My math teacher gave us an equality involving trigonometric functions and told us to "verify" them. I tried making the two sides equal something simple such as "1 = 1" but kept getting stuck. I would highly appreciate if someone could show me (step by step) how to verify or solve this problem.
$$\frac{\cos{2\theta}}{1 + \sin{2\theta}} = \frac{\cot{\theta} – 1}{\cot{\theta} + 1}$$
Best Answer
Let us look at the RHS only: (It is fine to manipulate both sides but it is usually preferred if you manage to make one into the other.)
$$ \frac{\cot{\theta}-1}{\cot\theta +1} = \frac{\frac{\cos\theta}{\sin\theta}-1}{\frac{\cos\theta}{\sin\theta} +1} $$ Now we multiply the numerator and denominator by $\sin\theta$ to get, \begin{equation} \frac{\cos\theta-\sin\theta}{\cos\theta +\sin\theta}. \end{equation} If we consider the double angle formulas for $\sin$ and $\cos$ i.e. $$ \cos{2\theta} = \cos^2\theta-\sin^2\theta, \\ \sin 2\theta= 2\sin\theta\cos\theta, $$ we can see that we can get a $\cos2\theta$ in the numerator of $\frac{\cos\theta-\sin\theta}{\cos\theta +\sin\theta}$ by multiplying by $\cos\theta+\sin\theta$ as the $\cos2\theta$ formula is a difference of two squares. Proceeding with this idea we get; $$ \left(\frac{\cos\theta-\sin\theta}{\cos\theta +\sin\theta}\right)\left(\frac{\cos\theta+\sin\theta}{\cos\theta+\sin\theta}\right) = \frac{\cos^2\theta - \sin^2\theta}{\cos^2\theta+2\sin\theta\cos\theta +\sin^2\theta}. $$ Now the numerator is equal to $\cos2\theta$ so that is done and we only need to fix the denominator, by a quick inspection we see that we have a term $2\sin\theta\cos\theta$ which gives our $\sin2\theta$ term we desired and using the fact $\cos^2\theta+\sin^2\theta=1$ we find we get $$ \frac{\cos2\theta}{1+\sin2\theta} $$ which is what we wanted!