Question

Suppose that $f$ is a function such that $f\left(\cos x\right)=\cos 17x$. Which one of the following function $g$ has the property that $g\left(\sin x\right)=\sin 17x$?

• A. $g\left(x\right)=f\left(\sqrt{1-x^2}\right)$
• B. $g\left(x\right)=f(x-\frac{\pi }{4})$
• C. $g\left(x\right)=\sqrt{1-f(x{)}^2}$
• D. $g\left(x\right)=f\left(x\right)$

Answer 1

The correct option is D $g\left(x\right)=f\left(x\right)$

Given $f\left(\cos x\right)=\cos 17x⟹f\left(x\right)=\cos \left(17{\text{cos}}^{-1}x\right)$

$⟹f\left(x\right)=\cos \left(17\right(\frac{\pi }{2}-{\text{sin}}^{-1}x\left)\right)=\cos (8\pi +\frac{\pi }{2}-17{\text{sin}}^{-1}x)=\sin \left(17{\text{sin}}^{-1}x\right)$

Given $g\left(\sin x\right)=\sin 17x⟹g\left(x\right)=\sin \left(17{\text{sin}}^{-1}x\right)$

So $g\left(x\right)=f\left(x\right)$