# If $(f\circ g)(x)=\tan^2x$ and $g(x)=\sqrt{\cos2x}$ then find $f(x)=?$

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

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

and

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

obviously

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

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 $$\tan^2 x=\frac{\sin^2 x}{1-\sin^2 x}$$

• I don't see how this gives $f(x)$. – Shaun Dec 8 '18 at 18:23
• $f(y)=$ some expression in terms of $y$ only. Just replace $y$ with $x$ (at the end) – Andrei Dec 8 '18 at 18:25
• 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 :) – Shaun Dec 8 '18 at 18:28
• Just look at the other answers. They explicitly follow the steps that I've described – Andrei Dec 8 '18 at 18:30
• Ah, okay; thanks again :) – Shaun Dec 8 '18 at 18:31