# Solve for $\theta$ in $h \cos^2 \theta = 2v_0 \sin \theta \cos \theta + 2a$

Given $$h$$, $$v_0$$ and $$a$$, I want to solve for $$\theta$$ in the following equation: $$h \cos^2 \theta = 2v_0 \sin \theta \cos \theta + 2a.$$ My approach was to rearrange and use $$\sin 2\theta = 2 \sin \theta \cos \theta$$ to get $$h \cos^2 \theta - v_0 \sin 2\theta = 2a$$ and then do the following: $$h \cos^2 \theta - \frac{h}{2} + \frac{h}{2} - v_0 \sin 2\theta = 2a$$ $$\frac{h}{2}(2 \cos^2 \theta - 1) + \frac{h}{2} - v_0 \sin 2\theta = 2a.$$ Then using $$2 \cos^2 \theta - 1 = \cos 2\theta$$, rearranging and multiplying everything by $$2$$ leaves me with $$h \cos 2\theta - 2 v_0 \sin 2\theta = 4a - h.$$

At this point, however, I do not have any idea on how to procede to get $$\theta$$. I already tried Photomath to maybe get a solution with the steps to solve, but that, unfortunately, didn't work.

Help is much appreciated!

Hint

The expression can be written as $$h{1+\cos 2\theta \over 2}=v_0\sin2\theta +2a$$which leads to $$\cos 2\theta={2v_0\over h}\sin2\theta +{4a\over h}-1$$or $$1-\sin^22 \theta=\left({2v_0\over h}\sin2\theta +{4a\over h}-1\right)^2$$as a quadratic equation of $$\sin2\theta$$.

I would use the following trigonometric identities:

$$\cos^2\theta \;\; =\;\; \frac{1+\cos(2\theta)}{2} \hspace{2pc} 2\sin\theta\cos\theta \;\; =\;\; \sin(2\theta) \hspace{2pc} \sin(2\theta) \;\; =\;\; \sqrt{1-\cos^2(2\theta)}.$$

From this you can obtain the equation

$$\frac{h}{2}(1 + \cos(2\theta)) \;\; =\;\; v_0\sqrt{1 - \cos^2(2\theta)} + 2a.$$

Isolate the square-root and square both sides to obtain

$$v_0^2\left (1 - \cos^2(2\theta)\right ) \;\; =\;\; \frac{h^2}{4}\left (1 + \cos(2\theta) \right )^2 - 2ah (1 + \cos(2\theta)) + 4a^2.$$

Move everything to one side and obtain the quadratic equation

$$\left (v_0^2 + \frac{h^2}{4}\right )\cos^2(2\theta) + \frac{h}{2}(h-4a)\cos(2\theta) + \left (4a^2-v_0^2 + \frac{h^2}{4} - 2ah\right ) \;\; =\;\; 0.$$

Hint

Whenever we have something like $$a\cos^2t+2h\sin t\cos t+b\sin^2t=c$$

Divide both sides by $$\cos^2t$$ to form a quadratic equation in $$\tan t$$

Or divide both sides by $$\sin^2t$$

Do a change of variables and make $$2\theta = \phi$$ to transform your equation into the form

$$A \cos \phi + B \sin \phi = C \tag{1}$$

with $$A= h$$, $$B=-2 v_0$$ and $$C = 4a-h$$

I can think of two ways to solve (1) for $$\phi$$.

## 1. In parts

One is to guess that the solution involves two parts, an to substitute $$\phi = \alpha+\beta$$, as well as \begin{aligned} \cos(\alpha+\beta) &= \cos \alpha \cos\beta - \sin \alpha \sin \beta \\ \sin(\alpha+\beta) & = \cos\alpha \sin \beta + \sin\alpha \cos \beta \end{aligned} \tag{2}

After simplifications this transforms (1) into

$$\cos \alpha \left( A \cos \beta + B \sin \beta \right) + \sin \alpha \left( B \cos \beta - A \sin \beta \right) = C \tag{3}$$

Consider the case where $$B \cos \beta - A \sin \beta =0$$, then the above is solved by

$$\beta = \tan^{-1} \left( \frac{B}{A} \right) \tag{4}$$ and \begin{aligned} \cos \alpha \left( A \frac{A}{\sqrt{A^2+B^2}} + B \frac{B}{\sqrt{A^2+B^2}} \right) + 0 & = C \\ \cos \alpha \left( \sqrt{A^2+B^2} \right) & = C \\ \end{aligned} \tag{4}

$$\alpha = \cos^{-1} \left( \frac{C}{\sqrt{A^2+B^2}} \right) \tag{5}$$

The final solution is then

$$\theta = \tfrac{1}{2} \phi = \tfrac{1}{2} ( \alpha + \beta)$$

$$\theta = \frac{1}{2} \left( \cos^{-1} \left( \frac{C}{\sqrt{A^2+B^2}} \right) + \tan^{-1} \left( \frac{B}{A} \right) \right) \tag{6}$$

$$\boxed{ \theta = \frac{1}{2} \left( \cos^{-1} \left( \frac{4a-h}{\sqrt{h^2-4 v_0^2}} \right) - \tan^{-1}\left( \frac{2v_0}{h} \right) \right) } \tag{7}$$

## 2. Tangent Half Angle Substitution

If you consider the substitution $$t = \tan \left( \tfrac{ \phi}{2} \right)$$ which leads to the following expressions (wiki)

\begin{aligned} \sin \phi & = \frac{2 t}{1+t^2} \\ \cos \phi & = \frac{1-t^2}{1+t^2} \end{aligned} \tag{2}

and the re-write of equation (1) as

$$A \left( \frac{1-t^2}{1+t^2} \right) + B \left( \frac{2 t}{1+t^2} \right) = C \tag{3}$$

with some re-arranging the above is solved for $$t$$ as a quadratic

$$A (t^2-1) -2 B t + C (t^2+1) = 0 \tag{4}$$

And since $$\phi = 2 \tan^{-1}(t)$$

$$\boxed{ \theta = -\tan^{-1} \left( \frac{v_0 \pm \sqrt{v_0^2+2 a h-4 a^2}}{2 a} \right) } \tag{5}$$

Continue with $$h \cos 2\theta - 2 v_0 \sin 2\theta = 4a - h$$ and write the equation as $$\sqrt{h^2+4v_0^2}\>\cos( 2\theta+\alpha)=4a - h$$ or, $$\cos( 2\theta+\alpha)=\cos\beta$$, where $$\alpha = \cos^{-1}\frac{h}{\sqrt{h^2+4v_0^2}}, \>\>\>\>\> \beta=\cos^{-1}\frac{4a - h}{\sqrt{h^2+4v_0^2}}$$ Then, we get $$2\theta +\alpha = 2\pi k\pm\beta$$ and the solutions

$$\theta =\pi k - \frac{\alpha\pm\beta}2$$