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if sinA+sinB+sinC=0 and cosA+cosB+cosC=0 then find the value of Sin(sqr)A+Sin(sqr)B+Sin(sqr)C.
please pleaseplease pleasepleaseplease
Hi Alok,
SinA + SinB = -Sinc ---------(1)
CosA + CosB = -CosC ---------(2)
Sq and add (1),(2)
2+2Cos(A-B) = 1 ------------(3)
Sq and Sub (2),(1)
Cos2A + Cos2B + 2Cos(A+B) = Cos2C
2Cos(A+B)Cos(A-B) + 2Cos(A+B) = Cos2C
or Cos(A+B){2Cos(A-B) + 2} = Cos2C
or Cos(A+B) = Cos2C ----------- [Using (3)]
Now SinA+SinB+SinC = 0
Squaring, Sin2A+Sin2B+Sin2C = -2∑SinASinB ----------(4)
Now Cos(A+B) = Cos2C
Similarly Cos(B+C) = Cos2A
and Cos(C+A) = Cos2B
So adding these, ∑CosACosB - ∑SinASinB = ∑Cos2A
Also Cos(A-B) = -1/2 [from (3)]
similarly Cos(B-C) = -1/2
and Cos(C-A) = -1/2
Adding ∑CosACosB + ∑SinASinB = -3/2
So -2∑SinASinB = ∑Cos2A + 3/2
Hence from (4) Sin2A+Sin2B+Sin2C = -2∑SinASinB = ∑Cos2A + 3/2
Now Cos2A+Cos2B+Cos2C = 2Cos(A+B)Cos(A-B) + Cos2C = -Cos(A+B) + Cos2C = 0 ------ [because Cos(A+B) = Cos2C; and Cos(A-B) = -1/2 which was proved previously]
Hence ∑Cos2A = 0.
Hence Sin2A+Sin2B+Sin2C = 3/2.
Note that you can have so many results from the above conditions, like
Sin2A+Sin2B+Sin2C = 3/2
∑Cos2A = 0
∑SinASinB = -3/4
∑CosACosB = -3/4
Hope this helps.
Best Regards,
Ashwin (IIT Madras).
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