Trig Problem with 2 Times the Angle

[Jason76]

Find the angle or angles:

$\displaystyle 2\tan\theta = \sqrt{3}$

$\displaystyle \tan\theta = \dfrac{\sqrt{3}/2}{1/2}$ - because $\displaystyle \tan\theta = \dfrac{\sin\theta}{\cos\theta}$ so $\displaystyle \tan\theta = \dfrac{\sin \sqrt{3}/2}{\cos 1/2}$

Now we must which angles correspond to $\displaystyle \sin \sqrt{3}/2$ and $\displaystyle \cos 1/2$

After we find those angles, we have to take into account again the double angle.

So we find the $\displaystyle \sin \sqrt{3}/2$ and $\displaystyle \cos 1/2$ = $\displaystyle \dfrac{\pi}{3}$

But we have to take into the account the double angle. :?: How do we do this?

 

[pka]

Find the angle or angles:
$\displaystyle 2\tan\theta = \sqrt{3}$

Let $\displaystyle \theta = \arctan \left( {\dfrac{{\sqrt 3 }}{2}} \right)$.

Now $\displaystyle \theta$ is one solution, what is the other?

 

[Jason76]

Let $\displaystyle \theta = \arctan \left( {\dfrac{{\sqrt 3 }}{2}} \right)$.

Now $\displaystyle \theta$ is one solution, what is the other?

(trying to figure out) is one solution. It's the $\displaystyle \arctan$ of $\displaystyle \dfrac{{\sqrt 3 }}{2}$

That could have been figured out with realizing tan = sin over cos.

 

[pka]

$\displaystyle \dfrac{\pi}{3}$ is one solution. It's the $\displaystyle \arctan$ of $\displaystyle \dfrac{{\sqrt 3 }}{2}$

That could have been figured out with realizing tan = sin over cos.

$\displaystyle {\dfrac{\pi}{3}}$ NO absolutely not.

$\displaystyle \arctan \left( {\dfrac{{\sqrt 3 }}{2}} \right) \ne \dfrac{\pi }{3}$.

The answer is not one of those well known angles.

It is approx $\displaystyle 0.7137243789447656308181237057415665579998227432770442$

 

[Jason76]

$\displaystyle {\dfrac{\pi}{3}}$ NO absolutely not.

$\displaystyle \arctan \left( {\dfrac{{\sqrt 3 }}{2}} \right) \ne \dfrac{\pi }{3}$.

The answer is not one of those well known angles.

It is approx $\displaystyle 0.7137243789447656308181237057415665579998227432770442$

That's what I was thinking since on my "unit circle" graph I could see no commonly known angle that had a tan of $\displaystyle \dfrac{\sqrt {3 }}{2}$

I wonder how we could come up with the same unusual angle by looking at it in terms of $\displaystyle \tan = \dfrac{\sin}{\cos}$

 

[pka]

I wonder how we could come up with the same unusual angle by looking at it in terms of $\displaystyle \tan = \dfrac{\sin}{\cos}$

$\displaystyle \sin(\theta)=\dfrac{\sqrt{3}}{\sqrt{7}}~\&~\cos( \theta)=\dfrac{2}{\sqrt{7}}$.