Question

已知sin(α+π/2)=-√5/5,α∈(0,π)

求1):[sin(α-π/2)-cos(3π/2+α)]/[sin(π-α)+cos(3π+α)]
2):cos(2α-3π/4)

Answer 1

sin(α+π/2)=cosa=-√5/5,所以a∈(π/2,π)
所以sina=2√5/5
[sin(α-π/2)-cos(3π/2+α)]/[sin(π-α)+cos(3π+α)]
=[cosa+sina]/[sina-cosa]
=（√5/5）/（3√5/5）
=1/3
cos(2α-3π/4)
=cos2acos3π/4+sin2asin3π/4
={1-2（sina）^2}*｛(-根号2）/2｝+2sinacosa*｛（根号2）/2｝
=（3根号2）/10-（4根号2）/10
=(-根号2)/10
如有不明白，可以追问！！
谢谢采纳！