Prove that $\displaystyle sin\theta+sin2\theta+sin3\theta+sin4\theta=4cos\fr ac{\theta}{2}cos{\theta}sin\frac{5\theta}{2}$

attempt

$\displaystyle 2sin\frac{3\theta}{2}cos\frac{-\theta}{2}+2sin\frac{7\theta}{2}cos\frac{-\theta}{2}$

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- Feb 24th 2010, 03:22 AMPunchProve(10)
Prove that $\displaystyle sin\theta+sin2\theta+sin3\theta+sin4\theta=4cos\fr ac{\theta}{2}cos{\theta}sin\frac{5\theta}{2}$

attempt

$\displaystyle 2sin\frac{3\theta}{2}cos\frac{-\theta}{2}+2sin\frac{7\theta}{2}cos\frac{-\theta}{2}$ - Feb 24th 2010, 03:42 AMSudharaka
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? - Feb 25th 2010, 01:47 AMPunch
- Feb 25th 2010, 11:18 PMSudharaka
- Feb 26th 2010, 02:17 AMPunch
Yes, I do have a question. From $\displaystyle sin\theta+sin2\theta+sin3\theta+sin4\theta=4cos\fr ac{\theta}{2}cos{\theta}sin\frac{5\theta}{2}$, how did you know you had to group it as $\displaystyle (sin\theta+sin4\theta)+(sin2\theta+sin3\theta)$?

- Feb 26th 2010, 04:16 AMSudharaka
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?