Prove that: \[ \left| \cos x\cos\left(\frac{\pi}{3}-x\right) \cos\left(\frac{\pi}{3}+x\right) \right| \le \frac14 \] for all values of \(x\).
Solution
Consider
\[
\cos\left(\frac{\pi}{3}-x\right)
\cos\left(\frac{\pi}{3}+x\right)
\]
Using the identity
\[
\cos(A-B)\cos(A+B)
=
\cos^2A-\sin^2B
\]
with
\[
A=\frac{\pi}{3}, \qquad B=x
\]
we get
\[
\cos\left(\frac{\pi}{3}-x\right)
\cos\left(\frac{\pi}{3}+x\right)
=
\cos^2\frac{\pi}{3}-\sin^2x
\]
\[
=
\frac14-\sin^2x
\]
Using
\[
\sin^2x=1-\cos^2x
\]
therefore,
\[
\frac14-\sin^2x
=
\frac14-(1-\cos^2x)
\]
\[
=
\cos^2x-\frac34
\]
Hence,
\[
\cos x
\cos\left(\frac{\pi}{3}-x\right)
\cos\left(\frac{\pi}{3}+x\right)
=
\cos x\left(\cos^2x-\frac34\right)
\]
\[
=
\cos^3x-\frac34\cos x
\]
Using the identity
\[
\cos3x
=
4\cos^3x-3\cos x
\]
Divide both sides by \(4\):
\[
\frac14\cos3x
=
\cos^3x-\frac34\cos x
\]
Thus,
\[
\cos x
\cos\left(\frac{\pi}{3}-x\right)
\cos\left(\frac{\pi}{3}+x\right)
=
\frac14\cos3x
\]
Taking modulus on both sides,
\[
\left|
\cos x
\cos\left(\frac{\pi}{3}-x\right)
\cos\left(\frac{\pi}{3}+x\right)
\right|
=
\frac14|\cos3x|
\]
Since
\[
|\cos3x|\le1
\]
therefore,
\[
\left|
\cos x
\cos\left(\frac{\pi}{3}-x\right)
\cos\left(\frac{\pi}{3}+x\right)
\right|
\le\frac14
\]
Hence proved.