# Finding solutions for sin on a unit circle.

• Sep 10th 2012, 03:29 PM
Chaim
Finding solutions for sin on a unit circle.
Hi!
I was confused with this.
I have to find all the solutions for 0<x<360o if sinethreex=(square root2)/2

So I was confused with this.
I know (square root2)/2 occurs in 45o, also pi/4
Then it occurs at 135o, also 3pi/4

I'm not sure if these 2 are the correct solutions.
I understand where sin (the y coordinate) is (sqrt2)/2

But I'm not sure what the sinethreex means.
Can someone explain.
• Sep 10th 2012, 03:38 PM
MarkFL
Re: Finding solutions for sin on a unit circle.
If $\displaystyle 0^{\circ}\le x\le360^{\circ}$ then $\displaystyle 0^{\circ}\le3x\le1080^{\circ}$

So, you will have six solutions, two for each of the 3 revolutions about the unit circle.

$\displaystyle 3x=45^{\circ}\:\therefore\:x=15^{\circ}$

$\displaystyle 3x=135^{\circ}\:\therefore\:x=45^{\circ}$

Can you find the remaining 4 solutions?
• Sep 10th 2012, 03:44 PM
Chaim
Re: Finding solutions for sin on a unit circle.
Quote:

Originally Posted by MarkFL2
If $\displaystyle 0^{\circ}\le x\le360^{\circ}$ then $\displaystyle 0^{\circ}\le3x\le1080^{\circ}$

So, you will have six solutions, two for each of the 3 revolutions about the unit circle.

$\displaystyle 3x=45^{\circ}\:\therefore\:x=15^{\circ}$

$\displaystyle 3x=135^{\circ}\:\therefore\:x=45^{\circ}$

Can you find the remaining 4 solutions?

Um, are we supposed to multiply the 0<x<360 by 3? o.o
Cause the problem says "Find all solutions for 0<x<360" if sin 3x = (square root of 2)/2

But yeah, the other 4 is pretty obvious then :)
390o, 495o, 750o, 855o
Thanks :O
• Sep 10th 2012, 03:46 PM
MarkFL
Re: Finding solutions for sin on a unit circle.
Multiplying the restriction on x by 3 shows you what the corresponding restriction on 3x is, which is the argument for the sine function.
• Sep 10th 2012, 03:54 PM
MarkFL
Re: Finding solutions for sin on a unit circle.
Quote:

Originally Posted by Chaim
...
But yeah, the other 4 is pretty obvious then :)
390o, 495o, 750o, 855o
Thanks :O

No, for the next 2, add $\displaystyle 360^{\circ}$ to $\displaystyle 45^{\circ}$ and $\displaystyle 135^{\circ}$, then divide by 3 to find x:

$\displaystyle 3x=(45+360)^{\circ}=405^{\circ}\:\therefore\:x=135 ^{\circ}$

$\displaystyle 3x=(135+360)^{\circ}=495^{\circ}\:\therefore\:x=16 5^{\circ}$

And then for the last 2, add $\displaystyle 720^{\circ}$ to $\displaystyle 45^{\circ}$ and $\displaystyle 135^{\circ}$ and then divide by 3.
• Sep 10th 2012, 04:14 PM
Chaim
Re: Finding solutions for sin on a unit circle.
Quote:

Originally Posted by MarkFL2
No, for the next 2, add $\displaystyle 360^{\circ}$ to $\displaystyle 45^{\circ}$ and $\displaystyle 135^{\circ}$, then divide by 3 to find x:

$\displaystyle 3x=(45+360)^{\circ}=405^{\circ}\:\therefore\:x=135 ^{\circ}$

$\displaystyle 3x=(135+360)^{\circ}=495^{\circ}\:\therefore\:x=16 5^{\circ}$

And then for the last 2, add $\displaystyle 720^{\circ}$ to $\displaystyle 45^{\circ}$ and $\displaystyle 135^{\circ}$ and then divide by 3.

Oh wow, I accidently added the wrong degrees to some of them xD, but thanks :)
• Sep 10th 2012, 04:59 PM
MaxJasper
Re: Finding solutions for sin on a unit circle.
Quote:

Originally Posted by Chaim
I have to find all the solutions for 0<x<360o if sin3x=(sqrt2)/2

Substituting cos(3x) by its equivalent in complex plane we need to solve:

$\displaystyle \frac{1}{2i}\left(z^3-\frac{1}{z^3}\right)=\frac{\sqrt{2}}{2}$

for z=x+i y and then they are all the principal solutions 0-2pi. We obtain 6 roots as follows:

$\displaystyle \left\{\frac{3 \pi }{4},\frac{\pi }{4},\pi +\tan ^{-1}\left(\frac{\sqrt{6}-\sqrt{2}}{-\sqrt{2}-\sqrt{6}}\right),\tan ^{-1}\left(\frac{-\sqrt{2}-\sqrt{6}}{\sqrt{2}-\sqrt{6}}\right)-\pi ,\tan ^{-1}\left(\frac{-\sqrt{2}-\sqrt{6}}{\sqrt{6}-\sqrt{2}}\right),\tan ^{-1}\left(\frac{\sqrt{6}-\sqrt{2}}{\sqrt{2}+\sqrt{6}}\right)\right\}$