Question

sin[π/3 -sin inverse (-1/2)] :​

Answer 1

Answer:

The answer is 1

Step-by-step explanation:

         $\sin[\frac{\pi}{3}-\sin^{-1}(-\frac{1}{2})]=\sin[\frac{\pi}{3}-(-\frac{\pi}{6})]\\=\sin(\frac{\pi}{3}+\frac{\pi}{6})=\sin\frac{\pi}{2}=1$

• Since  $\sin(-\frac{\pi}{6})=-\frac{1}{2}\Rightarrow \sin^{-1}(-\frac{1}{2})=-\frac{\pi}{6}$