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

Question

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!

Answer 1

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

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

Answer 2

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

Answer 3

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