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solution-B(sq) minus A(sq) by 2
Hi Nanda,
Let SinA + SinB = a ------------(1)
and CosA + CosB = b ------------(2)
Sq the two eqns, and subtract (2)^2 - (1)^2
You will get, Cos2A+Cos2B+2Cos(A+B) = b2-a2.
So, 2Cos(A+B)Cos(A-B)+2Cos(A+B) = b2-a2.
or Cos(A+B){2Cos(A-B)+2} = b2-a2 ----------------(3)
Now Sq the two eqns and add, ie (2)^2+(1)^2
You will get 2+2Cos(A-B) = a2+b2 -----------[Use this in (3)]
You'd get Cos(A+B){a2+b2) = (b2-a2).
or Cos(A+B) = (b2-a2)/{a2+b2)
which is the answer.
Best Regards,
Ashwin (IIT Madras).
(sinA + sinB)2=a2 __1 (cosA + Cos B)2 =b2 ___ 2
add 1 & 2 to get value of cos(A-B)
subtract 1 from 2 to get a relation between cos(A+B) & cos(A-B) then substitute value of cos(A-b)
cos(A+B)=b2 - a2 / b2 + a2
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