Why is $\displaystyle \cos^{-1} \left( \sin 2\theta \right) = \frac{\pi}{2} - 2\theta$?

Question

In the end of an exercise I reach the point that I state that the answer is $\cos^{-1} \left( \sin 2\theta \right)$. The book gives, as an answer, $\displaystyle \frac{\pi}{2} - 2\theta$.

By plotting in Wolfram I coud see that, for in interval considered ($0 < \theta < \frac{\pi}{4}$) they are equivalent. But whay is that? How can I show that?

Answer 1

$$\displaystyle \cos^{-1} \left( \sin 2\theta \right) = \frac{\pi}{2} - 2\theta$$

Take cosine of both sides to get the equivalent statement for $0 < \theta < \frac{\pi}{4}$

$$ \sin 2\theta = \cos(\frac{\pi}{2} - 2\theta) $$

Which is true for every $\theta$

Answer 2

It is based on the trigonometric identity: $\sin(x)=\cos(\frac{\pi}{2}-x)$, assume $0<x<\frac{\pi}{2}$ $\sin2\theta=\cos(\frac{\pi}{2}-2\theta)$

Take the inverse function of cosine from both side

$\cos^{-1}\sin(2\theta)=\cos^{-1}\cos(\frac{\pi}{2}-2\theta)$

$\cos^{-1}(\sin2\theta)=\frac{\pi}{2}-2\theta$ , as $0<\theta<\frac{\pi}{4}$

Answer 3

$$\begin{align} \sin \rho &= \sin \rho \\ &=\cos\left(\frac{\pi}{2}-\rho\right) && \text{the sine-cosine complementary relation}\\ \cos^{-1}(\sin \rho)&=\frac{\pi}{2}-\rho && \text{taking inverse cosine of both sides} \\ \cos^{-1}(\sin 2\theta)&=\frac{\pi}{2}-2\theta && \text{substitute} \qquad \rho = 2 \theta \end{align}$$

Answer 4

If $0 \le \phi \le \frac \pi 2$ then

$\cos^{-1} \cos \phi = \phi$

If $0 \le \phi \le \frac \pi 2$ then $0 \le \frac \pi 2 - \phi \le \frac \pi 2$ and $\sin \phi = \cos (\frac \pi 2 - \phi)$

and $\cos^{-1}(\sin \phi) =$

$\cos^{-1}(\cos \frac \pi 2 - \phi) =$

$\frac \pi 2 - \phi$.

And if $0\le \theta \le \frac \pi 4$ and $\phi = 2\theta$ then $0\le \phi \le \frac \pi 2$ and

$\cos^{-1}(\sin 2\theta) = $

$\cos^{-1}(\sin \phi) =$

$\cos^{-1}(\cos \frac \pi 2 - \phi) =$

$\frac \pi 2 - \phi$

$\frac \pi 2 - 2\theta$.