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 \(\displaystyle \tan = \dfrac{\sin}{\cos}\)

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