)] / [1+tan^2(x/2)] = cosx


I don't want to bother you, but do you know of any trig identity site that discusses the rules of sin or cos or tan of (x/2)?

thanks

Originally Posted by Mr_Green

[1-tan^2(x/2)] / [1+tan^2(x/2)] = cosx

Exactly like before,

$\displaystyle \cos x = \frac{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2} }{1} = \frac{\cos^2 \frac{x}{2} - \sin^2 \frac{x}{2}}{\cos^2 \frac{x}{2} + \sin^2 \frac{x}{2}} = \frac{1- \tan^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}$

don't you want to change the more complex side into a simplified state? is there a specific rule or is that jsut my teacher's preference?

Originally Posted by Mr_Green

don't you want to change the more complex side into a simplified state? is there a specific rule or is that jsut my teacher's preference?

I did it in a complete different way. I just developed it out how I remember it.