# Thread: Finding sinx formed by 2 rays.

1. ## Finding sinx formed by 2 rays.

Hello, can some one please check my work, I believe I did it correctly!

Also can you please check if I did Angular velocity correct.

(Sorry for bothering you >.<! I have a test on Tuesday!)

2. ## Re: Finding sinx formed by 2 rays.

You did both correctly.

For the first part, you can figure out most if it using only geometry. A right triangle with two equal sides is a 45-45-90 triangle. A right triangle with sides $x, x\sqrt{3}$ is a 30-60-90 triangle (where $\theta_2$ is the 30 degree angle). So, $\theta_1 = \dfrac{\pi}{4}, \theta_2 = \dfrac{\pi}{6}$. This means $\theta = \pi - \dfrac{\pi}{4} - \dfrac{\pi}{6} = \dfrac{7\pi}{12}$.

You get the same answer either way.

Thanks! !

4. ## Re: Finding sinx formed by 2 rays.

Originally Posted by SlipEternal
This means $\theta = \pi - \dfrac{\pi}{4} - \dfrac{\pi}{6} = \dfrac{7\pi}{12}$.

You get the same answer either way.
Yes, but would you figure out that $\sin(7\pi/12)=\frac{\sqrt{2}(1+\sqrt{3})}{4}$? It looks like this problem asked to find the exact value of $\sin\theta$. It is probably not a coincidence that both vectors have integer lengths.

5. ## Re: Finding sinx formed by 2 rays.

Can you please explain how a triangle with x, xroot3 yields a 30 , 60, 90 triangle and how yoh figured out that the other angle is pi /4

6. ## Re: Finding sinx formed by 2 rays.

I was taught that the sides of a 30-60-90 triangle will have lengths $x, x\sqrt{3}, 2x$ where $x$ is the length of the side opposite the 30 degree angle, $x\sqrt{3}$ is the length of the side opposite the 60 degree angle, and $2x$ is the length of the hypotenuse in geometry class. I assumed it was still taught. As for the other angle, convert degrees to radians: $45\dfrac{\pi}{180} = \dfrac{\pi}{4}$.

7. ## Re: Finding sinx formed by 2 rays.

Oh I see, its the special triangle. Thanks!

8. ## Re: Finding sinx formed by 2 rays.

Originally Posted by emakarov
Yes, but would you figure out that $\sin(7\pi/12)=\frac{\sqrt{2}(1+\sqrt{3})}{4}$? It looks like this problem asked to find the exact value of $\sin\theta$. It is probably not a coincidence that both vectors have integer lengths.
I got a different result when applying the half-angle formula. I got:

$\sin\left(\dfrac{7\pi}{12}\right) = \sin\left(\dfrac{\pi}{2}\right)\cos\left(\dfrac{ \pi}{12}\right) + \sin\left(\dfrac{\pi}{12}\right) \cos\left(\dfrac{\pi}{2}\right) = \cos\left(\dfrac{\pi}{12}\right)$.

Since $\cos(2\theta) = 2\cos^2\theta-1$, the half-angle formula gives:

$\cos \theta = \sqrt{\dfrac{\cos(2\theta)+1}{2}}$

Plugging in $\theta = \dfrac{\pi}{12}$ you have

$\cos \left( \dfrac{\pi}{12}\right) = \sqrt{\dfrac{\cos\left(\dfrac{\pi}{6}\right)+1}{2} } = \sqrt{\dfrac{\dfrac{\sqrt{3}}{2}+1}{2}} = \dfrac{1}{2}\sqrt{\sqrt{3}+2}$

So, I got a different result.

Edit: I just checked wolframalpha. Apparently, they are the same.

9. ## Re: Finding sinx formed by 2 rays.

I calculated mine using vector product: $\sin\theta=\frac{|u\times v|}{|u|\cdot|v|}$. It's also possible to use dot product to find cosine first. If the angles are arbitrary, I would guess the problem authors' intention was to use the formula

$\sin(\pi-\theta_1-\theta_2)= \sin(\theta_1+\theta_2)=\sin(\theta_1)\cos(\theta_ 2)+ \cos(\theta_1)\sin(\theta_2)$

The values in the right-hand side can be found by definition.