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manoj jangra Grade: 12
        

1.If cos(b-c).cos(c-a).cos(a-b) + 1 = 0.


the prove that at least one of the angles (a-b),(b-c) and (c-a) is and odd multiple of ∏


2.A + B + C = 180 prove that cos A/2 + cos B/2 + cos C/2 = 4cos(∏-A)/4 . cos(∏-B)/4 . cos(∏-C)/4

7 years ago

Answers : (1)

Badiuddin askIITians.ismu Expert
147 Points
										

Dear manoj


cos(b-c).cos(c-a).cos(a-b) + 1 = 0.


or   cos(b-c).cos(c-a).cos(a-b) =-1


only possible if


 cos(b-c)=cos(c-a)=cos(a-b) =-1


or


  cos(b-c)=cos(c-a)=1   and  cos(a-b) =-1


  cos(b-c)=1   , cos(c-a)=-1  and  cos(a-b) =1


  cos(b-c)=-1   ,  cos(c-a)=  cos(a-b) =1


in any case atleast one of three term will be eqaul to -1


 let cos(b-c) =-1


             b-c = (2n+1)


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7 years ago
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