[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
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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.
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