<s>[INST] Simplifying a trig relation My physics teacher wanted to prove Young's equation (my problem is pure mathematics), and in the derivation, he got the following expression:
$$\tag{1}
\begin{align}
\left(\gamma_{s l}-\gamma_{s o}\right) \cdot\left(\sin ^{2} \theta+\cos \theta\left(-\frac{(1-\cos \theta)(2+\cos \theta)}{(1+\cos \theta)}\right)\right) \\[6pt]
\quad+\;\gamma \cdot\left(2(1-\cos \theta)-\frac{(1-\cos \theta)(2+\cos \theta)}{(1+\cos \theta)}\right)=0
\end{align}$$
Which he says leads to:
$$\tag{2}
\gamma_{s l}+\gamma \cos \theta=\gamma_{s o}$$
It is easy to see from eq. (1) and (2) that we must have:
$$\tag{3}
\left(2(1-\cos \theta)-\frac{(1-\cos \theta)(2+\cos \theta)}{1+\cos \theta}\right)\left(\sin ^{2}(\theta)-\cos \theta \frac{(1-\cos \theta)(2+\cos \theta)}{1+\cos \theta}\right)^{-1}=\cos \theta
$$
But I have a hard time showing this (my trig-skills might not be up to par). Will someone help me show this?
 [/INST] The coefficient of $\gamma$ can be simplified as,
$\begin{align}(1-\cos\theta)\frac{2+2\cos\theta - 2- \cos\theta}{1+\cos\theta} = \boxed{\cos\theta\ \left[\frac{1-\cos\theta}{1+\cos\theta}\right]} \end{align}$
Now, the coefficient of $\gamma_{sl} - \gamma_{so}$ can be simplified as,
$\begin{align}\sin^2\theta -\cos\theta\left(\frac{(1-\cos\theta)(2+\cos\theta)}{1+\cos\theta}\right) &= (1-\cos\theta)(1+\cos\theta) -\cos\theta\left(\frac{(1-\cos\theta)(2+\cos\theta)}{1+\cos\theta}\right) \\
&= (1-\cos\theta)\left[\frac{(1+\cos^2\theta +2\cos\theta) - (2\cos\theta+\cos^2\theta)}{1+\cos\theta}\right] \\& = \boxed{\frac{1-\cos\theta}{1+\cos\theta}}\end{align}$
Now the coefficients differ by a factor $\cos\theta$.
Assuming the coefficient $\frac{1-\cos\theta}{1+\cos\theta}$ is non-zero,
$\gamma_{sl} - \gamma_{so} + \gamma \cos\theta = 0$
or
$$\boxed{\gamma_{so} = \gamma_{sl} + \gamma \cos\theta}$$
</s>