求学霸请教,要写过程哦,谢谢！/
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解：(1)
原式=(sin²1°+sin²89°)+(sin²2°+sin²88°)+
(sin²3°+sin²87°)+…+(sin²44°+sin²46°)
+sin²45°
=(sin²1°+cos²1°)+(sin²2°+cos²2°)+
(sin²3°+cos²3°)+…(sin²44°+cos²44°)
+(√2/2)²
=44+1/2=89/2
(2)∵cos(π/6-θ)=α
∴sin(π/3+θ)=α
则sin(2π/3-θ)=sin[π-(π/3+θ)]=sin(π/3+θ)=α
cos(5π/6+θ)=-cos[π-(π/6-θ)]
=-cos(π/6-θ)=-α
注：sinα=cos(π/2-α)
sin²α+cos²α=1
原式=(sin²1°+sin²89°)+(sin²2°+sin²88°)+
(sin²3°+sin²87°)+…+(sin²44°+sin²46°)
+sin²45°
=(sin²1°+cos²1°)+(sin²2°+cos²2°)+
(sin²3°+cos²3°)+…(sin²44°+cos²44°)
+(√2/2)²
=44+1/2=89/2
(2)∵cos(π/6-θ)=α
∴sin(π/3+θ)=α
则sin(2π/3-θ)=sin[π-(π/3+θ)]=sin(π/3+θ)=α
cos(5π/6+θ)=-cos[π-(π/6-θ)]
=-cos(π/6-θ)=-α
注：sinα=cos(π/2-α)
sin²α+cos²α=1
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