证明：(sinθ)^2+(sinφ)^2-2sinθsinφcos(θ-φ)=[sin(θ-φ]^2

波斯猫猫 · 发表于 2023-4-19 20:55

证明：(sinθ)^2+(sinφ)^2-2sinθsinφcos(θ-φ)=[sin(θ-φ)]^2.

(sinθ)^2+(sinφ)^2-2sinθsinφcos(θ-φ)-[sin(θ-φ)]^2

=(sinθ)^2+(sinφ)^2+cos(θ-φ)[cos(θ-φ)-2sinθsinφ]-1

=(sinθ)^2+(sinφ)^2+cos(θ-φ)cos(θ+φ)-1

=(sinθ)^2+(sinφ)^2+(cosθcosφ)^2-(sinθsinφ)^2-1

=(sinθ)^2+(sinφ)^2(cosθ)^2+(cosθcosφ)^2-1

=(sinθ)^2+(cosθ)^2-1=0

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证明：(sinθ)^2+(sinφ)^2-2sinθsinφcos(θ-φ)=[sin(θ-φ]^2

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