Trig identities again. I don’t get this?? The one with 2 stars is apparently harder but I don’t even get the easy one.

Guest Oct 1, 2018

#1**+2 **

This doesn't involve *proving* trig identities, but *using* them to find the value(s) of \(\theta\) and x that make the equations true.

By definition: \(\sec\theta=\frac{1}{\cos\theta}\)

Using this in a and rearranging: \(\frac{1}{\cos^2\theta}=2\tan\theta\)

Multiply both sides by \(\cos^2\theta\) and using the fact that \(\tan\theta=\frac{\sin\theta}{\cos\theta}\) we get, on simplifying: \(1=2\sin\theta\cos\theta\)

The RHS is just \(\sin{2\theta}\) so we have \(1=\sin{2\theta}\)

This results in \(2\theta=\frac{\pi}{2}\text{ or }\theta=\frac{\pi}{4}\)

For b rewrite x as \(2\theta\) and use \(\cos{2\theta}=\cos^2\theta-\sin^2\theta\text{ and }\sin{}2\theta=2\sin\theta\cos\theta\text{ together with }\cos^2\theta+\sin^2\theta=1\)

You should eventually find there are two values of x between 0 and 2pi that satisfy the equation.

Alan Oct 2, 2018