已知:tan(-5π-θ)•cos(θ-2π)•sin(-3π-θ)tan(7π2+θ)•sin(-4π+θ)•cot(-θ-π2)+2tan(6π-θ)•cos(-π+θ)=2,则sin(θ+3π)

题目简介

已知:tan(-5π-θ)•cos(θ-2π)•sin(-3π-θ)tan(7π2+θ)•sin(-4π+θ)•cot(-θ-π2)+2tan(6π-θ)•cos(-π+θ)=2,则sin(θ+3π)

题目详情

已知:
tan(-5π-θ)•cos(θ-2π)•sin(-3π-θ)
tan(
2
+θ)•sin(-4π+θ)•cot(-θ-
π
2
)
+2tan(6π-θ)•cos(-π+θ)
=2,则sin(θ+3π)=______.
题型:填空题难度:中档来源:不详

答案

已知等式变形得:class="stub"-tanθcosθsinθ
-cotθsinθtanθ
+2tanθcosθ=sinθ+2sinθ=2,
∴sinθ=class="stub"2
3

则sin(θ+3π)=-sinθ=-class="stub"2
3

故答案为:-class="stub"2
3

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