1. ## Prove(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

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

2. Originally Posted by Punch
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?

3. Originally Posted by Sudharaka
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?
If I am right, solving it will require another use of the factor formula on them.

However, I am a little puzzled at deciding which terms to group together for use of the factor formula in step one.

4. Originally Posted by Punch
If I am right, solving it will require another use of the factor formula on them.

However, I am a little puzzled at deciding which terms to group together for use of the factor formula in step one.
Dear Punch,

Yes, you are correct. Using the factor formula again would give you the answer. However if you have any problems regarding this question please don't hesitate to reply.

5. 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 $\displaystyle (sin\theta+sin4\theta)+(sin2\theta+sin3\theta)$?

6. Originally Posted by Punch
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 $\displaystyle (sin\theta+sin4\theta)+(sin2\theta+sin3\theta)$?
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.