Find the values of \(\sin(\frac{\pi}{3}-\sin^{-1}(-\frac{1}{2}))\) is equal to
\(\frac{1}{2}\)
\(\frac{1}{3}\)
\(\frac{1}{4}\)
1
Let \(\sin^{-1}(-\frac{1}{2})=x\)
Then, sin x=\(-\frac{1}{2}=-sin\frac{\pi}{6}=\sin(-\frac{\pi}{6}).\)
We know that the range of the principal value branch of sin-1=[-π/2,π/2].
sin-1(-1/2)=-π/6
therefore sin(π/3-sin1(-1/2))=sin(π/3+π/6)=sin(3π/6)=sin(π/2)=1.
The correct answer is D.
The elementary properties of inverse trigonometric functions will help to solve problems. Here are a few important properties related to inverse trigonometric functions:
Tan−1x + Tan−1y = π + tan−1 (x+y/ 1-xy), if xy > 1
Tan−1x + Tan−1y = tan−1 (x+y/ 1-xy), if xy < 1
Tan−1x + Tan−1y = tan−1 (x+y/ 1-xy), if xy < 1
Tan−1x + Tan−1y = -π + tan−1 (x+y/ 1-xy), if xy > 1
= x, if x∈[−π/2, π/2]
= π−x, if x∈[π/2, 3π/2]
=−2π+x, if x∈[3π/2, 5π/2] And so on.
= −x, ∈[−π,0]
= x, ∈[0,π]
= 2π−x, ∈[π,2π]
=−2π+x, ∈[2π,3π]
= x, (−π/2, π/2)
= x−π, (π/2, 3π/2)
= x−2π, (3π/2, 5π/2)