Is $\cos(x) \times \cos(2x)$ the same as $\cos(3x)$
By Sophia Bowman
When you multiply $\cos(x) \times \cos(2x)$ the same as $\cos(3x)$ or do you have to treat each differently?
$\endgroup$ 44 Answers
$\begingroup$The sum formula for $\cos$ is $\cos(A+B)=\cos(A)\cos(B)-\sin(A)\sin(B)$. Therefore, $$\cos(3x)=\cos(x+2x)=\cos(x)\cos(2x)-\sin(x)\sin(2x)$$
$\endgroup$ $\begingroup$Hint
You have $$\cos(a+b)=\cos(a)\cos(b)-\sin(a)\sin(b)$$ $$\cos(a-b)=\cos(a)\cos(b)+\sin(a)\sin(b)$$ Adding both $$\cos(a+b)+\cos(a-b)=2\cos(a)\cos(b)$$ or $$\cos(a)\cos(b)=\frac 12 \big(\cos(a+b)+\cos(a-b)\big)$$ Applied to the case you posted,$$\cos(x)\cos(2x)=\frac 12 \big(\cos(3x)+\cos(x)\big)$$
$\endgroup$ 1 $\begingroup$If $\cos3x=\cos2x\cdot\cos x$
$4\cos^3x-3\cos x=(2\cos^2x-1)\cos x$
$\iff\cos x[4\cos^2x-3-(2\cos^2x-1)]=0$
$\iff\cos x[\sin^2x]=0\iff\cos x=0$ or $\sin x=0\implies\sin2x=0$
$\implies x$ has to be a multiple of $\dfrac\pi2$
So, $\cos3x=\cos2x\cdot\cos x$ is an equation, not an identity
$\endgroup$ $\begingroup$Nope. It doesn't check out. Here is a diagram which will explain the so-called angle-sum formula for cosine. If the two angles are the same, this is an angle doubling, yielding another nice formula. With the angle double formula, you can use the angle sum again. If $\theta$ is the angle you want to triple, then use $\theta$ and $2\theta$ as your summands. This will give an angle tripling. There is a similar formula for sine as well. Can you find it?
(Yes, I'm aware the proof for sine is above, but try to work it through yourself! This is always the most instructive way to learn new mathematics).
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