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Andrew
10.05.11, 20:28
As, A+B+C = π, a[sin(π+(2C+A))] + b[sin(π+(2A+B))] +
c[sin(π+(2B+C))] = 0 Solving for just first part. Solve similarly for
rest. a[sin(π+(2C+A))] => a sinπ cos(2C+A) + a cosπ sin(2C+A) As,
Cosπ and sinπ both equal zero. The whole statement is zero.
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As, A+B+C = π, a[sin(π+(2C+A))] + b[sin(π+(2A+B))] + c[sin(π+(2B+C))] = 0 Solving for just first part. Solve similarly for rest. a[sin(π+(2C+A))] => a sinπ cos(2C+A) + a cosπ sin(2C+A) As, Cosπ and sinπ both equal zero. The whole statement is zero.
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