Prove that: \[ \sin 51^\circ + \cos 81^\circ = \cos 21^\circ \]
Solution
Convert cosine into sine form using:
\[
\cos \theta = \sin(90^\circ-\theta)
\]
\[
\cos 81^\circ
=
\sin 9^\circ
\]
Therefore,
\[
\sin 51^\circ + \cos 81^\circ
=
\sin 51^\circ + \sin 9^\circ
\]
Using the identity:
\[
\sin A + \sin B
=
2\sin\frac{A+B}{2}\cos\frac{A-B}{2}
\]
Taking
\[
A=51^\circ,\qquad B=9^\circ
\]
Then,
\[
=
2\sin\frac{51^\circ+9^\circ}{2}
\cos\frac{51^\circ-9^\circ}{2}
\]
\[
=
2\sin30^\circ\cos21^\circ
\]
\[
=
2\times\frac{1}{2}\times\cos21^\circ
\]
\[
=
\cos21^\circ
\]
Hence,
\[
\boxed{
\sin 51^\circ + \cos81^\circ
=
\cos21^\circ
}
\]