Q240 / 269 · NDA 2025 (Phase 1) · Mathematics
Consider the following for the two (02) items that follow:
Let $y = \sin^{-1}\left(x - \dfrac{4x^3}{27}\right)$.
What is $y$ equal to?
Let $y = \sin^{-1}\left(x - \dfrac{4x^3}{27}\right)$.
What is $y$ equal to?
- A$\sin^{-1} x$
- B$\sin^{-1}\left(\dfrac{x}{3}\right)$
- C$3\sin^{-1} x$
- Correct answerD$3\sin^{-1}\left(\dfrac{x}{3}\right)$
Solution
Given $y = \sin^{-1}\left(x - \dfrac{4x^3}{27}\right)$
Substitute $x = 3\sin\theta$, which gives $\sin\theta = \dfrac{x}{3}$ or $\theta = \sin^{-1}\left(\dfrac{x}{3}\right)$
The expression becomes:
$y = \sin^{-1}\left(3\sin\theta - \dfrac{4(3\sin\theta)^3}{27}\right)$
$y = \sin^{-1}\left(3\sin\theta - \dfrac{4(27\sin^3\theta)}{27}\right)$
$y = \sin^{-1}(3\sin\theta - 4\sin^3\theta)$
Using the trigonometric identity $\sin(3\theta) = 3\sin\theta - 4\sin^3\theta$:
$y = \sin^{-1}(\sin(3\theta))$
$y = 3\theta$
Substituting back the value of $\theta$:
$y = 3\sin^{-1}\left(\dfrac{x}{3}\right)$
Answer: $3\sin^{-1}\left(\dfrac{x}{3}\right)$
Substitute $x = 3\sin\theta$, which gives $\sin\theta = \dfrac{x}{3}$ or $\theta = \sin^{-1}\left(\dfrac{x}{3}\right)$
The expression becomes:
$y = \sin^{-1}\left(3\sin\theta - \dfrac{4(3\sin\theta)^3}{27}\right)$
$y = \sin^{-1}\left(3\sin\theta - \dfrac{4(27\sin^3\theta)}{27}\right)$
$y = \sin^{-1}(3\sin\theta - 4\sin^3\theta)$
Using the trigonometric identity $\sin(3\theta) = 3\sin\theta - 4\sin^3\theta$:
$y = \sin^{-1}(\sin(3\theta))$
$y = 3\theta$
Substituting back the value of $\theta$:
$y = 3\sin^{-1}\left(\dfrac{x}{3}\right)$
Answer: $3\sin^{-1}\left(\dfrac{x}{3}\right)$
Hence, the correct answer is option: 'D'.