已知sin(5π-θ)+sin(5π/2-θ)=根号7/2求[sin(π/2+θ）]^2-[cos(3π/2-θ)]^2的值
问题描述：
已知sin(5π-θ)+sin(5π/2-θ)=根号7/2
求[sin(π/2+θ）]^2-[cos(3π/2-θ)]^2的值
答
解:sin(5pai-0)＋sin(5pai/2-0)=sin0＋cos0=√7/2,因为sin^20＋cos^20-1,所以sin0-cos0=±1/2所以[sin(pai/2＋0)]^2-[cos(3pai/2-0)]^2=cos^20-sin^20=(sin0＋cos0)(sin0-cos0)=±√7/4。
答
sin(5π-θ)+sin(5π/2-θ)=√7/2
sinθ+cosθ=√7/2,1+sin2θ=7/4,sin2θ=3/4 ,cos2θ=±√7/4
sin(π/2+θ）]^2-[cos(3π/2-θ)]^2=cos^2θ-sin^2θ=cos2θ=±√7/4