Prove that: \[ \sin 80^\circ – \cos 70^\circ = \cos 50^\circ \]
Solution
Convert cosine into sine form using:
\[
\cos \theta = \sin(90^\circ-\theta)
\]
\[
\cos 70^\circ
=
\sin 20^\circ
\]
Therefore,
\[
\sin 80^\circ – \cos 70^\circ
=
\sin 80^\circ – \sin 20^\circ
\]
Using the identity:
\[
\sin A – \sin B
=
2\cos\frac{A+B}{2}\sin\frac{A-B}{2}
\]
Taking
\[
A=80^\circ,\qquad B=20^\circ
\]
Then,
\[
=
2\cos\frac{80^\circ+20^\circ}{2}
\sin\frac{80^\circ-20^\circ}{2}
\]
\[
=
2\cos50^\circ\sin30^\circ
\]
\[
=
2\cos50^\circ\times\frac{1}{2}
\]
\[
=
\cos50^\circ
\]
Hence,
\[
\boxed{
\sin 80^\circ – \cos70^\circ
=
\cos50^\circ
}
\]