Prove that: \[ 4(\cos^3 10^\circ + \sin^3 20^\circ) = 3(\cos 10^\circ + \sin 20^\circ) \]
Solution
We use the identity
\[
4a^3 – 3a = \cos 3\theta
\quad \text{when } a=\cos\theta
\]
So,
\[
4\cos^3 10^\circ
=
3\cos 10^\circ + \cos 30^\circ
\]
Also, using
\[
4\sin^3\theta
=
3\sin\theta – \sin 3\theta
\]
we get
\[
4\sin^3 20^\circ
=
3\sin 20^\circ – \sin 60^\circ
\]
Adding both equations,
\[
4(\cos^3 10^\circ + \sin^3 20^\circ)
=
3\cos 10^\circ + \cos 30^\circ
+3\sin 20^\circ – \sin 60^\circ
\]
Since
\[
\cos 30^\circ = \sin 60^\circ
\]
their terms cancel out:
\[
4(\cos^3 10^\circ + \sin^3 20^\circ)
=
3\cos 10^\circ + 3\sin 20^\circ
\]
\[
=
3(\cos 10^\circ + \sin 20^\circ)
\]
Hence proved.