Show that \( \sin25^\circ\cos115^\circ=\frac{1}{2}(\sin140^\circ-1) \)
Solution
Using the identity:
\[
2\sin A\cos B=\sin(A+B)+\sin(A-B)
\]
\[
2\sin25^\circ\cos115^\circ
\]
\[
= \sin(25^\circ+115^\circ)+\sin(25^\circ-115^\circ)
\]
\[
= \sin140^\circ+\sin(-90^\circ)
\]
\[
= \sin140^\circ-1
\]
Dividing both sides by \(2\),
\[
\sin25^\circ\cos115^\circ
=
\frac{1}{2}(\sin140^\circ-1)
\]
Hence Proved
\[
\sin25^\circ\cos115^\circ
=
\frac{1}{2}(\sin140^\circ-1)
\]