Solving $\cos\phi\cos(2\theta - \phi)+\sin(\theta - \phi)\sin(\theta + \phi)=0$ for $\theta$

Question

Let $$ d = \frac{2v_i \cos(\theta)\sin(\theta - \phi)}{g\cos^2\phi} $$ Now I have to equate $\frac{d}{d\theta}$ to $0$ and find $\theta$. $$\frac{(2(v_i)^2\sin{\theta}\sin(\theta - \phi)+2(v_i)^2\cos{\theta}\cos(\theta - \phi))g\cos^2\theta+4(v_i)^2\cos{\theta}\sin(\theta - \phi)g\cos{\theta}\sin\phi}{g^2\cos^4\phi}=0$$ Then I get to the following expression: $$-\cos{\phi}\sin{\theta}\sin({\theta - \phi})+\cos{\theta}\cos(\theta - \phi)\cos\phi+2\cos{\theta}\sin(\theta - \phi)\sin \theta=0$$

$$\cos\phi\cos(2\theta - \phi)+\sin(\theta - \phi)\sin(\theta + \phi)=0$$ I need to get to $\theta = \frac{\pi}{4}-\frac{\phi}{2}$ but I don't know how to continue. I tried different ways but I never get to the desired solution.

Answer 1

use Mathematica Solve

Solve[Cos[\[Phi]]*Cos[2 \[Theta] - \[Phi]] + 
    Sin[\[Theta] - \[Phi]]*Sin[\[Theta] + \[Phi]] == 0, \[Theta], 
  Assumptions -> {\[Theta] \[Element] Reals, \[Phi] \[Element] 
     Reals}] // FullSimplify

$$ \left\{\left\{\theta \to \fbox{$\frac{1}{2} \pi (4 c_1-1)\text{ if }c_1\in \mathbb{Z}$}\right\},\left\{\theta \to \fbox{$\frac{1}{2} (4 \pi c_1+\pi )\text{ if }c_1\in \mathbb{Z}$}\right\},\left\{\theta \to \fbox{$2 \phi +\pi \left(2 c_1-\frac{1}{2}\right)\text{ if }c_1\in \mathbb{Z}$}\right\},\left\{\theta \to \fbox{$\frac{1}{2} (4 \phi +4 \pi c_1+\pi )\text{ if }c_1\in \mathbb{Z}$}\right\}\right\} $$

Answer 2

Since you want the critical points, then want the local maximum or local minimum of $d$ as a function of $\theta$. This can be tackled directly from the expression for $d$. $\phi$ is fixed, so what we should concentrate on is

$ f(\theta) = \cos(\theta) \sin(\theta - \phi) $

Using the identity

$ cos(A) sin(B) = 1/2 ( \sin(B + A) + \sin(B - A) ) $

We get

$ f(\theta) = 1/2 ( \sin( 2 \theta - \phi) + \sin(- \phi) ) $

The second term is constant, and does not depend on $\theta$. So the maximum of $f$ is when

$2 \theta - \phi = \dfrac{\pi}{2} $

This leads to $ \theta = \dfrac{\phi}{2} + \dfrac{\pi}{4} $

If you looking for the minimum then this occurs when

$ 2 \theta - \phi = - \dfrac{\pi}{2} $

And this leads to $ \theta = \dfrac{\phi}{2} - \dfrac{ \pi}{4} $

Answer 3

I got the expression $\cos \theta \cos ( \theta- \varphi) = \sin \theta \sin ( \theta-\varphi) $ from the derivative to solve for $\theta$. This gets expanded out to

$$ \cos \theta ( \cos \varphi \cos \theta + \sin \varphi \sin \theta) = \sin \theta ( \cos \varphi \sin \theta - \sin \varphi \cos \theta) \tag{1} $$

Now apply the tangent-half-angle substitution(s) of

$$ \begin{array}{c|c} z = \tan \left( \tfrac{\varphi}{2} \right) & t = \tan \left( \tfrac{\theta}{2} \right) \\ \varphi = 2 \tan^{-1}(z) & \theta = 2 \tan^{-1}(t) \\ \hline \cos(\varphi) = \tfrac{1-z^2}{1+z^2} & \cos(\theta) = \tfrac{1-t^2}{1+t^2} \\ \sin(\varphi) = \tfrac{2 z}{1+z^2} & \sin(\theta) = \tfrac{2 t}{1+t^2} \end{array} \tag{2}$$

The result is

$$\small \frac{1-t^2}{1+t^2} \left( \frac{(1-z^2)(1-t^2)}{(1+z^2)(1+t^2)} + \frac{(2 z)(2 t)}{(1+z^2)(1+t^2)} \right) = \frac{2 t}{1+t^2} \left( \frac{(2 t)(1-z^2)}{(1+z^2)(1+t^2)} - \frac{(1-t^2)(2 z)}{(1+z^2)(1+t^2)} \right) $$

which is simplified by eliminating common terms down to

$$ (1-t^2)(z^2 (1-t^2)-4 t z + t^2-1) = 4 t (t z^2 + (1-t)^2 z -t) $$

Finally, bring both sides together to get

$$ (1-z^2) t^4 -8 z t^3 - 6 (1-z^2) t^2 + (8 z) t + (1-z^2) = 0 \tag{3} $$

which is a solvable quatric polynomial in terms of $t$ and hence $\theta = 2 \tan^{-1}(t)$

If you restict $\varphi = 0 \ldots \pi/2$ then $z (1-z^2) \ge 0$ and the four solutions of $t$ are

$$ \large t = \begin{cases} \text{-} \frac{ \sqrt{2} (1-z) \sqrt{1+z^2}-z^2-2z-1}{1-z^2} \\ \text{+} \frac{ \sqrt{2} (1-z) \sqrt{1+z^2}+z^2+2z+1}{1-z^2} \\ \text{-} \frac{ \sqrt{2} (1+z) \sqrt{1+z^2}+z^2-2z+1}{1-z^2} \\ \text{+} \frac{ \sqrt{2} (1+z) \sqrt{1+z^2}-z^2+2z-1}{1-z^2} \\ \end{cases} \tag{4} $$

Expanded out, the above is also written as

$$ \large \tan \left( \frac{\theta}{2} \right) = \begin{cases} \frac{(1+\sin(\varphi))\sqrt{1+\cos(\varphi)}-\cos(\varphi)+\sin(\varphi)-1}{\cos(\varphi)\sqrt{1+\cos(\varphi)}} \\ \frac{(1+\sin(\varphi))\sqrt{1+\cos(\varphi)}+\cos(\varphi)-\sin(\varphi)+1}{\cos(\varphi)\sqrt{1+\cos(\varphi)}} \\ \frac{(\sin(\varphi)-1)\sqrt{1+\cos(\varphi)}-\cos(\varphi)-\sin(\varphi)-1}{\cos(\varphi)\sqrt{1+\cos(\varphi)}} \\ \frac{(\sin(\varphi)-1)\sqrt{1+\cos(\varphi)}+\cos(\varphi)+\sin(\varphi)+1}{\cos(\varphi)\sqrt{1+\cos(\varphi)}} \\ \end{cases}$$