Prove that: \[ \cos\frac{\pi}{15} \cos\frac{2\pi}{15} \cos\frac{4\pi}{15} \cos\frac{7\pi}{15} = \frac1{16} \]
Solution
Using the identity
\[
2\sin A\cos A=\sin2A
\]
we successively write:
\[
\sin\frac{2\pi}{15}
=
2\sin\frac{\pi}{15}\cos\frac{\pi}{15}
\]
\[
\sin\frac{4\pi}{15}
=
2\sin\frac{2\pi}{15}\cos\frac{2\pi}{15}
\]
\[
\sin\frac{8\pi}{15}
=
2\sin\frac{4\pi}{15}\cos\frac{4\pi}{15}
\]
\[
\sin\frac{16\pi}{15}
=
2\sin\frac{8\pi}{15}\cos\frac{8\pi}{15}
\]
Multiply all these equations:
\[
\sin\frac{2\pi}{15}
\sin\frac{4\pi}{15}
\sin\frac{8\pi}{15}
\sin\frac{16\pi}{15}
\]
\[
=
2^4
\sin\frac{\pi}{15}
\sin\frac{2\pi}{15}
\sin\frac{4\pi}{15}
\sin\frac{8\pi}{15}
\]
\[
\qquad\times
\cos\frac{\pi}{15}
\cos\frac{2\pi}{15}
\cos\frac{4\pi}{15}
\cos\frac{8\pi}{15}
\]
Cancel the common sine terms:
\[
\sin\frac{16\pi}{15}
=
16\sin\frac{\pi}{15}
\cos\frac{\pi}{15}
\cos\frac{2\pi}{15}
\cos\frac{4\pi}{15}
\cos\frac{8\pi}{15}
\]
Now,
\[
\sin\frac{16\pi}{15}
=
-\sin\frac{\pi}{15}
\]
Hence,
\[
-\sin\frac{\pi}{15}
=
16\sin\frac{\pi}{15}
\cos\frac{\pi}{15}
\cos\frac{2\pi}{15}
\cos\frac{4\pi}{15}
\cos\frac{8\pi}{15}
\]
Cancel
\[
\sin\frac{\pi}{15}
\]
\[
-1
=
16
\cos\frac{\pi}{15}
\cos\frac{2\pi}{15}
\cos\frac{4\pi}{15}
\cos\frac{8\pi}{15}
\]
Using
\[
\cos\frac{8\pi}{15}
=
-\cos\frac{7\pi}{15}
\]
therefore,
\[
1
=
16
\cos\frac{\pi}{15}
\cos\frac{2\pi}{15}
\cos\frac{4\pi}{15}
\cos\frac{7\pi}{15}
\]
\[
\therefore
\quad
\cos\frac{\pi}{15}
\cos\frac{2\pi}{15}
\cos\frac{4\pi}{15}
\cos\frac{7\pi}{15}
=
\frac1{16}
\]
Hence proved.