Q.Show that

Concept: Inverse Sine Principal Value — The identity $\sin^{-1}(\sin \theta) = \theta$ holds only when $\theta$ lies in the principal branch $[-\pi/2, \pi/2]$. The substitution $x = \sin\theta$ or $x = \cos\theta$ must respect the given domain so that the angle after simplification stays within this range.

Proof for (i): Let $x = \sin\theta$, where $\theta \in [-\pi/4, \pi/4]$ because $x \in [-1/\sqrt{2}, 1/\sqrt{2}]$. Then $2x\sqrt{1-x^2} = 2\sin\theta \cos\theta = \sin 2\theta$. Since $2\theta \in [-\pi/2, \pi/2]$, we have $\sin^{-1}(\sin 2\theta) = 2\theta = 2\sin^{-1}x$.

Proof for (ii): Let $x = \cos\theta$, where $\theta \in [0, \pi/4]$ because $x \in [1/\sqrt{2}, 1]$. Then $2x\sqrt{1-x^2} = 2\cos\theta \sin\theta = \sin 2\theta$. Here $2\theta \in [0, \pi/2]$, so $\sin^{-1}(\sin 2\theta) = 2\theta = 2\cos^{-1}x$.

The Core Idea

The expression $2x\sqrt{1-x^{2}}$ looks like $\sin 2\theta$ if we set $x = \sin\theta$ or $x = \cos\theta$. Recall:

If $x = \sin\theta$, then $\sqrt{1-x^{2}} = \cos\theta$ (taking the non-negative root, since $\sqrt{\cdot}$ denotes the principal square root). So:

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

But $\sin^{-1}(\sin y) = y$ only when $y$ lies in the principal range of $\sin^{-1}$, which is $[-\pi/2, \pi/2]$. If $y$ is outside this interval, $\sin^{-1}(\sin y)$ gives the principal value — the unique angle in $[-\pi/2, \pi/2]$ whose sine equals $\sin y$.

So the problem reduces to: for a given $x$, choose $\theta$ such that $x = \sin\theta$ or $x = \cos\theta$, then check whether $2\theta$ falls inside $[-\pi/2, \pi/2]$. If it does, the identity is direct; if not, we adjust.

Step-by-Step Derivation

1. Set $x = \sin\theta$ and express the argument.

Let $\theta = \sin^{-1}x$. Then $x = \sin\theta$, and by definition $\theta \in [-\pi/2, \pi/2]$. For such $\theta$, $\cos\theta \ge 0$, so $\sqrt{1-x^{2}} = \sqrt{1-\sin^{2}\theta} = |\cos\theta| = \cos\theta$.

Hence:

Therefore:

2. Determine when $2\theta$ lies in $[-\pi/2, \pi/2]$.

Since $\theta \in [-\pi/2, \pi/2]$, $2\theta \in [-\pi, \pi]$. The principal range of $\sin^{-1}$ is $[-\pi/2, \pi/2]$. So $\sin^{-1}(\sin 2\theta) = 2\theta$ exactly when $2\theta \in [-\pi/2, \pi/2]$.

Solve for $\theta$:

Since $\theta = \sin^{-1}x$, this means:

Taking sine (which is increasing on $[-\pi/2, \pi/2]$):

For $x$ in this interval, $\sin^{-1}(\sin 2\theta) = 2\theta = 2\sin^{-1}x$. This proves part (i).

3. For part (ii), use $x = \cos\theta$ instead.

Let $\theta = \cos^{-1}x$. Then $x = \cos\theta$, and $\theta \in [0, \pi]$. For $\theta$ in this range, $\sin\theta \ge 0$, so $\sqrt{1-x^{2}} = \sqrt{1-\cos^{2}\theta} = |\sin\theta| = \sin\theta$.

Thus:

So again:

4. Find when $2\theta$ lies in $[-\pi/2, \pi/2]$ for $\theta = \cos^{-1}x$.

Here $\theta \in [0, \pi]$, so $2\theta \in [0, 2\pi]$. The principal range $[-\pi/2, \pi/2]$ intersects $[0, 2\pi]$ in $[0, \pi/2]$. So we need $2\theta \in [0, \pi/2]$, i.e.:

Since $\theta = \cos^{-1}x$, this means:

Taking cosine (which is decreasing on $[0, \pi]$):

For $x$ in this interval, $\sin^{-1}(\sin 2\theta) = 2\theta = 2\cos^{-1}x$. This proves part (ii).

Method: Proving a double-angle inverse-trig identity by substitution

Use this for "show that" identities where the argument of an inverse function looks like a double-angle expression (e.g. $2x\sqrt{1-x^2}=\sin 2\theta$, or $\tfrac{2x}{1+x^2}$).

Steps

Step 1: Substitute so the messy argument collapses to a single trig ratio.

Choose $x=\sin\theta$ or $x=\cos\theta$ so that $\sqrt{1-x^2}$ becomes a clean cosine or sine. With $x=\sin\theta$, $\sqrt{1-x^2}=\cos\theta$ and

The outer inverse then reads $\sin^{-1}(\sin 2\theta)$.

Step 2: Apply $\sin^{-1}(\sin y)=y$ ONLY after checking $y$ is in the principal range.

This is the crux, not a formality. $\sin^{-1}(\sin y)=y$ holds only when $y\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$. Translate that condition on $2\theta$ back into a condition on $x$; it is exactly the domain the problem states.

Step 3: Pick the substitution that matches the given domain.

For $x\in\left[-\tfrac{1}{\sqrt2},\tfrac{1}{\sqrt2}\right]$ use $x=\sin\theta$ (giving $2\sin^{-1}x$); for $x\in\left[\tfrac{1}{\sqrt2},1\right]$ use $x=\cos\theta$ (giving $2\cos^{-1}x$), because that keeps $2\theta$ inside the principal range. The two domains in the question are precisely where each substitution is legal.

Common Mistakes

Mistake 1: Writing $\sin^{-1}(\sin 2\theta)=2\theta$ without checking the range.

Why it's wrong: this cancellation is valid only when $2\theta\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$; ignoring that gives the identity on the wrong domain. Correct approach: translate $2\theta\in\left[-\tfrac{\pi}{2},\tfrac{\pi}{2}\right]$ into a condition on $x$ — that is exactly why each part is stated for its own interval.

Mistake 2: Using the same substitution $x=\sin\theta$ for both parts.

Why it's wrong: for $x\in\left[\tfrac{1}{\sqrt2},1\right]$, $2\sin^{-1}x$ leaves the principal range, so $x=\sin\theta$ fails part (ii). Correct approach: switch to $x=\cos\theta$ there, which keeps $2\theta$ in range and yields $2\cos^{-1}x$.

Mistake 3: Taking $\sqrt{1-x^2}=-\cos\theta$ or dropping the modulus.

Why it's wrong: the principal square root is non-negative, and on the chosen branch $\cos\theta\ge 0$, so $\sqrt{1-x^2}=\cos\theta$. A sign slip here breaks $2x\sqrt{1-x^2}=\sin 2\theta$. Correct approach: confirm the cosine (or sine) is non-negative on the substitution's interval before dropping the root.