Question

The value of \( \sin^{-1} \left( \frac{2\sqrt{2}}{3} \right) + \sin^{-1 \left( \frac{1}{3} \right) \) is equal to:}

• A. \( \frac{\pi}{6} \)
• B. \( \frac{\pi}{4} \)
• C. \( \frac{\pi}{2} \)
• D. \( \frac{2\pi}{3} \)
• E. \( 0 \)

Hint:

Always check if the squares of the two arguments in a \( \sin^{-1} \) sum add up to 1. If \( a^2 + b^2 = 1 \), the answer is almost always \( \pi/2 \).

Answer 1

The Correct Option is C

Concept: Recall the identity \( \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \). We can convert one \( \sin^{-1} \) term into a \( \cos^{-1} \) term to see if they share the same argument.

Step 1: Convert \( \sin^{-1}(1/3) \) to \( \cos^{-1} \).
Let \( \alpha = \sin^{-1} \left( \frac{1}{3} \right) \). Then \( \sin \alpha = \frac{1}{3} \). Using the identity \( \sin^2 \alpha + \cos^2 \alpha = 1 \): \[ \cos^2 \alpha = 1 - \left( \frac{1}{3} \right)^2 = 1 - \frac{1}{9} = \frac{8}{9} \] \[ \cos \alpha = \sqrt{\frac{8}{9}} = \frac{2\sqrt{2}}{3} \] So, \( \alpha = \cos^{-1} \left( \frac{2\sqrt{2}}{3} \right) \).

Step 2: Substitute back into the expression.
The original expression is: \[ \sin^{-1} \left( \frac{2\sqrt{2}}{3} \right) + \sin^{-1} \left( \frac{1}{3} \right) \] Substituting our result from
Step 1: \[ \sin^{-1} \left( \frac{2\sqrt{2}}{3} \right) + \cos^{-1} \left( \frac{2\sqrt{2}}{3} \right) \]

Step 3: Apply the identity.
Using \( \sin^{-1}(x) + \cos^{-1}(x) = \frac{\pi}{2} \): \[ \frac{\pi}{2} \]