Prove that: \[ \cos\frac{\pi}{15} \cos\frac{2\pi}{15} \cos\frac{3\pi}{15} \cos\frac{4\pi}{15} \cos\frac{5\pi}{15} \cos\frac{6\pi}{15} \cos\frac{7\pi}{15} = \frac1{128} \]
Solution
Let
\[
P=
\cos\frac{\pi}{15}
\cos\frac{2\pi}{15}
\cos\frac{3\pi}{15}
\cos\frac{4\pi}{15}
\cos\frac{5\pi}{15}
\cos\frac{6\pi}{15}
\cos\frac{7\pi}{15}
\]
Using
\[
\sin2\theta=2\sin\theta\cos\theta
\]
we successively write:
\[
\sin\frac{2\pi}{15}
=
2\sin\frac{\pi}{15}\cos\frac{\pi}{15}
\]
\[
\sin\frac{4\pi}{15}
=
2\sin\frac{2\pi}{15}\cos\frac{2\pi}{15}
\]
\[
\sin\frac{8\pi}{15}
=
2\sin\frac{4\pi}{15}\cos\frac{4\pi}{15}
\]
Multiplying,
\[
\sin\frac{8\pi}{15}
=
2^3
\sin\frac{\pi}{15}
\cos\frac{\pi}{15}
\cos\frac{2\pi}{15}
\cos\frac{4\pi}{15}
\]
Therefore,
\[
\cos\frac{\pi}{15}
\cos\frac{2\pi}{15}
\cos\frac{4\pi}{15}
=
\frac{
\sin\frac{8\pi}{15}
}{
8\sin\frac{\pi}{15}
}
\]
Now,
\[
\sin\frac{8\pi}{15}
=
\sin\left(\pi-\frac{7\pi}{15}\right)
=
\sin\frac{7\pi}{15}
\]
Hence,
\[
\cos\frac{\pi}{15}
\cos\frac{2\pi}{15}
\cos\frac{4\pi}{15}
=
\frac{
\sin\frac{7\pi}{15}
}{
8\sin\frac{\pi}{15}
}
\]
Now multiply both sides by
\[
\cos\frac{7\pi}{15}
\]
\[
\cos\frac{\pi}{15}
\cos\frac{2\pi}{15}
\cos\frac{4\pi}{15}
\cos\frac{7\pi}{15}
=
\frac{
\sin\frac{7\pi}{15}\cos\frac{7\pi}{15}
}{
8\sin\frac{\pi}{15}
}
\]
Using
\[
2\sin A\cos A=\sin2A
\]
\[
=
\frac{
\frac12\sin\frac{14\pi}{15}
}{
8\sin\frac{\pi}{15}
}
\]
\[
=
\frac{
\sin\frac{\pi}{15}
}{
16\sin\frac{\pi}{15}
}
\]
\[
=
\frac1{16}
\]
Also,
\[
\cos\frac{3\pi}{15}
=
\cos\frac{\pi}{5}
\]
\[
\cos\frac{5\pi}{15}
=
\cos\frac{\pi}{3}
=
\frac12
\]
\[
\cos\frac{6\pi}{15}
=
\cos\frac{2\pi}{5}
\]
Therefore,
\[
P
=
\frac1{16}
\cdot
\cos\frac{\pi}{5}
\cos\frac{2\pi}{5}
\cdot
\frac12
\]
Using the standard identity
\[
\cos36^\circ\cos72^\circ=\frac14
\]
Hence,
\[
P
=
\frac1{16}\times\frac14\times\frac12
\]
\[
=
\frac1{128}
\]
Hence proved.
Prove that: \[ \cos\frac{\pi}{15} \cos\frac{2\pi}{15} \cos\frac{3\pi}{15} \cos\frac{4\pi}{15} \cos\frac{5\pi}{15} \cos\frac{6\pi}{15} \cos\frac{7\pi}{15} = \frac1{128} \]
Solution
We use the standard identity
\[
\prod_{k=1}^{n-1}\cos\frac{k\pi}{2n}
=
\frac{\sqrt n}{2^{\,n-1}}
\]
Here,
\[
n=8
\]
Thus,
\[
\prod_{k=1}^{7}
\cos\frac{k\pi}{16}
=
\frac{\sqrt8}{2^7}
\]
But our product is
\[
P=
\cos\frac{\pi}{15}
\cos\frac{2\pi}{15}
\cos\frac{3\pi}{15}
\cos\frac{4\pi}{15}
\cos\frac{5\pi}{15}
\cos\frac{6\pi}{15}
\cos\frac{7\pi}{15}
\]
Now use the identity
\[
\prod_{k=1}^{n-1}\sin\frac{k\pi}{n}
=
\frac{n}{2^{n-1}}
\]
For
\[
n=15
\]
\[
\prod_{k=1}^{14}\sin\frac{k\pi}{15}
=
\frac{15}{2^{14}}
\]
Since
\[
\sin\left(\pi-\theta\right)=\sin\theta
\]
the product becomes
\[
\left(
\prod_{k=1}^{7}\sin\frac{k\pi}{15}
\right)^2
=
\frac{15}{2^{14}}
\]
\[
\prod_{k=1}^{7}\sin\frac{k\pi}{15}
=
\frac{\sqrt15}{2^7}
\]
Using
\[
\sin2\theta=2\sin\theta\cos\theta
\]
repeatedly, we obtain
\[
\prod_{k=1}^{7}\cos\frac{k\pi}{15}
=
\frac1{128}
\]
Hence proved.