Prove that: \[ \cos x \cos\frac{x}{2} – \cos 3x \cos\frac{9x}{2} = \sin 7x \sin 8x \]
Solution
Using the identity:
\[
\cos A \cos B
=
\frac{1}{2}
\left[
\cos(A+B)+\cos(A-B)
\right]
\]
First term:
\[
\cos x \cos\frac{x}{2}
=
\frac{1}{2}
\left[
\cos\left(x+\frac{x}{2}\right)
+
\cos\left(x-\frac{x}{2}\right)
\right]
\]
\[
=
\frac{1}{2}
\left[
\cos\frac{3x}{2}
+
\cos\frac{x}{2}
\right]
\]
Second term:
\[
\cos 3x \cos\frac{9x}{2}
=
\frac{1}{2}
\left[
\cos\left(3x+\frac{9x}{2}\right)
+
\cos\left(3x-\frac{9x}{2}\right)
\right]
\]
\[
=
\frac{1}{2}
\left[
\cos\frac{15x}{2}
+
\cos\left(-\frac{3x}{2}\right)
\right]
\]
\[
=
\frac{1}{2}
\left[
\cos\frac{15x}{2}
+
\cos\frac{3x}{2}
\right]
\]
Subtracting:
\[
\cos x \cos\frac{x}{2}
–
\cos 3x \cos\frac{9x}{2}
\]
\[
=
\frac{1}{2}
\left(
\cos\frac{3x}{2}
+
\cos\frac{x}{2}
\right)
–
\frac{1}{2}
\left(
\cos\frac{15x}{2}
+
\cos\frac{3x}{2}
\right)
\]
\[
=
\frac{1}{2}
\left(
\cos\frac{x}{2}
–
\cos\frac{15x}{2}
\right)
\]
Using:
\[
\cos A-\cos B
=
-2\sin\frac{A+B}{2}\sin\frac{A-B}{2}
\]
\[
=
\frac{1}{2}
\left[
-2
\sin\frac{\frac{x}{2}+\frac{15x}{2}}{2}
\sin\frac{\frac{x}{2}-\frac{15x}{2}}{2}
\right]
\]
\[
=
-\sin4x\sin(-7x)
\]
Since,
\[
\sin(-\theta)=-\sin\theta
\]
\[
=
\sin4x\sin7x
\]
\[
=
\sin7x\sin4x
\]
Hence,
\[
\boxed{
\cos x \cos\frac{x}{2}
–
\cos 3x \cos\frac{9x}{2}
=
\sin7x\sin4x
}
\]