Why does taking the square root to the independent variable (x) cause the squaring of x in each point?

Question

Why does the graph of $\text{sin}(\sqrt{x})$ look like this why is the maximum at ($(\frac{\pi}{2})^2,1)$ and not ($\frac{\sqrt\pi}{\sqrt2},1)$ why did we square and not take the square root and does the apply also to cubic roots , fourth roots ...etc ?

Answer 1

$\sin \theta$ achieves a maximum when $\theta = \frac{\pi}{2}$. For $\sin \sqrt{x}$ to achieve a maximum we need $\sqrt{x} = \frac{\pi}{2}$. As you noticed, this happens when $x = (\frac{\pi}{2})^2$.

Answer 2

Firstly, it's easier to use $\ \sqrt{\frac{\pi}{2}}\ $ rather than $\frac{\sqrt\pi}{\sqrt2}.$

$f:\mathbb{R}^+\to[-1,1];\quad f(x) =\sin\left(\sqrt{x}\right).$

$f\left(\sqrt{\frac{\pi}{2}}\right)= \sin \sqrt{\sqrt{\frac{\pi}{2}}}\approx 0.90\neq 1.$

$f\left(\left(\frac{\pi}{2}\right)^2\right) = \sin\left( \sqrt{\left(\frac{\pi}{2}\right)^2} \right) = \sin\left(\frac{\pi}{2}\right)=1.$

Answer 3

If you substitute $\color{blue}{x}=\color{blue}{\sqrt{\pi/2}}$ to $\color{red}{f}(x)=\color{red}{\sin\sqrt{\color{black}{x}}}$, you will actually get: $$ y = \color{red}{\sin \sqrt{\color{blue}{\sqrt{\frac\pi2}}}} = \sin \sqrt[4]{\frac{\pi}2}. $$

So instead of applying root twice, you need to undo it in your value. In other words, you need to square the value of $\mathop{\mathrm{argmax}}\sin(x)$