已知sin(3π+θ)=1/3,求cos(π+θ)cosθ[cos(π−θ)−1]+cos(θ−2π)sin(θ−3π2)cos(θ−π)−sin(3π2+θ)的值.

问题描述:

已知sin(3π+θ)=

1
3
,求
cos(π+θ)
cosθ[cos(π−θ)−1]
+
cos(θ−2π)
sin(θ−
2
)cos(θ−π)−sin(
2
+θ)
的值.

∵sin(3π+θ)=-sinθ=

1
3

∴sinθ=-
1
3

原式=
−cosθ
cosθ(−cosθ−1)
+
cos(2π−θ)
−sin(
2
−θ)cos(π−θ)+cosθ

=
1
1+cosθ
+
cosθ
cos2θ +cosθ
=
1
1+cosθ
+
1
1−cosθ
=
2
1−cos2θ
=
2
sin2θ
=
2
(−
1
3
)
2
=18.