Miscellaneous · Q13: $y=\sin^{-1}x+\sin^{-1}\sqrt{1-x^2}$ is constant on $(0,1)$

Basics you need

Solution

1. Check that $y$ is defined on $0 < x < 1$

   For $0 < x < 1$ we have $0 < x^2 < 1$⁠, so $0 < 1 - x^2 < 1$ and $0 < \sqrt{1-x^2} < 1$⁠. Both inputs, $x$ and $\sqrt{1-x^2}$⁠, lie in $[-1, 1]$⁠, so both inverse sines exist.

   $\sin^{-1}$ only accepts inputs between $-1$ and 1. Before differentiating anything we make sure the function we are differentiating actually exists on the stated interval.
   Domain of $\sin^{-1}$ is $[-1, 1]$ (Chapter 2, Section 2.2; Basic 1).

2. Substitute $x = \sin\theta$⁠, with $\theta = \sin^{-1}x$

   Let $\theta = \sin^{-1} x$⁠. Since $0 < x < 1$⁠, the angle $\theta$ lies in $\left(0, \frac{\pi}{2}\right)$ (the sine is positive and less than 1 there, and $\sin^{-1}$ takes values in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$⁠). Then $x = \sin\theta$ and the first term is simply

   $$\sin^{-1} x = \theta.$$
   Defining $\theta$ as $\sin^{-1} x$ (rather than just writing "put $x = \sin\theta$") makes the substitution automatically legitimate: every $x$ in $(0,1)$ gives exactly one such $\theta$⁠, and we know precisely which interval it is in. That interval is what the next step needs.
   $\sin(\sin^{-1}x) = x$ for $x \in [-1,1]$ (Chapter 2, Section 2.3). Basics: trigonometric identities

3. Rewrite $\sqrt{1-x^2}$ as $\sin\left(\frac{\pi}{2} - \theta\right)$

   $$\begin{aligned} \sqrt{1 - x^2} &= \sqrt{1 - \sin^2\theta} \\ &= \sqrt{\cos^2\theta} = |\cos\theta| \\ &= \cos\theta \\ &= \sin\left(\tfrac{\pi}{2} - \theta\right). \end{aligned}$$
   The step $|\cos\theta| = \cos\theta$ needs $\cos\theta \ge 0$⁠. That holds because $\theta \in \left(0, \frac{\pi}{2}\right)$⁠, where cosine is positive. The last line is the complementary-angle identity (Basic 2).
   $\sin^2\theta + \cos^2\theta = 1$⁠; $\sqrt{u^2} = |u|$ (Basic 3). Basics: trigonometric identities

4. Undo the sine, checking the principal range

   Since $0 < \theta < \frac{\pi}{2}$⁠, the angle $\frac{\pi}{2} - \theta$ also lies in $\left(0, \frac{\pi}{2}\right)$⁠, which is inside the principal range $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$⁠. Therefore

   $$\sin^{-1}\sqrt{1 - x^2} = \sin^{-1}\left(\sin\left(\tfrac{\pi}{2} - \theta\right)\right) = \tfrac{\pi}{2} - \theta.$$
   $\sin^{-1}(\sin\phi) = \phi$ is true only for $\phi$ in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$⁠. We have just checked that $\phi = \frac{\pi}{2} - \theta$ is there, so the inverse really does undo the sine. (If $\phi$ were outside that interval, $\sin^{-1}(\sin\phi)$ would be some other angle with the same sine.)
   Principal value branch of $\sin^{-1}$ (Chapter 2, Section 2.2) and $\sin^{-1}(\sin\phi) = \phi$ for $\phi \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ (Chapter 2, Section 2.3). Lesson: derivatives of inverse trig functions

5. Add: $y$ is the constant $\frac{\pi}{2}$⁠, so $\frac{dy}{dx} = 0$

   $$y = \theta + \left(\tfrac{\pi}{2} - \theta\right) = \frac{\pi}{2} \quad \text{for every } x \in (0,1).$$

   A constant function has derivative 0, so

   $$\frac{dy}{dx} = 0, \qquad 0 < x < 1.$$
   The $\theta$'s cancel, leaving no $x$ at all: the value of $y$ does not change as $x$ moves through $(0,1)$⁠. The triangle shows why: the two acute angles of a right triangle always add up to a right angle.
   The derivative of a constant function is 0 (Class XI, Limits and Derivatives). Basics: derivative rules

6. Confirm with the chain rule

   Differentiate the second term with outer function $\sin^{-1}u$ and inner function $u = \sqrt{1-x^2}$⁠:

   $$\frac{du}{dx} = \frac{1}{2\sqrt{1-x^2}}\cdot(-2x) = \frac{-x}{\sqrt{1-x^2}}.$$

   The outer derivative is $\frac{1}{\sqrt{1-u^2}}$⁠, and $1 - u^2 = 1 - (1 - x^2) = x^2$⁠, so $\sqrt{1-u^2} = \sqrt{x^2} = |x| = x$ (because $x > 0$⁠). Hence

   $$\begin{aligned} \frac{d}{dx}\sin^{-1}\sqrt{1-x^2} &= \frac{1}{x}\cdot\frac{-x}{\sqrt{1-x^2}} \\ &= \frac{-1}{\sqrt{1-x^2}}, \end{aligned}$$ $$\frac{dy}{dx} = \frac{1}{\sqrt{1-x^2}} - \frac{1}{\sqrt{1-x^2}} = 0. \ ✓$$
   This is a fully independent route, and it shows exactly where $x > 0$ is used: $\sqrt{x^2} = x$⁠. The derivative formula for $\sin^{-1}u$ needs $-1 < u < 1$⁠, which holds since $0 < u < 1$ (Step 1).
   $\frac{d}{dx}\sin^{-1}x = \frac{1}{\sqrt{1-x^2}}$ (Example 24); chain rule (Theorem 4, Section 5.3.1); $\frac{d}{dx}\sqrt{v} = \frac{1}{2\sqrt v}\cdot\frac{dv}{dx}$⁠. Lesson: chain rule

7. Why the interval $0 < x < 1$ matters

   For $-1 < x < 0$ the same steps break at two places. Then $\theta \in \left(-\frac{\pi}{2}, 0\right)$⁠, so $\frac{\pi}{2} - \theta$ lies in $\left(\frac{\pi}{2}, \pi\right)$⁠, outside the principal range; and in Step 6 $\sqrt{x^2} = -x$⁠. Redoing Step 6 with $|x| = -x$ gives $\frac{dy}{dx} = \frac{2}{\sqrt{1-x^2}}$ there, and indeed $y = \frac{\pi}{2} + 2\sin^{-1}x$ for $-1 \le x \le 0$⁠.

   So the answer 0 is a property of the stated interval, not of the formula everywhere. For example at $x = -0.6$⁠, $y = \sin^{-1}(-0.6) + \sin^{-1}(0.8)$⁠, which is $\approx 0.2837941$⁠, not $\frac{\pi}{2}$⁠. This is the same lesson as in Chapter 2: check that every angle lies in the principal range before cancelling $\sin^{-1}$ against $\sin$⁠.
   For $\frac{\pi}{2} - \theta \in \left(\frac{\pi}{2}, \pi\right)$⁠: $\sin^{-1}\left(\sin\left(\frac{\pi}{2} - \theta\right)\right) = \frac{\pi}{2} + \theta$⁠, since $\sin(\pi - \phi) = \sin\phi$⁠.

Every fact used, and where it comes from