Prove that: \[ \cos 3A + \cos 5A + \cos 7A + \cos 15A = 4\cos 4A \cos 5A \cos 6A \]
Solution
Group the terms:
\[
(\cos 3A + \cos 15A)
+
(\cos 5A + \cos 7A)
\]
Using the identity:
\[
\cos A + \cos B
=
2\cos\frac{A+B}{2}\cos\frac{A-B}{2}
\]
For
\(\cos 3A + \cos 15A\):
\[
\cos 3A + \cos 15A
=
2\cos\frac{3A+15A}{2}
\cos\frac{3A-15A}{2}
\]
\[
=
2\cos 9A \cos(-6A)
\]
Since,
\[
\cos(-\theta)=\cos\theta
\]
\[
=
2\cos 9A \cos 6A
\]
For
\(\cos 5A + \cos 7A\):
\[
\cos 5A + \cos 7A
=
2\cos\frac{5A+7A}{2}
\cos\frac{5A-7A}{2}
\]
\[
=
2\cos 6A \cos(-A)
\]
\[
=
2\cos 6A \cos A
\]
Adding both results:
\[
=
2\cos 9A \cos 6A
+
2\cos 6A \cos A
\]
\[
=
2\cos 6A(\cos 9A+\cos A)
\]
Again using:
\[
\cos A + \cos B
=
2\cos\frac{A+B}{2}\cos\frac{A-B}{2}
\]
\[
\cos 9A+\cos A
=
2\cos\frac{9A+A}{2}
\cos\frac{9A-A}{2}
\]
\[
=
2\cos 5A \cos 4A
\]
Therefore,
\[
=
2\cos 6A \times 2\cos 5A \cos 4A
\]
\[
=
4\cos 4A \cos 5A \cos 6A
\]
Hence,
\[
\boxed{
\cos 3A + \cos 5A + \cos 7A + \cos 15A
=
4\cos 4A \cos 5A \cos 6A
}
\]