Prove that: \[ \sin 47^\circ + \cos 77^\circ = \cos 17^\circ \]
Solution
Convert cosine into sine form using:
\[
\cos \theta = \sin(90^\circ-\theta)
\]
\[
\cos 77^\circ
=
\sin 13^\circ
\]
Therefore,
\[
\sin 47^\circ + \cos 77^\circ
=
\sin 47^\circ + \sin 13^\circ
\]
Using the identity:
\[
\sin A + \sin B
=
2\sin\frac{A+B}{2}\cos\frac{A-B}{2}
\]
Taking
\[
A=47^\circ,\qquad B=13^\circ
\]
Then,
\[
=
2\sin\frac{47^\circ+13^\circ}{2}
\cos\frac{47^\circ-13^\circ}{2}
\]
\[
=
2\sin30^\circ\cos17^\circ
\]
\[
=
2\times\frac{1}{2}\times\cos17^\circ
\]
\[
=
\cos17^\circ
\]
Hence,
\[
\boxed{
\sin47^\circ + \cos77^\circ
=
\cos17^\circ
}
\]