# Prove the following…

$$\cos{3x}=2\cos{2x}\cos{x}-\cos{x}$$ I have no idea where to start or what to do... I've tried adding and subtracting $\cos{x}$ from $\cos{3x}$, like this:$$\cos{3x}+\cos{x}-\cos{x}$$so I get that $$\cos{3x}+\cos{x}=2\cos{2x}\cos{x}$$ But I have no idea how these equal.

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Do you know the angle addition formulas? If so, hint: $\cos(3x) = \cos(2x+x)$... –  Alex Wertheim Jun 17 '14 at 16:12

Hints:

$$\cos(x+y)=\cos x\cos y-\sin x\sin y$$

$$\sin2x=2\sin x\cos x$$

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

With the above you're done.

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Use Prosthaphaeresis Formula, $\displaystyle\cos C+\cos D=\cdots$ on $\cos3x+cosx$

Alternatively, $$\cos3x+cosx=\cos(2x+x)+\cos(2x-x)=?$$

Alternatively use Werner Formula $$2\cos2x\cos x=\cos(2x+x)+cos(2x-x)$$

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Hint

$$\cos(3x)=\cos(2x+x)=\cos(2x)\cos(x)-\sin(2x)\sin(x)$$

and use the formula for $\sin(2x)$.

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