Question

If sin A + sin B + sin C = 0, show that
sin 3A + sin 3B + sin 3C + 12 sin A sin B sin C = 0.​

Answer 1

Given Question :

If sin A + sin B + sin C = 0, show that

sin 3A + sin 3B + sin 3C + 12 sin A sin B sin C = 0.

Identities Used :-

We know,

If a + b + c = 0, then

$\sf \:  {a}^{3} + {b}^{3} + {c}^{3} = 3abc$

Also we know,

$\sf \:  sin3x = 3 \: sinx - 4 {sin}^{3} x$

Step by step explanation :-

☆ Since, sinA + sinB + sinC = 0

$\tt\implies \: {sin}^{3} A + {sin}^{3} B + {sin}^{3} C = 3sinA \: sinB \: sinC - - - (1)$

$\tt \:  Consider \: sin3A + sin3B + sin3C + 12sinAsinBsinC$

$\sf \:  = 3sinA - 4 {sin}^{3} A + 3sinB - 4 {sin}^{3} B + \\ \sf \: 3sinC - 4 {sin}^{3} C + 12sinAsinBsinC$

$\sf \:  = 3(sinA + sinB + sinC) - 4( {sin}^{3} A + {sin}^{3} B + {sin}^{ 3} C) \\ \sf \: \sf \: 12sin \: A \: sin \: B \: sinC \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \:$

$\tt \:  = 3 \times 0 - 4(3sinAsinBsinC) + 12sinAsinBsinC$

$\tt \:  = 0 - 12sinAsinBsinC + 12sinAsinBsinC$

$\tt \:  = \: 0$

$\large{\boxed{\boxed{\bf{Hence, Proved}}}}$

Answer 2

Answer:

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