Question

Solve: $\sin 10+\sin 20+\sin 40+\sin 50$

Answer 1

Step 1: Use $\sin C+\text{sin}D$ rule

The given expression is $\sin 10+\sin 20+\sin 40+\sin 50$

$=\sin 10+\sin 20+\sin 40+\sin 50$

$=(\sin 10+\sin 50)+(\sin 20+\sin 40)$

$=(2\sin 30\cos 20)+\left(2\sin 30\cos 10\right)$ $\left[\left(SinC+SinD\right)=2Sin\left(\frac{C+D}{2}\right)cos\left(\frac{C-D}{2}\right)\right]$

$=2\sin 30(\cos 10+\cos 20)$

Step 2: Use the identity of $\mathrm{Cos}(90-\theta )$

$=2\times \frac{1}{2}\left(cos\right(90-80)+cos(90-70\left)\right)$ $\left[Cos\left(90-\theta \right)=Sin\theta \right]$

$=\sin 80+\sin 70$

Hence, $\sin 10+\sin 20+\sin 40+\sin 50=\sin 80+\sin 70$