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)=
,求1 3
+cos(π+θ) cosθ[cos(π−θ)−1]
的值. cos(θ−2π) sin(θ−
)cos(θ−π)−sin(3π 2
+θ)3π 2
答
∵sin(3π+θ)=-sinθ=
,1 3
∴sinθ=-
,1 3
原式=
+−cosθ cosθ(−cosθ−1)
cos(2π−θ) −sin(
−θ)cos(π−θ)+cosθ3π 2
=
+1 1+cosθ
=cosθ −cos2θ +cosθ
+1 1+cosθ
=1 1−cosθ
=2 1−cos2θ
=2
sin2θ
=18.2 (−
)21 3