Question

已知sin(a/2+π/6)=1/4求sin(π/6+a)

Answer 1

∵sin(½a+π/6)＝¼,

∴(½a+π/6)∈(2kπ,2kπ+π/6)U(2kπ+5π/6,2kπ+π),

(a+π/3)∈(4kπ,4kπ+π/3)U(4kπ+5π/3π,4kπ+2π)

∴cos(a+π/3)＝1-2sin²(½a+π/6)＝7/8,

sin(a+π/3)=±√1-cos²(a+π/3)=±√15/8.

∴sin(a+π/6)=sin[(a+π/3)-π/6]

=sin(a+π/3)cos(π/6)-cos(a+π/3)sin(π/6)

=(±√15/8)·(√3/2)-(7/8)·(1/2)

=(±3√5-7)/16.