Question

If $\sin^{-1}(4x) + \sin^{-1}(4\sqrt{3}x) = -\frac{\pi}{2}$, then the value of $x$ is ______.

• A. $\pm \frac{1}{8}$
• B. $\pm \frac{1}{6}$
• C. $\pm \frac{2}{3}$
• D. $\pm \frac{1}{4}$

Hint:

Using $\sin(\sin^{-1} x)$ triangles to convert $\cos(\sin^{-1} x)$ into $\sqrt{1-x^2}$ instantly removes the scary nested inverse trig functions and turns the problem into basic high school algebra! Always check for extraneous roots.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
Rearrange the equation into the form $\sin^{-1} A = -\frac{\pi}{2} - \sin^{-1} B$ and take the sine of both sides.

Step 2: Formula Application:
$\sin^{-1}(4x) = -\frac{\pi}{2} - \sin^{-1}(4\sqrt{3}x)$. Taking sine: $4x = \sin(-\frac{\pi}{2} - \sin^{-1}(4\sqrt{3}x)) = -\cos(\sin^{-1}(4\sqrt{3}x))$.

Step 3: Explanation:
$4x = -\sqrt{1 - (4\sqrt{3}x)^2}$. Squaring both sides: $16x^2 = 1 - 48x^2 \implies 64x^2 = 1 \implies x^2 = 1/64$. $x = \pm 1/8$. Since the sum is $-\pi/2$, $x$ must be negative to satisfy the range. $x = -1/8$.

Step 4: Final Answer:
The value of $x$ is $\pm 1/8$ (mathematically $x = -1/8$ is the valid solution).