Question

求助，谢谢，麻烦说答案的同时说一下过程

Answer 1

解：
cos(α+π/2)=4/5
⇒cos(π-α-π/2)=-4/5
⇒cos(π/2-α)=-4/5
⇒sinα=-4/5
⇒cosα=±3/5
∴
sin2α=2sinαcosα=±24/25
cos2α=2cos²α-1=-7/25
又∵
sin(π/12)
=sin(π/4-π/6)
=sin(π/4)cos(π/6)-cos(π/4)sin(π/6)
=(√6-√2)/4
cos(π/12)
=cos(π/4-π/6)
=cos(π/4)cos(π/6)+sin(π/4)sin(π/6)
=(√6+√2)/4
sin(2α+π/12)
=sin2αcos(π/12)+cos2αsin(π/12)
=(±24/25)*(√6+√2)/4+(-7/25)(√6-√2)/4
=(17√6+31√2)/100
或(-31√6-17√2)/100