Question

已知cosA = cosθ×sinC,cosB = sinθ×sinc,求(sinA)^2+(sinB)^2+(sinC)^2的值

Answer 1

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

Answer 2

(sinA)^2=1-(cosA)^2 (sinB)^2=1-(cosB)^2
原式=1-(cosθ×sinC)^2+1-(sinθ×sinc)^2+(sinC)^2
=2-(cos²θ+sin²θ)×(sinC)^2+(sinC)^2
=2-(sinC)^2+(sinC)^2
=2

Answer 3

(sina)^2+(sinb)^2+(sinc)^2
=1-(cosa)^2+1-(cosb)^2+(sinc)^2
=2-(cosθ×sinC)^2-(sinθ×sinc)^2+(sinC)^2
=2-(sinC)^2[(cosθ)^2+(sinθ)^2-1]
=2-0
=2

Answer 4

(sinA)^2+(sinB)^2+(sinC)^2=1-（cosA）^2+1-(sinB)^2+(sinC)^2=2-(cosθ×sinC)^2-(sinθ×sinc)^2+(sinC)^2=2