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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

6 years ago


Answers : (1)


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


  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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6 years ago

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