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