Show that \( \sin 50^\circ \cos 85^\circ = \dfrac{1-\sqrt{2}\sin35^\circ}{2\sqrt{2}} \)
Solution
Using the identity:
\[
2\sin A\cos B=\sin(A+B)+\sin(A-B)
\]
\[
2\sin50^\circ\cos85^\circ
\]
\[
= \sin(50^\circ+85^\circ)+\sin(50^\circ-85^\circ)
\]
\[
= \sin135^\circ+\sin(-35^\circ)
\]
\[
= \frac{1}{\sqrt2}-\sin35^\circ
\]
Dividing both sides by \(2\),
\[
\sin50^\circ\cos85^\circ
=
\frac{1}{2}\left(\frac{1}{\sqrt2}-\sin35^\circ\right)
\]
\[
=
\frac{1-\sqrt2\sin35^\circ}{2\sqrt2}
\]
Hence Proved
\[
\sin50^\circ\cos85^\circ
=
\frac{1-\sqrt2\sin35^\circ}{2\sqrt2}
\]