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How would I prove the following trig identity?

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

I am not sure where to begin any help would be useful.

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Hint: $\cos x = \cos \left(\frac x2 + \frac x2\right)$ ... and you now a formula for $\cos(a+b)$, don't you? –  martini Jul 28 '12 at 15:45
    
Another hint: $\cos(x/2) = \frac{1}{2}(e^{ix/2}+e^{-ix/2})$. –  marlu Jul 28 '12 at 15:46
    
I have never seen the second one marlu commented but I know cos(a+b) is cosAcosB-cosAsinB. –  Fernando Martinez Jul 28 '12 at 15:47
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2 Answers

up vote 3 down vote accepted

I am sure you know the formula $\cos(a+b) = \cos a \cos b - \sin a \sin b$. Let $a=b = \frac{x}{2}$, which gives $\cos x = (\cos \frac{x}{2})^2 - (\sin \frac{x}{2})^2$. Since $(\cos \frac{x}{2})^2 + (\sin \frac{x}{2})^2 = 1$, this gives $\cos x = (\cos \frac{x}{2})^2 + (\cos \frac{x}{2})^2 -1$, which is your formula above.

The other follows a similar approach, except you replace the $(\cos \frac{x}{2})^2$ term instead of the $(\sin \frac{x}{2})^2$ term.

Here is the second part explicitly:

We already have $\cos x = (\cos \frac{x}{2})^2 - (\sin \frac{x}{2})^2$. Since $(\cos \frac{x}{2})^2 + (\sin \frac{x}{2})^2 = 1$, this gives $(\cos \frac{x}{2})^2 = 1-(\sin \frac{x}{2})^2$. Substituting gives $\cos x = 1-(\sin \frac{x}{2})^2 - (\sin \frac{x}{2})^2 = 1-2 (\sin \frac{x}{2})^2$.

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For my second one I have sin(x/2)cos(x/2)+cos(x/2)sin(x/2) but how would I proceed do I need to add them or something? –  Fernando Martinez Jul 28 '12 at 16:18
    
Just replace $(\cos \frac{x}{2})^2$ by $1-(\sin \frac{x}{2})^2$. –  copper.hat Jul 28 '12 at 16:19
    
Your question doesn't have a $\sin x$ expansion? –  copper.hat Jul 28 '12 at 16:21
    
Now I have sin(x/2)(1-sin(x/2)+(1-sinx/2)sin(x/2) –  Fernando Martinez Jul 28 '12 at 16:24
    
Still I am unsure what to.... –  Fernando Martinez Jul 28 '12 at 16:26
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Paint is here to help you! Just stare at the image below long enough and realize everything is ok. After, use Pythagoras' theorem:

$$ \begin{aligned} (1 + \cos(x))^2 + \sin^2(x) \\ = (2\cos(x/2))^2 \\ = 1 + 2\cos(x) + (\cos^2(x) + \sin^2(x))\\ = 2 + 2\cos(x) \\ = 4\cos^2(x/2). \end{aligned}$$

Now, you can write

$$1+\cos(x) = 2\cos^2(x/2)\\\\ \Longrightarrow \cos(x) = 2\cos^2(x/2) - 1.$$

Paint

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