Prove that: \[ \left| \sin x\sin\left(\frac{\pi}{3}-x\right) \sin\left(\frac{\pi}{3}+x\right) \right| \le \frac14 \] for all values of \(x\).
Solution
Consider
\[
\sin\left(\frac{\pi}{3}-x\right)
\sin\left(\frac{\pi}{3}+x\right)
\]
Using the identity
\[
\sin(A-B)\sin(A+B)
=
\sin^2A-\sin^2B
\]
with
\[
A=\frac{\pi}{3}, \qquad B=x
\]
we get
\[
\sin\left(\frac{\pi}{3}-x\right)
\sin\left(\frac{\pi}{3}+x\right)
=
\sin^2\frac{\pi}{3}-\sin^2x
\]
\[
=
\frac34-\sin^2x
\]
Therefore,
\[
\sin x
\sin\left(\frac{\pi}{3}-x\right)
\sin\left(\frac{\pi}{3}+x\right)
=
\sin x\left(\frac34-\sin^2x\right)
\]
\[
=
\frac34\sin x-\sin^3x
\]
Using the identity
\[
\sin3x
=
3\sin x-4\sin^3x
\]
Divide both sides by \(4\):
\[
\frac14\sin3x
=
\frac34\sin x-\sin^3x
\]
Hence,
\[
\sin x
\sin\left(\frac{\pi}{3}-x\right)
\sin\left(\frac{\pi}{3}+x\right)
=
\frac14\sin3x
\]
Taking modulus on both sides,
\[
\left|
\sin x
\sin\left(\frac{\pi}{3}-x\right)
\sin\left(\frac{\pi}{3}+x\right)
\right|
=
\frac14|\sin3x|
\]
Since
\[
|\sin3x|\le1
\]
therefore,
\[
\left|
\sin x
\sin\left(\frac{\pi}{3}-x\right)
\sin\left(\frac{\pi}{3}+x\right)
\right|
\le\frac14
\]
Hence proved.