# Analytically find the value of sin(x)

• May 18th 2011, 01:52 PM
JustMeAgain
Analytically find the value of sin(x)
Hi all,
I'm having a little math problem, and I know a topic with the same name is already on this forum, but I want to try a different approach. The other topic uses Taylor series to calculate the value of a sine, but I want to try it though using normal geometry. I have done some work already but am quite stuck, and my math teacher doesn't know how to go on either, so I'm hoping one of you can help me out. I have tried to attach a word document, but that didn't really work out, I had to change the extension to .doc. If you want to read it, you will have to change it back to .docx. I'm not posting the whole document in this thread because it would then become a very long thread.
Thanks in advance for any responses.

JustMeAgain

EDIT: turns out that if you're using word 2010, you can just open the file without change the extension. It opens up fine without any problems then.
• May 18th 2011, 03:36 PM
pickslides
$\displaystyle \int \frac{1}{\sqrt{1-x^2}}~dx = \sin^{-1}x+C$

you should be able to prove this using a trig substitution.
• May 19th 2011, 06:59 AM
JustMeAgain
I have looked at trig substitution as you suggested, but have not come to the same conclusion. I found that:
$\displaystyle \int \frac{1}{\sqrt{1-x^2}}~dx = \int \cos(t)~dx = sin(t) + C$
where $t=\tan^{-1}x$
but this may still need some work, and I'm not sure whether I'm stuck on that approach.
However, I am unclear about what you mean with your x in $\sin^{-1}x$. Do you mean the k I used in the word document? or do you mean something completely different?

JustMeAgain
• May 19th 2011, 07:24 AM
Mondreus
Try the substitution $x = \sin u \implies dx = \cos u du$

Please try to learn some basic LaTeX if you're posting questions. I and many others refuse to open Word files and other strange attachments.
• May 19th 2011, 07:26 AM
TheEmptySet
Quote:

Originally Posted by JustMeAgain
I have looked at trig substitution as you suggested, but have not come to the same conclusion. I found that:
$\displaystyle \int \frac{1}{\sqrt{1-x^2}}~dx = \int \cos(t)~dx = sin(t) + C$
where $t=\tan^{-1}x$
but this may still need some work, and I'm not sure whether I'm stuck on that approach.
However, I am unclear about what you mean with your x in $\sin^{-1}x$. Do you mean the k I used in the word document? or do you mean something completely different?

JustMeAgain

I think you have made an error in the substitution.

If

$x=\sin(t) \implies dx=\cos(t)dt$

This gives

$\int \frac{\cos(t)}{\sqrt{1-\sin^2(t)}}dt=\int dt=t+C$

Now just solve for t and back substitute.
• May 19th 2011, 09:44 AM
JustMeAgain
Ow, i see. I have chosen the wrong function to substitute x indeed... Thanks for pointing that out, I'll report any progress or further problems.
• May 19th 2011, 09:45 AM
JustMeAgain
I know, I didn't use LaTeX in the first post, because as I explained earlier, it would have been a very long post! The word document about 2 pages long, and includes annotations and images. Which would've been impossible in a normal, decent sized post.
• May 19th 2011, 10:55 AM
JustMeAgain
Ok, I have looked at your ideas, and it has helped quite a lot so far, I just don't understand one step. How do you substitute 1 through cos(t)? Plus, I don't know if this brings me to the solution. Asuming what you told me t would become
$\sin^{-1}(x)$
But this again brings a sine into the equasion, and it kinda defeats the purpose of analytically determining the sine though using the inverse sine... is it what is seems, and is this approach impossible and will I have to find my way though a Taylor series?
So I guess what this boils down to my last question before abandoning this and trying a different approach: is it possible to determine the prime of the following function:
$\frac{1}{\sqrt{1-x^2}}$?
Or maybe determine the difference between the two points used to calculate the value of
$\int\frac{1}{\sqrt{1-x^2}}~dx$
when the outcome is know?
Because I'm totally stuck here...
• May 19th 2011, 11:05 AM
Mondreus
$1-\sin^2t=\cos^2 t\implies \sqrt{1-\sin^2 t}=\sqrt{\cos^2t}=|\cos t|$

When you have the substitution $x=\sin t$, you also need to specify that $-\pi/2 \leq t \leq \pi/2\implies |\cos t|=\cos t$

Since $dx=\cos t dt$ we have $\int \frac{1}{\sqrt{1-x^2}}dx=\int \frac{\cos t}{\cos t}dt=\int dt$
• May 19th 2011, 11:26 AM
JustMeAgain
Thanks for explaining, now I get it :). But this would end in the prime of the function being: $\sin^{-1}(x)$
This again would involve a sine, so is there a way to get around a trig function like sine or cosine? Because I seem to think that I need to have a look at Taylor Series. Do you agree with me?
• May 19th 2011, 11:34 AM
Mondreus
I don't know what the entire problem is since I haven't looked at the Word file (don't even have Word on this computer).
• May 19th 2011, 12:21 PM
JustMeAgain
Ow, appareantly noone looks at the word file here, so I'll post it's contents here, but it's going to be a very long post.
so, the formula of a circle is: $r^2 = y^2 + x^2$
now for this situation i'll use r = 1 because I want to analytically determine the value of a sine of cosine.
So we can write the function for a circle as two functions: $y = \sqrt{1-x^2}$ and $y = -\sqrt{1-x^2}$
I'll use this illustration to show what I mean:
http://img846.imageshack.us/img846/4450/graphwj.jpg
So what we want to know is k, because that's the answer the sine will give you. To determine that we can use Pythagoras, but we'll need the value of m for that.
Since we know that length of the circle between the point (1,0) and (L,k) is the same as the angle a I thought we could use the formula for the length of a function and re-engineer it to find the value of m
The standard formula for determining the length of a piece of graph would be:
$\int\sqrt{f(x)'^2 + 1}~dx$
filling in the circle formula, it doesn't matter which one because of the power, we get:
$\int\frac{1}{\sqrt{1-x^2}}~dx$
m should be the difference between the values of x used to calculate $\int\frac{1}{\sqrt{1-x^2}}~dx = a$ (I'm using radians not degrees here!)
This is my progress on the problem, and from here on I'm stuck, and it seems like this approach is impossible, I hope this is clear enough?
• May 19th 2011, 01:32 PM
Mondreus
That will just lead straight back to the definition of sin and cos.

$\arcsin x |_L^1=\arcsin 1 - \arcsin L = a \implies L = \sin(a+\pi/2)=\cos a$
• May 19th 2011, 10:38 PM
JustMeAgain
Ok :( then I guess this won't help me any further, since I would need the calculator again to calculate the arcsin.. :S
Thanks anyway