Question 15 of 42intermediate🔧 ApplyLong Answer4 marks

Prove the following:

3sin⁡−1x=sin⁡−1(3x−4x3),x∈[−12,12]3\sin^{-1}x = \sin^{-1}(3x - 4x^3), \quad x \in \left[-\frac{1}{2}, \frac{1}{2}\right]
Correct Answer

Let x=sin⁡θx = \sin \theta. Then, sin⁡−1x=θ\sin^{-1}x = \theta.

LHS:

3sin⁡−1x=3θ3\sin^{-1}x = 3\theta

RHS:

sin⁡−1(3x−4x3)\sin^{-1}(3x - 4x^3)

Substituting x=sin⁡θx = \sin \theta, we get:

sin⁡−1(3sin⁡θ−4sin⁡3θ)\sin^{-1}(3\sin \theta - 4\sin^3 \theta)

Using the identity sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin 3\theta = 3\sin \theta - 4\sin^3 \theta, the expression becomes:

sin⁡−1(sin⁡3θ)\sin^{-1}(\sin 3\theta)

For this to equal 3θ3\theta, the angle 3θ3\theta must lie within the principal range of sin⁡−1\sin^{-1}, which is [−π2,π2]\left[-\frac{\pi}{2}, \frac{\pi}{2}\right].

Given x∈[−12,12]x \in \left[-\frac{1}{2}, \frac{1}{2}\right], we have sin⁡θ∈[−12,12]\sin \theta \in \left[-\frac{1}{2}, \frac{1}{2}\right].

This implies θ∈[−π6,π6]\theta \in \left[-\frac{\pi}{6}, \frac{\pi}{6}\right].

Multiplying by 3, we get 3θ∈[−π2,π2]3\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right].

Since 3θ3\theta lies in the principal branch, sin⁡−1(sin⁡3θ)=3θ\sin^{-1}(\sin 3\theta) = 3\theta.

Therefore, RHS =3θ== 3\theta = LHS.

∴ Hence, 3sin⁡−1x=sin⁡−1(3x−4x3)\mathbf{3\sin^{-1}x = \sin^{-1}(3x - 4x^3)}.

Exercise: EXERCISE 2.2 | Q: 1 | (Page 12)
For More Understanding

Explanation

This solution follows the substitution method demonstrated in Example 3 of the provided text. We substitute xx with sin⁡θ\sin \theta to convert the inverse trigonometric function into a trigonometric one.

The critical step, emphasized in the textbook examples, is verifying the domain. The condition x∈[−12,12]x \in \left[-\frac{1}{2}, \frac{1}{2}\right] ensures that the resulting angle 3θ3\theta falls within the principal range of sin⁡−1\sin^{-1}, allowing the simplification sin⁡−1(sin⁡3θ)=3θ\sin^{-1}(\sin 3\theta) = 3\theta.

Solution Steps

  1. Step 1: Substitute x=sin⁡θx = \sin \theta, implying sin⁡−1x=θ\sin^{-1}x = \theta.

  2. Step 2: Simplify LHS to 3θ3\theta.

  3. Step 3: Substitute xx into RHS to get sin⁡−1(3sin⁡θ−4sin⁡3θ)\sin^{-1}(3\sin \theta - 4\sin^3 \theta).

  4. Step 4: Apply the identity sin⁡3θ=3sin⁡θ−4sin⁡3θ\sin 3\theta = 3\sin \theta - 4\sin^3 \theta to get sin⁡−1(sin⁡3θ)\sin^{-1}(\sin 3\theta).

  5. Step 5: Verify domain: x∈[−12,12]x \in \left[-\frac{1}{2}, \frac{1}{2}\right] implies θ∈[−π6,π6]\theta \in \left[-\frac{\pi}{6}, \frac{\pi}{6}\right], so 3θ∈[−π2,π2]3\theta \in \left[-\frac{\pi}{2}, \frac{\pi}{2}\right].

  6. Step 6: Conclude sin⁡−1(sin⁡3θ)=3θ\sin^{-1}(\sin 3\theta) = 3\theta since 3θ3\theta is in the principal range, proving LHS = RHS.

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