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

来源：学生作业帮助网编辑：作业帮时间：2024/05/06 09:24:48

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

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

(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

已知cosA = cosθ×sinC,cosB = sinθ×sinc,求(sinA)^2+(sinB)^2+(sinC)^2的值已知cosA=cosθ*sinC,cosB=sinθ*sinC,求sin²A+sin²B+sin²C如题求得是sinA平方+sinB平方+sinC平方的值已知cosA=cosθ*sinC,cosB=sinθ*sinC,求sin²A+sin²B+sin²C求得是sinA平方+sinB平方+sinC平方的值 cos^2A - cos^2B + sin^2C=2cosA *sinB *sinC证明已知sinA+sinB=sinC,cosA+cosB=cosC,求cos(A-B)的值已知sinA+sinB=sinC,cosA+cosB=cosC,求cos(A-B)的值 sinA+sinB+sinC=0; cosA+cosB+cosC=0,求cos(B-C)的值? sina+sinb+sinc=0,cosa+cosb+cosc=0,求cos(B-C)的值? sinA+sinB+sinc=0 cosA+cosB+cosC=0 cos(B-C) cos^2A - cos^2B + sin^2C=2cosA *sinB *sinCABC为三角形,利用(cosA)平方-(cosB)平方=sin(A+B)*sin(B-A)证明cosA平方 - cosB平方 + sinC平方=2cosA *sinB *sinC 已知sinA+sinB+sinC=0,cosA+cosB+cosC=0,求cos(A－B)的值已知sina+sinb+sinc=0且cosa+cosb+cosc=0 求cos(a-b)的值已知sina+sinb+sinc=0,cosa+cosb+cosc=0,则cos(a-b)的值是? 三角形ABC中,已知sinC=(sinA+sinB)/(cosA+cosB),求C 三角形ABC中 ,已知a/ √ 3cosA=c/sinC .(1 求A的大小 sina+sinb+sinc=0 cosa+cosb+cosc=0求证cos*2a+cos*2b+cos*2c=3|2 在△ABC中,sinA+sinB+sinC=0,cosA+cosB+cosC=0,求证：cos²A+cos²B+cos²C=3/2 在△ABC中,sinA+sinB+sinC=0,cosA+cosB+cosC=0.求证：cos²A+cos²B+cos²c=3／2.