# Find theta as a function of phi

• May 10th 2011, 04:33 PM
fysikbengt
Find theta as a function of phi
I have found the equation 2\cos \theta \sin \theta \tan \phi +2\sin \theta ^2 -1=0 and want to solve out \theta as a function of \phi .

The history of this equation is from a physics problem where a ball is thrown at angle \theta down a slope of angle \phi . The question is what angle \theta to throw the ball so that it has the greatest range. I have checked the equation numerically but cannot solve it.
• May 10th 2011, 06:33 PM
Soroban
Hello, fysikbengt!

Quote:

$2\cos\theta\sin\theta\tan\phi + 2\sin^2\!\theta-1\:=\:0$

$\text{Solve for }\theta\text{ as a function of }\phi.$

$\text{We have: }\;\underbrace{2\cos\theta\sin\theta}_{\sin2\theta }\tan\phi + \underbrace{2\sin^2\!\theta - 1}_{-\cos2\theta} \:=\:0$

. . $\sin2\theta\tan\phi - \cos2\theta \:=\:0 \quad\Rightarrow\quad \sin2\theta\tan\phi \:=\:\cos2\theta$

. . $\frac{\sin2\theta}{\cos2\theta} \:=\:\frac{1}{\tan\phi} \quad\Rightarrow\quad \tan2\theta \:=\:\cot\phi$

. . $2\theta \:=\:\tan^{-1}(\cot\phi) \:=\: \left(\tfrac{\pi}{2} - \phi\right)$

. . $\theta \:=\:\tfrac{1}{2}\left(\tfrac{\pi}{2}-\phi\right)$

• May 11th 2011, 05:00 AM
fysikbengt
Thanks
Quote:

Originally Posted by Soroban

. . $2\theta \:=\:\tan^{-1}(\cot\phi) \:=\: \left(\tfrac{\pi}{2} - \phi\right)$

So easy!

I really had to think about that last step but now it is obvious to me.

Thanks for help.