Question

sin[ \(\frac { π }{ 3 }\)-sin^-1(-\(\frac { 1 }{ 2 }\)) ] is equal to
(A) \(\frac { 1 }{ 2 }\)
(B) \(\frac { 1 }{ 3 }\)
(C) \(\frac { 1 }{ 4 }\)
(D) 1

Answer 1

(B) \(\frac { 1 }{ 3 }\)

---

sin^-1(-\(\frac { 1 }{ 2 }\)) = sin^-1sin ( – \(\frac { π }{ 6 }\) ) = – \(\frac { π }{ 6 }\)
∴ sin[\(\frac { π }{ 3 }\)-sin^-1(-\(\frac { 1 }{ 2 }\)) ] = sin[ \(\frac { π }{ 3 }\) – (-\(\frac { π }{ 6 }\) )
= sin( \(\frac { π }{ 3 }\) + \(\frac { π }{ 6 }\))
= sin \(\frac { π }{ 2 }\) = 1.