Obel1x wrote: It is given the function $f:\mathbb{R} \to \mathbb{R}$ such that it holds $f(\sin x)=\sin (2011x)$. Find the value of $f(\cos x)$. $f(\cos x)=f(\sin(\frac{\pi}2-x)))=\sin (2011(\frac{\pi}2-x))$ $=-\cos (2011x)$

@ pco : How did you get $ -cos(2011x) $ ? I got $cos(2011x)$ This is further equal to cos (211x) if x is measured in degrees (reducing modulo 360)

Pascal96 wrote: @ pco : How did you get $ -cos(2011x) $ ? I got $cos(2011x)$ This is further equal to cos (211x) if x is measured in degrees (reducing modulo 360) $\sin(2011(\frac{\pi}2-x))$ $=\sin(1006\pi-\frac{\pi}2-2011x))$ $=\sin(-\frac{\pi}2-2011x))$ $=-\cos(2011x)$