Prove that: \[ \sin^2 24^\circ-\sin^2 6^\circ = \frac{\sqrt5-1}{8} \]
Solution
Using the identity
\[
\sin^2A-\sin^2B
=
\sin(A+B)\sin(A-B)
\]
Let
\[
A=24^\circ,\qquad B=6^\circ
\]
Then
\[
A+B=30^\circ
\]
\[
A-B=18^\circ
\]
Therefore,
\[
\sin^2 24^\circ-\sin^2 6^\circ
=
\sin30^\circ\sin18^\circ
\]
\[
=
\frac12\sin18^\circ
\]
Now use the standard value
\[
\sin18^\circ
=
\frac{\sqrt5-1}{4}
\]
Hence,
\[
\sin^2 24^\circ-\sin^2 6^\circ
=
\frac12\cdot\frac{\sqrt5-1}{4}
\]
\[
=
\frac{\sqrt5-1}{8}
\]
Hence proved.