if X+y =60 and coax + cosy = 1 then cos(X-y) =?
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given that,
cosx+cosy=1
2{cos(x+y)/2.cos(x-y)/2}=1
cos60/2.cos(x-y)/2=1/2 [x+y=60]
√3/2.cos(x-y)/2=1/2
cos(x-y)/2=1/√3
{cos(x-y)/2}^2=(1/√3)^2 [both sides
make square]
2{cos(x-y)/2}^2=2(1/√3)^2 [both sides
multiple by 2]
[2{cos(x-y)/2}^2]-1=[2(1/√3)^2]-1 [both sides
subtract by 1]
cos(x-y)=2/3-1
cos(x-y)=-1/3
cosx+cosy=1
2{cos(x+y)/2.cos(x-y)/2}=1
cos60/2.cos(x-y)/2=1/2 [x+y=60]
√3/2.cos(x-y)/2=1/2
cos(x-y)/2=1/√3
{cos(x-y)/2}^2=(1/√3)^2 [both sides
make square]
2{cos(x-y)/2}^2=2(1/√3)^2 [both sides
multiple by 2]
[2{cos(x-y)/2}^2]-1=[2(1/√3)^2]-1 [both sides
subtract by 1]
cos(x-y)=2/3-1
cos(x-y)=-1/3
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