已知cosA = cosθ×sinC,cosB = sinθ×sinc,求(sinA)^2+(sinB)^2+(sinC
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已知cosA = cosθ×sinC,cosB = sinθ×sinc,求(sinA)^2+(sinB)^2+(sinC)^2的值
cosA = cosθ×sinC ∴(cosA )^2= cos²θ×sin²C
∴(sinA)^2=1-(cosA )^2=1- cos²θ×sin²C
同理(sinB)^2=1-sin²θ×sin²C
∴(sinA)^2+(sinB)^2+(sinC)^2
=1- cos²θ×sin²C+1-sin²θ×sin²C+(sinC)^2
=2-(sinC)^2[1-sin²θ-cos²θ]
=2-(sinC)^2*0
=2
∴(sinA)^2=1-(cosA )^2=1- cos²θ×sin²C
同理(sinB)^2=1-sin²θ×sin²C
∴(sinA)^2+(sinB)^2+(sinC)^2
=1- cos²θ×sin²C+1-sin²θ×sin²C+(sinC)^2
=2-(sinC)^2[1-sin²θ-cos²θ]
=2-(sinC)^2*0
=2
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