Why does simplifying $\arcsin(2x\sqrt{1-x^2})$ two ways give different results?

Question

Find derivative of $y=\arcsin(2x \sqrt{1-x^2}) $ in domain $\frac{-1}{\sqrt{2}} < x < \frac{1}{\sqrt{2}}$

If you put $x=\sin\theta$ then

$$ y= \arcsin(2sin(\theta) \sqrt{1- sin^2 ( \theta)})$$ $$y= \arcsin( sin2\theta)$$ $$y= 2\theta$$

$$y = 2 \arcsin(x)$$

But, if you put $x=\cos\theta$ then , again,

$$y=2\theta$$

But, resubstituting

$$ y= 2 \arccos(x)$$

But derivatives of both are different.

Now where's the mistake? Is it something related to the original domain I took?

Answer 1

Notice, your first substitution is correct. When you substitute $x=\cos\theta$ $$\sin^{-1}(2\cos\theta\sqrt{1-\cos^2\theta})$$ $$=\sin^{-1}(2\cos\theta\sqrt{\sin^2\theta})$$ $$=\sin^{-1}(2\cos\theta|\sin\theta|)$$ $$|\sin\theta|=\begin{cases}-\sin\theta\ \ \forall \ \ \ -\frac{\pi}{4}<\theta<0\\ \sin\theta\ \ \ \forall \ \ \ 0\le \theta<\frac{\pi}{4}\end{cases}$$

Answer 2

It is a good habit of writing the interval for the parameter when you make trigonometric substitutions. In the first case, $$x=\sin\theta, \qquad\theta\in\left[-\dfrac\pi2,\dfrac\pi2\right]$$ Now, since $x\in\left(-\dfrac1{\sqrt2},\dfrac1{\sqrt2}\right)$, hence, $\theta\in\left(-\dfrac\pi4,\dfrac\pi4\right)$ and therefore, $$\sin^{-1}\Big(\sin(2\theta)\Big)=2\theta$$ In the second case, $$x=\cos\theta,\qquad\theta\in[0,\pi]$$ Now, since by the given domain of $x$, we infer $\theta\in\left(\dfrac\pi4,\dfrac{3\pi}4\right)$. Hence, $$\sin^{-1}\Big(\sin(2\theta)\Big)=\pi-2\theta$$ Hope it helps.

Answer 3

There is nothing wrong with your algebra. It’s just that you need to take extra care when dealing with the domain. Note that $\arcsin(\sin \theta=\theta$ only when $-\frac{\pi}{2} \le \theta\le \frac{\pi}{2}$.

Substituting $x=\sin\theta$ gives $$y= \arcsin(2\sin\theta|\cos\theta|)=\arcsin|\sin 2\theta| $$ Now, $$-\frac{1}{\sqrt 2} \lt \sin\theta \lt \frac{1}{\sqrt 2} \\ \implies -\frac{\pi}{4} \le \theta\le \frac{\pi}{4} \\\implies -\frac{\pi}{2} \lt2\theta \lt \frac{\pi}{2}$$ and we can safely say that $$y=2\theta=2\arcsin x$$

But when $x=\cos \theta$, we have $$ -\frac{1}{\sqrt 2} \lt \cos\theta \lt \frac{1}{\sqrt 2} \\ \implies \frac{\pi}{4} \lt \theta \lt \frac{3\pi}{4} \\ \implies \frac{\pi}{2} \lt 2\theta \lt \frac{3\pi}{2}$$ This time, we need to shift by $\pi$ so that $$-\frac{\pi}{2}\lt 2\theta-\pi\lt\frac{\pi}{2}$$So, $$ y =\arcsin(\sin 2\theta) =\arcsin\left(-\sin(2\theta-\pi)\right) = \pi-2\arccos x \\ = 2\left(\frac{\pi}{2}-\arccos x\right) \\ = 2\arcsin x$$