Question
Show that:
$\sin25^\circ\cos115^\circ=\frac{1}{2}(\sin140^\circ-1)$

Answer

$\text{LHS}=\ \sin25^\circ\cos115^\circ$
$=\ \frac{2\sin25^\circ\cos115^\circ}{2}$
We Know that
$2\sin\text{A}\cos\text{B}=\sin(\text{A+B})+\sin(\text{A}-\text{B})$
$=\ \frac{1}{2}[\sin(25^\circ+115^\circ)+\sin(25^\circ-115^\circ)]$
$=\ \frac{1}{2}[\sin140^\circ+\sin(-90^\circ)]$
$\sin(-\theta)=-\sin\theta$
$\text{And},\sin(90^\circ+\theta)=\cos\theta$
$\Rightarrow\ \frac{1}{2}[\sin(90^\circ+50^\circ)-\sin90^\circ]$
$=\ \frac{1}{2}[\cos50^\circ-1]$
Also,
$\cos\theta=\sin(90^\circ+\theta)$
$\cos50^\circ=\sin(90^\circ+50^\circ)=\sin140^\circ$
$\frac{1}{2}[\sin140^\circ-1]=\text{RHS}$

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