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I want to write $\cos(3x)\cos^2(x)$ as $\cos(3x)\cos^2(x)=\Sigma a_n\sin(n(x+\frac{\pi}{2}))$ where $n\in \mathbb{N}$.

I've tried to use some trigonometric identities such as $\cos(3x)=\cos(x)(2\cos(2x)-1)$ in order to get $\cos(3x)\cos^2(x)=\cos^3(x)(2\cos(2x)-1)=\frac{1}{4}(3\cos(x)+3\cos(3x))(2\cos(2x)-1)$

However, I'm stuck because here. (My idea is to get all in terms of cosine and then use $\cos(x)=\sin(x+\frac{\pi}{2})$).

I don't know how to continue (maybe my approach is wrong), I've tried other aproaches such as $\cos(3x)\cos^2(x)=\cos(3x)(1-\sin^2(x))$ without getting any result.

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  • $\begingroup$ I would use $\cos x=\frac12(e^{ix}+e^{-ix})$. $\endgroup$ Dec 16 '17 at 7:38
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$\cos(3x)\cos^2(x) = \cos(3x)(\frac{1}{2}\cos(2x)+\frac{1}{2}) = \frac{1}{2}\cos(3x)\cos(2x)+\frac{1}{2}\cos(3x)$

But $\cos(3x)\cos(2x) = \frac{1}{2}(\cos(x)+\cos(5x))$ and you are basically done.

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Note that $$\cos^2 x = \frac{1+\cos 2x}{2}$$ giving us, $$\cos 3x\cos^2 x = \frac{1}{2}[\cos 3x + \cos 2x \cos 3x]$$ And as $\cos 2x\cos 3x = \frac{1}{2}[\cos x + \cos 5x]$, we get, $$\cos 3x\cos^2 x = \frac{1}{4}[\cos x +2\cos 3x + \cos 5x]$$

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You can use the formula: $\cos x\cos y=\frac12(\cos (x+y)+\cos(x-y))$. So: $$\cos 3x\cos^2 x=(\cos3x\cos x)\cos x=\frac12(\cos 4x+\cos 2x)\cos x=$$ $$\frac12\cos 4x \cos x+\frac12\cos2x \cos x=\frac14(\cos 5x+\cos3x)+\frac14(\cos 3x+\cos x)=$$ $$\frac14(\cos 5x+2\cos 3x+\cos x)=\frac14\left(\sin(5x+\frac{\pi}{2})+2\sin (3x+\frac{\pi}{2})+\sin(x+\frac{\pi}{2})\right).$$

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Hint:

Observe that $$\sin(2m+1)\left(\dfrac\pi2+x\right)=\sin\left(m\pi+\dfrac\pi2+mx\right)=(-1)^m\sin\left(\dfrac\pi2+mx\right)=(-1)^m\cos(mx)$$

Set $m=0,1$

Use $\sin3y=3\sin y -4\sin^3y$

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