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As, A+B+C = π,a[sin(π+(2C+A))] + b[sin(π+(2A+B))] + c[sin(π+(2B+C))] = 0Solving 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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Andrew10.05.11, 20:28As, 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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