已知sin(5π-θ)+sin(5π/2-θ)=根号7/2
题目
已知sin(5π-θ)+sin(5π/2-θ)=根号7/2
求(1)[sin(π/2+θ)]^3-[cos(3π/2-θ)]^3
(2))[sin(π/2-θ)]^4-[cos(7π/2+θ)]^4
答案
根据诱导公式,sin(5π-θ)+sin(5π/2-θ)=sinθ+cosθ=根号7/2
(sinθ+cosθ)^2=7/4
(sinθ)^2+(cosθ)^2+2sinθcosθ=7/4
2sinθcosθ=3/4,sinθcosθ=3/8
(1)[sin(π/2+θ)]^3-[cos(3π/2-θ)]^3=(cosθ)^3+(sinθ)^3
=(sinθ+cosθ)[(sinθ)^2-sinθcosθ+(cosθ)^2]=(5倍根号7)/16
(2)[sin(π/2-θ)]^4-[cos(7π/2+θ)]^4=(cosθ)^4-(sinθ)^4
=[(sinθ)^2+(cosθ)^2][(sinθ)^2-(cosθ)^2]
=(sinθ+cosθ)(sinθ-cosθ)
=(sinθ+cosθ)*根号[(sinθ+cosθ)^2-4sinθcosθ]=(根号7)/4
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