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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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    I have never seen the second one marlu commented but I know cos(a+b) is cosAcosB-cosAsinB.2012-07-28

2 Answers 2

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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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    I understand. Do them one at a time. I make mistakes all the time; I often use a symbolic manipulation system to check my calculations (wxmaxima).2012-07-28
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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