sinX+sinY=1 cosX+cosY=0 求cos(2X)+cos(2Y)的值
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sinX+sinY=1 cosX+cosY=0 求cos(2X)+cos(2Y)的值
![sinX+sinY=1 cosX+cosY=0 求cos(2X)+cos(2Y)的值](/uploads/image/z/7049740-4-0.jpg?t=sinX%2BsinY%3D1+cosX%2BcosY%3D0+%E6%B1%82cos%282X%29%2Bcos%282Y%29%E7%9A%84%E5%80%BC)
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sinx=1-siny----(1 cosx=-cosy-----(2
(1)式^2+(2)式^2-------- 1=(1-siny)^2+cosy^2
1=1+sin^2y+cos^2y-2siny
1=1+1-2siny
siny=1/2
sinx=1-1/2=1/2
所以cos2x=1-2sin^2x=1-2(1/2)^2=1/2
同理cos2y=1/2
即cos2x+cos2y=1
sinx=1-siny----(1 cosx=-cosy-----(2
(1)式^2+(2)式^2-------- 1=(1-siny)^2+cosy^2
1=1+sin^2y+cos^2y-2siny
1=1+1-2siny
siny=1/2
sinx=1-1/2=1/2
所以cos2x=1-2sin^2x=1-2(1/2)^2=1/2
同理cos2y=1/2
即cos2x+cos2y=1
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