Question

已知sin(a-π/4）=7根号2/10，cos2a=7/25,求sina及tan(a+π/3）

Answer 1

sin(a-π/4）=7√2/10
sinacosπ/4-sinπ/4cosa=7√2/10
sina-cosa=7/5 (1)
cos2a=7/25
cos²a-sin²a=7/25
(cosa+sina)(cosa-sina)=7/25
sina+cosa=-1/5 (2)
(1)+(2)得2sina=6/5
sina=3/5
(1)-(2)得2cosa=8/5
cosa=4/5
∴tana=3/4
tan(a+π/3）=(tana+tanπ/3)/(1-tanatanπ/3)
=-(48+25√3)/11

Answer 2

1)cos2a=-sin(2a-π/2)=-2sin(a-π/4)cos(a-π/4)=7/25
sin(a-π/4)=7/5√2,∴cos(a-π/4)=-1/5√2,
tan(a-π/4)=-7
sina=sin(a-π/4+π/4)=sin(a-π/4)/√2+cos(a-π/4)/√2=3/5
tana=tan(a-π/4+π/4)=[tan(a-π/4)+1]/[1-tan(a-π/4)]=-6/8=-3/4
tan(a+π/3)=(tana+√3)/(1-√3tana)=(4√3-3)/(4+3√3)=(48-25√3)/11
2)2sin(x+π/3)=-a在(0,2π)内有2不同解
结合图形-2<-a<2且-a≠√3,
即-2<a<-√3或-√3<a<2
a和b关于x=π/6或5π/6对称,
∴a+b=2*π/6=π/3或a+b=2*5π/6=5π/3
tan(a+b)=tan(π/3)=√3,tan(a+b)=tan(5π/3)=-√3