Example 3: $\sin^{-1}\left(2x\sqrt{1-x^2}\right)$ in terms of $\sin^{-1}x$ and $\cos^{-1}x$

Basics you need

Part (i): $-\frac{1}{\sqrt2}\le x\le\frac{1}{\sqrt2}$

1. Substitute $x = \sin\theta$ and locate $\theta$

   Let $\theta = \sin^{-1}x$⁠. Then $x = \sin\theta$⁠, and $\theta$ is in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$⁠. We can say more. Since $\sin^{-1}$ is increasing, applying it to

   $$-\frac{1}{\sqrt2} \le x \le \frac{1}{\sqrt2}$$

   keeps the inequalities, and $\sin^{-1}\left(\pm\frac{1}{\sqrt2}\right) = \pm\frac{\pi}{4}$⁠. So

   $$-\frac{\pi}{4} \le \theta \le \frac{\pi}{4}.$$
   The textbook says "Let $x = \sin\theta$⁠. Then $\sin^{-1}x = \theta$⁠." That second sentence is only true if $\theta$ is chosen in $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$⁠, so we define $\theta$ as $\sin^{-1}x$⁠. Pinning $\theta$ down to $\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$ is the key fact for the whole part: Steps 2 and 4 both depend on it.
   An increasing function keeps the order of inequalities: if $a\le x\le b$ then $f(a)\le f(x)\le f(b)$⁠. $\sin^{-1}$ is increasing on $[-1,1]$ because sine is increasing on $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$⁠. Lesson: graphs of the inverse functions

2. Simplify the square root, with the correct sign

   $$\begin{aligned} \sqrt{1-x^2} &= \sqrt{1-\sin^2\theta} = \sqrt{\cos^2\theta} \\ &= |\cos\theta| = \cos\theta. \end{aligned}$$

   The last equality holds because $\theta\in\left[-\frac{\pi}{4}, \frac{\pi}{4}\right]$ is on the right half of the unit circle, where $\cos\theta\ge0$ (in fact $\cos\theta\ge\frac{1}{\sqrt2}$⁠).

   The textbook writes $\sqrt{1-\sin^2\theta} = \cos\theta$ without comment. That step needs $\cos\theta\ge0$⁠. If $\cos\theta$ were negative, the square root (which is never negative) would equal $-\cos\theta$ instead, and every later line would change sign.
   $\sqrt{a^2} = |a|$ (Basic 2), and $1-\sin^2\theta = \cos^2\theta$ (Basic 3).

3. Recognise the double-angle formula

   $$2x\sqrt{1-x^2} = 2\sin\theta\cos\theta = \sin2\theta.$$

   So the expression we must simplify is

   $$\sin^{-1}\left(2x\sqrt{1-x^2}\right) = \sin^{-1}(\sin 2\theta).$$
   This is the reason for choosing $x = \sin\theta$ in the first place: $x$ becomes $\sin\theta$⁠, $\sqrt{1-x^2}$ becomes $\cos\theta$⁠, and the product $2\sin\theta\cos\theta$ is a single trig function of a single angle. It also shows that the left-hand side is always defined: $\sin2\theta$ lies in $[-1,1]$⁠, the domain of $\sin^{-1}$⁠.
   $\sin2\theta = 2\sin\theta\cos\theta$ (Class XI, Basic 3).

4. Cancel $\sin^{-1}(\sin\,\cdot)$⁠, after checking $2\theta$

   Doubling $-\frac{\pi}{4}\le\theta\le\frac{\pi}{4}$ gives

   $$-\frac{\pi}{2} \le 2\theta \le \frac{\pi}{2}.$$

   So $2\theta$ is in the principal range of $\sin^{-1}$⁠, and by Basic 1

   $$\sin^{-1}(\sin2\theta) = 2\theta = 2\sin^{-1}x.$$

   This proves $\sin^{-1}\left(2x\sqrt{1-x^2}\right) = 2\sin^{-1}x$ for $-\frac{1}{\sqrt2}\le x\le\frac{1}{\sqrt2}$⁠.

   This is the step the interval in the question exists for. $\sin^{-1}$ can only return angles between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$⁠, so $\sin^{-1}(\sin2\theta)$ can only equal $2\theta$ when $2\theta$ is itself in that range. With $|\theta|\le\frac{\pi}{4}$⁠, it is. The final substitution $\theta = \sin^{-1}x$ comes straight from Step 1.
   $\sin^{-1}(\sin t) = t$ for $t\in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$ (Section 2.3). Lesson: sin⁻¹(sin x)

Part (ii): $\frac{1}{\sqrt2}\le x\le 1$

1. Substitute $x = \cos\theta$ and locate $\theta$

   Let $\theta = \cos^{-1}x$⁠. Then $x = \cos\theta$ and $\theta\in[0,\pi]$⁠. Since $\cos^{-1}$ is decreasing, applying it to $\frac{1}{\sqrt2}\le x\le1$ reverses the inequalities. With $\cos^{-1}1 = 0$ and $\cos^{-1}\frac{1}{\sqrt2} = \frac{\pi}{4}$⁠:

   $$0 \le \theta \le \frac{\pi}{4}.$$
   The textbook just says "Take $x = \cos\theta$⁠, then proceeding as above". We fill in what "as above" requires. We use cosine this time because the target answer is $2\cos^{-1}x$⁠, so we want $\theta$ to be $\cos^{-1}x$⁠. As in Part (i), we define $\theta = \cos^{-1}x$ so that $\theta$ is in $[0,\pi]$ and the step "$\cos^{-1}x = \theta$" is legitimate.
   A decreasing function reverses inequalities: larger $x$ gives smaller $\cos^{-1}x$⁠. On the unit circle, a point further right (larger $\cos$⁠) on the upper half is at a smaller angle. Lesson: principal values

2. Simplify the square root, with the correct sign

   $$\begin{aligned} \sqrt{1-x^2} &= \sqrt{1-\cos^2\theta} = \sqrt{\sin^2\theta} \\ &= |\sin\theta| = \sin\theta. \end{aligned}$$

   The last equality holds because $\theta\in\left[0, \frac{\pi}{4}\right]$ is on the upper half of the circle, where $\sin\theta\ge0$⁠.

   Same care as Part (i), Step 2. Here the sign is guaranteed by the range $[0,\pi]$ of $\cos^{-1}$ alone: sine is never negative on the upper half of the circle.
   $1-\cos^2\theta = \sin^2\theta$ and $\sqrt{a^2}=|a|$⁠.

3. Recognise the double-angle formula

   $$2x\sqrt{1-x^2} = 2\cos\theta\sin\theta = \sin2\theta,$$

   so $\sin^{-1}\left(2x\sqrt{1-x^2}\right) = \sin^{-1}(\sin2\theta)$⁠.

   The two factors have swapped roles compared with Part (i): now $x$ is the cosine and the square root is the sine. The product $2\sin\theta\cos\theta$ does not care about the order.

4. Cancel $\sin^{-1}(\sin\,\cdot)$⁠, after checking $2\theta$

   Doubling $0\le\theta\le\frac{\pi}{4}$ gives $0\le2\theta\le\frac{\pi}{2}$⁠, which is inside $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$⁠. Therefore

   $$\sin^{-1}(\sin2\theta) = 2\theta = 2\cos^{-1}x.$$

   This proves $\sin^{-1}\left(2x\sqrt{1-x^2}\right) = 2\cos^{-1}x$ for $\frac{1}{\sqrt2}\le x\le1$⁠.

   Again the interval in the question is exactly what is needed. For $\theta = \cos^{-1}x$⁠, the condition $2\theta\le\frac{\pi}{2}$ means $\theta\le\frac{\pi}{4}$⁠, which means $x\ge\frac{1}{\sqrt2}$⁠. For smaller $x$ this cancellation would be invalid.
   $\sin^{-1}(\sin t) = t$ for $t\in\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$⁠.

5. Why two different formulas are needed

   Try the Part (i) formula on the Part (ii) interval, at $x = 0.8$⁠. Then $2x\sqrt{1-x^2} = 2(0.8)(0.6) = 0.96$⁠, and

   $$\begin{aligned} \sin^{-1}(0.96) &\approx 1.287002, \\ 2\sin^{-1}(0.8) &\approx 1.854590, \\ 2\cos^{-1}(0.8) &\approx 1.287002. \end{aligned}$$

   So $2\sin^{-1}x$ is wrong here and $2\cos^{-1}x$ is right. The reason: for $x>\frac{1}{\sqrt2}$⁠, $\theta = \sin^{-1}x$ is more than $\frac{\pi}{4}$⁠, so $2\sin^{-1}x$ is more than $\frac{\pi}{2}$⁠, which $\sin^{-1}$ can never output. (Indeed $1.854590 > \frac{\pi}{2}\approx1.570796$⁠.)

   The two answers agree at the shared endpoint $x = \frac{1}{\sqrt2}$⁠: $2\sin^{-1}\frac{1}{\sqrt2} = \frac{\pi}{2} = 2\cos^{-1}\frac{1}{\sqrt2}$⁠. That is a useful check that the two parts fit together. On the far left, for $-1\le x<-\frac{1}{\sqrt2}$⁠, neither formula holds (the curve there is $-\pi - 2\sin^{-1}x$⁠), which is why the question does not mention that interval.
   The values $2\sin^{-1}x$ cover $[-\pi,\pi]$ but $\sin^{-1}$ only outputs $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$⁠; whenever a proposed answer can leave the principal range, it cannot be right everywhere. Lesson: the folding rules

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