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