We've given : $$(f\circ g)(x)=\tan^2x$$ and $$g(x)=\sqrt{\cos 2x}$$ Then how to find the function $f(x)$? I know that $$(f\circ g)(x)=f(g(x))= f( \sqrt{\cos2x})$$ But I do not know how to find $f(x)$! Please help me!


So you can find

$$\cos x=\frac{\cos2x+1}{2}=\frac{g^2(x)+1}{2}$$


$$f(g(x))=\tan^2 x=\frac{1}{\cos^2x}-1=\frac{1-g^2(x)}{1+g^2(x)}$$




Think "what do I have to do to $\sqrt{\cos(2x)}$ to get $\tan^2x$?".

Note that $$\cos(2x) = 2\cos^2x-1\qquad\mbox{and}\qquad \tan^2x=\sec^2x-1,$$so that $$\tan^2x=\frac{2}{\cos(2x)+1}-1$$

Meaning that if you start with $\sqrt{\cos(2x)}$, you have to first square it and then apply the above. $$f(x)=\frac{2}{x^2+1}-1. $$


You say $y=\sqrt{\cos 2x}$. Then you calculate $\tan^2 x$ in terms of $y$. You square $y$, then you calculate $\sin^2x$. Then $$\tan^2 x=\frac{\sin^2 x}{1-\sin^2 x}$$

  • $\begingroup$ I don't see how this gives $f(x)$. $\endgroup$ – Shaun Dec 8 '18 at 18:23
  • $\begingroup$ $f(y)=$ some expression in terms of $y$ only. Just replace $y$ with $x$ (at the end) $\endgroup$ – Andrei Dec 8 '18 at 18:25
  • $\begingroup$ Nah, I still don't see it. (I'm a little rusty on this stuff. Sorry.) I'll take your word for it for now, then perhaps think about it later when I have more time. Thank you anyway :) $\endgroup$ – Shaun Dec 8 '18 at 18:28
  • $\begingroup$ Just look at the other answers. They explicitly follow the steps that I've described $\endgroup$ – Andrei Dec 8 '18 at 18:30
  • $\begingroup$ Ah, okay; thanks again :) $\endgroup$ – Shaun Dec 8 '18 at 18:31

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.