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I need to prove this:

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

using only $\sin(a-b)$ and $\cos(a-b)$ formulas wich I already proved. I also proven this:

$$\cos^2(x) + \sin^2(x) = 1$$ $$\sin(-x) = -\sin(x)$$ $$\cos(2x) = \cos^2x -\sin^2(x)$$ $$\sin(2x) = 2\sin(x)\cos(x)$$ and the $\sin(a+b)$ and $\cos(a+b)$ formulas. Do I need something more so I can prove this identity? Could you guys give me some help?

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I recommend setting $x=2y$ and substituting the variable $y$ in for $x$. – NotNotLogical Apr 11 '14 at 3:18
up vote 6 down vote accepted

$$\cos2x=\cos^2x-\sin^2x=\frac{\cos^2x-\sin^2x}{\cos^2x+\sin^2x}=\frac{1-\tan^2x}{1+\tan^2x}$$ dividing the numerator & the denominator by $\cos^2x$

Dividing the numerator & the denominator by $\cos^2x$ $$\cos2x=\cos^2x-\sin^2x=\frac{1-\tan^2x}{\sec^2x}=\frac{1-\tan^2x}{1+\tan^2x}$$

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why did you divide by $cos^2(x) + \sin^2(x)$? – Marter Js Apr 11 '14 at 3:20
Because that’s equal to $1$. – Lubin Apr 11 '14 at 3:23

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