sinA+sinB+sinY=0 ,cosA+cosB+cosY=0,求cos(B-Y)的值.
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sinA+sinB+sinY=0 ,cosA+cosB+cosY=0,求cos(B-Y)的值.
sinA+sinB+sinY=0
-sinA=sinB+sinY
两边平方:
(sinA)^2=(sinB)^2+2sinBsinY+(sinY)^2--------1
cosA+cosB+cosY=0
-cosA=cosB+cosY
两边平方:
(cosA)^2=(cosB)^2+2cosBcosY+(cosY)^2--------2
1+2:
2(sinBsinY+cosBcosY)=-1
2cos(B-Y)=-1
cos(B-Y)=-1/2
-sinA=sinB+sinY
两边平方:
(sinA)^2=(sinB)^2+2sinBsinY+(sinY)^2--------1
cosA+cosB+cosY=0
-cosA=cosB+cosY
两边平方:
(cosA)^2=(cosB)^2+2cosBcosY+(cosY)^2--------2
1+2:
2(sinBsinY+cosBcosY)=-1
2cos(B-Y)=-1
cos(B-Y)=-1/2
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