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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?

D.R.
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3 Answers3

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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, $$p=\cos(\frac{\pi}{15})\cos(\frac{2\pi}{15})\ldots \cos(\frac{7\pi}{15})=\frac1{2^7}$$

Ananth Kamath
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$$\prod_{k=1}^7\cos\frac{k\pi}{15}= \cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\left(-\cos\frac{8\pi}{15}\right)\cos\frac{\pi}{5}\cos\frac{\pi}{3}\cos\frac{2\pi}{5}=$$ $$=-\frac{16\sin\frac{\pi}{15}\cos\frac{\pi}{15}\cos\frac{2\pi}{15}\cos\frac{4\pi}{15}\cos\frac{8\pi}{15}}{16\sin\frac{\pi}{15}}\cdot\frac{4\sin\frac{\pi}{5}\cos\frac{\pi}{5}\cos\frac{2\pi}{5}}{8\sin\frac{\pi}{5}}=$$ $$=-\frac{\sin\frac{16\pi}{15}}{16\sin\frac{\pi}{15}}\cdot\frac{\sin\frac{4\pi}{5}}{8\sin\frac{\pi}{5}}=\frac{1}{128}.$$

Michael Rozenberg
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Denoting $\zeta_{30} = e^{2\pi i/30}$ and $\zeta_{15} = e^{2\pi i/15}$ we see that your product is equal to $$\alpha = \frac{(\zeta_{30} + \zeta_{30}^{-1})(\zeta_{30}^2 + \zeta_{30}^{-2})...(\zeta_{30}^7 + \zeta_{30}^{-7})}{2^7} = -\frac{(1+\zeta_{15})(1+\zeta_{15}^2)...(1+\zeta_{15}^7)}{2^7\zeta_{15}}$$

We observe that since $\zeta_{15}^{15-n}=\left(\zeta_{15}^{n}\right)^*$ and $\zeta_{15}\zeta_{15}^* = 1$, we have $$\alpha \alpha^* = \frac{(1 + \zeta_{15})(1+\zeta_{15}^2)...(1+\zeta_{15}^{14})}{2^{14}}=\frac{(1+1)(1 + \zeta_{15})(1+\zeta_{15}^2)...(1+\zeta_{15}^{14})}{2^{15}}$$

The numerator is $f(1)$ where $f(X) = X^{15} + 1$, and is thus equal to $2$.

Since we also know $\alpha \in \mathbb{R}^+$ this tells us $\alpha^2 = \frac{1}{2^{14}}$ and so $\alpha = \frac{1}{2^7}$.

Tob Ernack
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