Prove that $\displaystyle sin\theta+sin2\theta+sin3\theta+sin4\theta=4cos\fr ac{\theta}{2}cos{\theta}sin\frac{5\theta}{2}$
attempt
$\displaystyle 2sin\frac{3\theta}{2}cos\frac{-\theta}{2}+2sin\frac{7\theta}{2}cos\frac{-\theta}{2}$
Dear Punch,
$\displaystyle sin\theta+sin2\theta+sin3\theta+sin4\theta$
$\displaystyle (sin\theta+sin4\theta)+(sin2\theta+sin3\theta)$
$\displaystyle 2sin\frac{5\theta}{2}cos\frac{3\theta}{2}+2sin\fra c{5\theta}{2}cos\frac{\theta}{2}$
$\displaystyle 2sin\frac{5\theta}{2}\left[cos\frac{3\theta}{2}+cos\frac{\theta}{2}\right]$
Can you do it from here?
Dear Punch,
As you could see the answer contains a $\displaystyle 5\theta$. So to get $\displaystyle 5\theta$ you have to group it like that. Suppose if you group $\displaystyle (sin\theta+sin3\theta)$ and $\displaystyle (sin2\theta+sin4\theta)$ then you can't get a $\displaystyle 5\theta $ term.
Is this answer cleared your problem?