Question

设a为锐角，若cos(a+π/6)=4/5.则sin(2a+π/12)=?

Answer 1

∵0<α<π/2 ∴π/6<α+π/6<2π/3
又cos(α+π/6)=4/5 ∴π/6<α+π/6<π/2
∴sin(α+π/6)=3/5
∴sin[2(α+π/6)]=2sin(α+π/6)cos(α+π/6)=2*3/5*4/5=24/25
cos[2(α+π/6)]=1-2sin²(α+π/6)=1-2*(3/5)²=7/25
又2α+π/12=2(α+π/6)-π/4
∴sin(2α+π/12)=sin[2(α+π/6)-π/4]=sin[2(α+π/6)]cosπ/4-cos[2(α+π/6)]sinπ/4
=(24/25)*(√2/2)-(7/25)*(√2/2)=(17√2)/50

Answer 2

sin（2a+π/12）=sin【2(a+π/6) — π/4】
=sin2(a+π/6)cos(π/4）—cos2(a+π/6)sin(π/4）
因为cos(a+π/6)=4/5,a为锐角,
所以0