Question

已知tana=2,则sin(3π/2-2a)=

Answer 1

∵tana=2
∴sin(3π/2-2a)=sin(3π/2-2a-π）=-sin(π/2-2a)
-sin(π/2-2a)=-（sinπ/2cos2a-cosπ/2sin2a)
∵sinπ/2=1,cosπ/2=0
∴-sin(π/2-2a)=-cos2a
=-2cos²a+1
∵tana=2,∴cosa=±√5/5
∴cos²a=1/5
∴-sin(π/2-2a)=-2cos²a+1
=3/5

Answer 2

解：∵tana=2
∴sin(3π/2-2a)
=-cos(2a)
=-(1-tan^2a)/（1+tan^2a）
=-(1-4)/(1+4)
=3/5

Answer 3

tana=2
sina/cosa=2
sina=2cosa
(sina)^2+(cosa)^2=5(cosa)^2=1
(cosa)^2=1/5

sin(3π/2-2a)
=-sin(2a+π/2)
=cos(2a)
=2(cosa)^2-1
=2/5-1
=-3/5