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I was requested to verify, with $t=\tan(\frac{x}{2})$, the following identity:

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

I'm quite rusty on my trigonometry, and hasn't been able to found the proof of this. I'm sure there may be some trigonometric property I should know to simplify the work. Could someone hint me or altotegher tell me how to solve this problem? I tried to simplify the RHS looking to get $\cos(x)$ out of it but failed.

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Write $y=x/2$. Then, multiplying by $\cos^2y$ on top and bottom, $$\frac{1-\tan^2y}{1+\tan^2y}=\frac{\cos^2y-\sin^2y}{\cos^2y+\sin^2y}=\frac{\cos2y}1=\cos x$$ The denominator simplifies by the Pythagorean identity $\cos^2x+\sin^2x=1$ and the numerator simplifies by $\cos2x=\cos^2x-\sin^2x$.

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$$\cos(x)=\cos^2(x/2)-\sin^2(x/2)=(1-t^2)/(\sec^2(x/2))=(1-t^2)/(1+t^2)$$

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Remember the high-school identities: $\;1+\tan ^2\theta=\frac1{\cos^2\theta}\;$ and $\;\cos2\theta=\cos^2\theta-\sin^2\theta$:

$$\frac{1-\tan^2\frac x 2}{1+\tan^2\frac x 2}=\cos^2\tfrac x2\bigl(1-\tan^2\tfrac x 2\bigr)=\cos^2\tfrac x2-\sin^2\tfrac x2=\cos x.$$

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