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

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
edited by Alan  Oct 2, 2018

23 Online Users


New Privacy Policy

We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive information about your use of our website.
For more information: our cookie policy and privacy policy.