# Sum to Infinity of a Geometric Series

1. Aug 17, 2011

### odolwa99

1. The problem statement, all variables and given/known data

Q.: A geometric series has first term 1 and common ratio $\frac{1}{2}$sin2$\theta$. Find the sum of the first 10 terms when $\theta$ = $\frac{\pi}{4}$, giving your answer in the form h - $\frac{1}{2^k}$, where h, k $\in$ N.

2. Relevant equations

Sn = $\frac{a(1 - r^n)}{1 - r}$, when IrI < 1

3. The attempt at a solution

S10 = $\frac{a(1 - r^n)}{1 - r}$

S10 = $\frac{1(1 - (1/2)sin2(\pi/ 4) to the power of 10}{1 - (1/2)sin2(\pi/4)}$ ... The tags did not take, for some reason, hence the 'to the power of 10'. Sorry about that.

I'm stuck here because if I solve for $\frac{1}{2}$sin2$\theta$, I get a decimal answer = 0.013706. The question wants the answer defined as whole numbers but I'm unable to work out the next step in the sequence. Can someone help and give me a tip on what to do next? Thanks.

Ans.: From text book: 2 - (1/ 29)

2. Aug 17, 2011

### micromass

How did you get that?

$\sin(2\theta)=\sin(\pi/2)$

has a nice answer, no?

3. Aug 17, 2011

### odolwa99

Ok, so the sum is now...

$\frac{1 - (1/2)sin(\pi/ 2)tothepowerof10}{1 - (1/2)sin(\pi/ 2)}$

so...

1 - (1/2)sin($\pi$/ 2)10 = 1 - (1/2)sin(\pi/ 2)
1 - 1 - (1/2)sin($\pi$/ 2)10 + (1/2)sin(\pi/ 2)
- (1/2)sin($\pi$/ 2)9

Ok, obviously I'm wrong but this is what my attempt looks like. Thanks for the help so far but can you help me out a little bit more...please? Thanks.

4. Aug 17, 2011

### micromass

What is $\sin(\pi /2)$?? Isn't that a nice value?

5. Aug 17, 2011

### odolwa99

Err...In a right angled triangle, Sin( $\theta$ ) = Opposite/ Hypotenuse...?

Last edited: Aug 17, 2011
6. Aug 17, 2011

### micromass

Go check your trigonometric formula's...

You should really know things like $\sin(0)$ and $\sin(\pi/2)$ by heart.

7. Aug 17, 2011

### odolwa99

Ok. Sorry about all this. It's literally been years since I last studied maths and I'm having to begin from scratch all over again. You'll have to be patient with this old man. I'll see what I can find out from your advice and show my work tomorrow. Good night.

8. Aug 17, 2011

### micromass

It's alright. This might be helpful: http://en.wikipedia.org/wiki/Trigonometric_identities

9. Aug 18, 2011

### odolwa99

Ok, so here is, hopefully, my final attempt...

S10 = $\frac{1(1 - (1/2)sin2(\pi/ 4)tothepowerof10)}{1 - (1/2)sin2(\pi/ 4)}$

S10 = $\frac{1 - (1/2)sin(2\pi/ 4)tothepowerof10}{1 - (1/2)sin(2\pi/ 4)}$

S10 = $\frac{1 - (1/2)sin(\pi/ 2)tothepowerof10}{1 - (1/2)sin(\pi/ 2)}$

S10 = $\frac{1 - (1/2)(\sqrt{4}/2)tothepowerof10}{1 - (1/2)(\sqrt{4}/2)}$

S10 = $\frac{1 - (2/ 4)tothepowerof10}{1 - (2/ 4)}$

S10 = $\frac{1 - (1/2)tothepowerof10}{(1/2)}$

S10 = 2 - $\frac{1}{2^9}$

10. Aug 18, 2011

### HallsofIvy

Are you talking about $sin^{10}(\pi/2)$ or $sin((\pi/2)^{10})$?

$sin(\pi/2)= 1$, but why in the world would you write that as "$\sqrt{4}/2$"?

11. Aug 18, 2011

### Staff: Mentor

I've seen some texts present the trig functions in this way as a mnemnonic device.

$sin(0)= \sqrt{0}/2$
$sin(\pi/6)= \sqrt{1}/2$
$sin(\pi/4)= \sqrt{2}/2$
$sin(\pi/3)= \sqrt{3}/2$
$sin(\pi/2)= \sqrt{4}/2$

The OP might have been referring to something like this, and didn't realize that $\sqrt{4}/2$ could and should be replaced with 1.

12. Aug 18, 2011

### odolwa99

Based on the answer, I solved the question as sin(($\pi$/ 2)10).

As regards, expressing sin($\pi$/ 4) as $\frac{\sqrt{4}}{2}$; yes, I was using the mnemonic but, as I indicated above, this is the first time I've dealt with these kinds of equations so I'll express it as 1, henceforth.

Thanks for helping, guys. I really do appreciate it.

Last edited: Aug 18, 2011
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