<s>[INST] Is there any way to simplify the product of cosines? I recently saw a problem: Estimate the following: $cos(\frac{\pi}{15})cos(\frac{2\pi}{15})\ldots cos(\frac{7\pi}{15})$, and the options were between different consecutive power of ten.
How would I do this, no calculator of course, and is there's any way to shorten any general products of sines or cosines or other trig functions?
 [/INST] We have $\cos(\frac{\pi}{15})\cos(\frac{2\pi}{15})\ldots \cos(\frac{7\pi}{15})$. Let's write this as $p=\cos(\frac{\pi}{15})\cos(\frac{2\pi}{15})\ldots \cos(\frac{7\pi}{15})$
We have to multiply both sides of the equation with
$$q= \sin(\frac{\pi}{15})\sin(\frac{2\pi}{15})\ldots \sin(\frac{7\pi}{15})$$
Now, $$p.q = \sin(\frac{\pi}{15})\sin(\frac{2\pi}{15})\ldots \sin(\frac{7\pi}{15})\cos(\frac{\pi}{15})\cos(\frac{2\pi}{15})\ldots \cos(\frac{7\pi}{15})$$
Multiply both sides with $2^7$
$$2^7 p.q =[2 \sin(\frac{\pi}{15})\cos(\frac{\pi}{15})][2\sin(\frac{2\pi}{15})\cos(\frac{2\pi}{15})]\ldots [2\sin(\frac{7\pi}{15})\cos(\frac{7\pi}{15})]$$
$$2^7 p.q=\sin(\frac{2\pi}{15})\sin(\frac{4\pi}{15})\ldots \sin(\frac{14\pi}{15})$$
Now we have to apply the identity $\sin\theta=\sin(\pi-\theta)$
$$2^7 p.q=\sin(\frac{\pi}{15})\sin(\frac{2\pi}{15})\ldots \sin(\frac{7\pi}{15})$$
Which means $$2^7 p.q=q$$
Therefore,
$$p=\cos(\frac{\pi}{15})\cos(\frac{2\pi}{15})\ldots \cos(\frac{7\pi}{15})=\frac1{2^7}$$
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