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The integral $\int_0^{\pi/8}\cos(5x)\cos(3x)\cos(4x) \,dx$ is equal to $k/24$. Find the constant $k$.

So far, I assume that the best way to solve this question is to solve the integral and compare the answer to find $k$.

I have thought about rewriting the integral by using the identity $\cos(2x) = \cos^2(x) - \sin^2(x)$ or $\cos(2x) = 1 - 2\sin^2(x)$, but that results in a very long and confusing function. I don't see any possible substitutions.

Please help!

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I suspect $$\cos 5x\cos 3x = \frac{1}{2} \left(\cos 8x + \cos 2x\right)$$ will make it bearable. May be worth a try. – Daniel Fischer Feb 23 '14 at 21:20

Adding together the following trig identities: $$\cos(A+ B) = \cos A \cos B - \sin A \sin B$$ $$\cos(A-B) = \cos A \cos B + \sin A \sin B$$ gives $$2\cos A \cos B = \cos(A+B) + \cos (A-B)$$ and hence $$\cos A \cos B = \dfrac{\cos(A+B) + \cos(A-B)}{2}$$ You can apply this identity to reduce your integral to a sum of cosines, which can be easily integrated.

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$$\cos 5x\cdot\cos 3x\cdot\cos 4x \\=\cos 5x\cdot\frac{(\cos 7x+\cos x)}{2} \\=\frac{\cos 5x\cdot\cos 7x+\cos 5x\cdot\cos x}{2} \\=\frac{\cos 13x+\cos 2x+\cos 6x+\cos 4x}{4}$$

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Sorry, how do you go from cos3xcos4x to (cos7x+cosx)/2 ? – Chrysanthemum Feb 23 '14 at 21:23
since $\cos x\cos y=0.5(\cos(x+y)+\cos(x-y))$ – Sameh Shenawy Feb 23 '14 at 21:24


Use the formula for $\cos(5x+3x)$ and $\cos(5x-3x)$ to show that

$$\cos 5x \cos 3x \equiv \tfrac{1}{2}(\cos 8x + \cos 2x)$$

Substitute this into $\cos 5x \cos 3x \cos 4x$, expand and then apply the same trick to the expansion.

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