Question

高一数学已知sin(π/2-b)*cos(a+b)-sin(π+b)*sin(a+b)=3/5其中a∈(3π/2,2π)求tan(π/4-a/2)

Answer 1

解
sin(π/2-b)*cos(a+b)-sin(π+b)*sin(a+b)=3/5
即
cosbcos(a+b)+sinbsin(a+b)
=cos[b-(a+b)]
=cos(-a)
=cosa
∴cosa=3/5
∵a∈(3π/2,2π)
∴sina<0
∴sina=-√1-(3/5)²=-4/5
∴tana=-4/3
又tana=(2tana/2)/(1-tan²a/2)=-4/3
∴4tan²a/2-4=6tana/2
即2tan²a/2-3tana/2-2=0
(2tana/2+1)(tana/2-2)=0
∴tana/2=2(舍去)
或tana/2=-1/2

∴tan(π/4-a/2)
=(1-tana/2)/(1+tana/2)
=(1+1/2)/(1-1/2)
=3/2×2
=3