Question:
\(\sin\left[\dfrac{\pi}{3} – \sin^{-1}\left(-\dfrac{1}{2}\right)\right]\) is equal to
\(\sin\left[\dfrac{\pi}{3} – \sin^{-1}\left(-\dfrac{1}{2}\right)\right]\) किसके बराबर है?
Options:
- \(\dfrac{1}{4}\)
- 1
- \(\dfrac{1}{2}\)
- \(\dfrac{1}{3}\)
- \(\dfrac{1}{4}\)
- 1
- \(\dfrac{1}{2}\)
- \(\dfrac{1}{3}\)
Correct Answer: b)
Explanation:
\(\sin^{-1}(-1/2) = -\dfrac{\pi}{6}\).
\(\sin\left[\dfrac{\pi}{3} – \left(-\dfrac{\pi}{6}\right)\right] = \sin\left(\dfrac{\pi}{3} + \dfrac{\pi}{6}\right) = \sin\left(\dfrac{\pi}{2}\right) = 1\).
\(\sin^{-1}(-1/2) = -\dfrac{\pi}{6}\)।
\(\sin\left[\dfrac{\pi}{3} + \dfrac{\pi}{6}\right] = \sin\left(\dfrac{\pi}{2}\right) = 1\)।