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2015-04-05T15:08:07+05:30

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3/5  Sin A  + 4/5  Cos A  = 1
we see that  (3/5)² + (4/5)² = 1
let  3/5 = Sin B            and      4/5 = Cos B
 So Sin A Sin B + Cos A Cos B = 1 = Cos 0°
       Cos (A - B) = cos 0°
       A = B    or    A - B = 2 π

   if A = B,  Sin A = Sin B = 3/5.
 or,  if A - B = 2π,  Sin A = Sin (2π+B) = Sin B = 3/5

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alternately:

  4 Cos A = 5 - 3 Sin A
  16 Cos² A = 25 + 9 Sin² A - 30 Sin A
   16 - 16 Sin² A = 25 + 9 Sin² A - 30 Sin A

      25 Sin² A  - 30 Sin A + 9 = 0

     Sin A = [ 30 +- √(900 - 900) ] / 50 = 3/5


1 5 1
Thanx a lot.. u helped me very much.. just one ques how did sinAsinB+ cosAcosB become cos(A-B)?
that is the formula for Cos (A - B).
Sin A Cos B + Cos A Sin B = Sin (A+B) is another formul
Kk.. thanx