Prove that: \[ \sin^2 42^\circ-\cos^2 78^\circ = \frac{\sqrt5+1}{8} \]
Solution
Using
\[
\cos(90^\circ-\theta)=\sin\theta
\]
we get
\[
\cos78^\circ=\sin12^\circ
\]
Therefore,
\[
\sin^2 42^\circ-\cos^2 78^\circ
=
\sin^2 42^\circ-\sin^2 12^\circ
\]
Using the identity
\[
\sin^2A-\sin^2B
=
\sin(A+B)\sin(A-B)
\]
Let
\[
A=42^\circ,\qquad B=12^\circ
\]
Then
\[
A+B=54^\circ
\]
\[
A-B=30^\circ
\]
Hence,
\[
\sin^2 42^\circ-\sin^2 12^\circ
=
\sin54^\circ\sin30^\circ
\]
\[
=
\sin54^\circ\cdot\frac12
\]
Now use the standard value
\[
\sin54^\circ
=
\cos36^\circ
=
\frac{\sqrt5+1}{4}
\]
Therefore,
\[
\sin^2 42^\circ-\cos^2 78^\circ
=
\frac12\cdot\frac{\sqrt5+1}{4}
\]
\[
=
\frac{\sqrt5+1}{8}
\]
Hence proved.