若a∈[5π/2,7π/2],则sqrt(1+sina)+sqrt(1-sina)=?A.2cos(a/2) B.-2cos(a/2) C.2sin(a/2) D.-2sin(a/2)sqrt根号

若a∈[5π/2,7π/2],则sqrt(1+sina)+sqrt(1-sina)=?A.2cos(a/2) B.-2cos(a/2) C.2sin(a/2) D.-2sin(a/2)sqrt根号
若a∈[5π/2,7π/2],则sqrt(1+sina)+sqrt(1-sina)=?
A.2cos(a/2) B.-2cos(a/2) C.2sin(a/2) D.-2sin(a/2)
sqrt根号

若a∈[5π/2,7π/2],则sqrt(1+sina)+sqrt(1-sina)=?A.2cos(a/2) B.-2cos(a/2) C.2sin(a/2) D.-2sin(a/2)sqrt根号
[sqrt(1+sina)+sqrt(1-sina)]^2=1+sina+2sqrt[(1+sina)(1-sina)]+1-sina=2+2sqrt[1-(sina)^2]=2+2sqrt(cosa)^2
a∈[5π/2,7π/2],cosa<0
2+2sqrt(cosa)^2=2-2cosa=2(1-cosa)=2{1-1+2[sin(a/2)]^2}=4[sin(a/2)]^2
a/2∈[5π/4,7π/4],sin(a/2)<0
所以,原式=-2sin(a/2)
选D