Draw the graph of $\sin(\pi/​2-​2x)$

Question

I tried to draw the graph of this function $\sin\left(​\frac{\pi}{2}​ - ​2​x\right).$

If I understand correctly, this means that we have to shrink the graph by $2$, shift the curve by $\frac{\pi}{2}$ to the left, and invert it, because we multiplied $x$ by a negative number ($-2$ in this case).

The curve I got this way is the same as the curve of $\sin(2x)$ function. Where did I make a mistake? What kind of transformations, and in which order we need to make, to transform $\sin(x)$ to $\sin\left(​\frac{\pi}{2}​ - ​2​x\right)?$

Answer 1

Thinking about it the way that you are saying, you can write $\sin(\pi/2-2x)=\sin(-(2x-\pi/2))=-\sin(2x-\pi/2)=-\sin(2(x-\pi/4))$, where in only the second to last step we have used a trig identity. So you shrink by a factor of $2$, shift to the right by $\pi/4$, and reflect through the $x$ axis. This winds up putting a maximum at $0$, which you might recognize as actually being a cosine, as you could derive using a different trig identity.

Answer 2

$\sin(\pi/2-X)=\cos(X)$ hence $\sin(\pi/2-2x)=\cos(2x)$ which you should be able to draw.

You can't be getting $\sin(2x)$ since $\sin(2*0)=0\neq\sin(\pi/2-2*0)=1$