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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```
7 years ago

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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