Prove my mathematics induction, trigonometry

**differentiate** · November 13th 2009, 06:23 PM

if u can do this, you are a genius

Prove by mathematical induction:

$\sum_{k=1}^{n}sink\theta = \frac{sin[\frac{1}{2}(n+1)\theta ]sin(\frac{1}{2}n\theta )}{sin(\frac{1}{2}\theta )}$

For n =1,
LHS = $sin\theta$
RHS = $\frac{sin(\frac{1}{2}(2\theta ))sin(\frac{1}{2})\theta }{sin(\frac{1}{2}\theta )} = sin\theta$
therefore, it is true for n =1

assume that it is true for n = a,
$\sum_{k=1}^{a}sina\theta = \frac{sin[\frac{1}{2}(a+1)\theta ]sin(\frac{1}{2}a\theta )}{sin(\frac{1}{2}\theta )}$

for n = a + 1

Prove: $\sum_{k=1}^{a}sina\theta = \frac{sin[\frac{1}{2}(a+2)\theta ]sin(\frac{1}{2}(a+1)\theta )}{sin(\frac{1}{2}\theta )}$

I don't know how to do the next step. I'm really bad with trigonometry.

HELP!

Thank you in advance

**CaptainBlack** · November 13th 2009, 10:21 PM

Originally Posted by differentiate

if u can do this, you are a genius

Prove by mathematical induction:

$\sum_{k=1}^{n}sink\theta = \frac{sin[\frac{1}{2}(n+1)\theta ]sin(\frac{1}{2}n\theta )}{sin(\frac{1}{2}\theta )}$

For n =1,
LHS = $sin\theta$
RHS = $\frac{sin(\frac{1}{2}(2\theta ))sin(\frac{1}{2})\theta }{sin(\frac{1}{2}\theta )} = sin\theta$
therefore, it is true for n =1

assume that it is true for n = a,
$\sum_{k=1}^{a}sina\theta = \frac{sin[\frac{1}{2}(a+1)\theta ]sin(\frac{1}{2}a\theta )}{sin(\frac{1}{2}\theta )}$

for n = a + 1

Prove: $\sum_{k=1}^{a}sina\theta = \frac{sin[\frac{1}{2}(a+2)\theta ]sin(\frac{1}{2}(a+1)\theta )}{sin(\frac{1}{2}\theta )}$

I don't know how to do the next step. I'm really bad with trigonometry.

HELP!

Thank you in advance

By assumption:

$S_a(\theta)=\sum_{k=1}^{a}\sin(k\theta) = \frac{\sin[\frac{1}{2}(a+1)\theta ]\sin(\frac{1}{2}a\theta )}{\sin(\frac{1}{2}\theta )}$

Now:

$S_{a+1}(\theta)=\sum_{k=1}^{a+1}\sin(k\theta)=\sum _{k=1}^{a}\sin(k\theta)+\sin[(a+1)\theta]$ $= \frac{\sin[\frac{1}{2}(a+1)\theta ]\sin(\frac{1}{2}a\theta )}{\sin(\frac{1}{2}\theta )} +\sin[(a+1)\theta]$

Now you simplify the right most expression.

CB

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