Question

(Sin3A+sinA)SinAi +(cos3A-cos) cosA=0

Answer 1

=> (sin3A + sinA)sinA + (Cos3A - cosA)CosA = 0

=> sin3A.sinA + sin²A + cos3A.cosA - cos²A = 0

=> sin3A.sinA + Cos3A.CosA -(cos²A-sin²A) = 0

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USIND TRIGONOMETRIC IDENTITY :-

=> cos(a-b) = sina.sinb + cosa.cosb

=> cos²a - sin²a = cos2a
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THEN,

=> cos(3A-A) - cos2A = 0

=> cos2A - cos2A = 0

=> 0 = 0
________[PROVED]

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_-_-_-_✌☆$BE \: \: \: \: BRAINLY$☆✌_-_-_-_

Answer 2

Heya!!
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➡Trigonometry ⬅
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$= > (sin \: 3a + sin \: a \: )sin \: a \: + (cos \: 3a - cos \: a)cos \: a = 0 \\ \\ = > > using \: identity \: ... \\ \\ |1| .sin \: 3a = 3sin \: a \: - 4sin {}^{3} a \\ \\ |2| .cos \: 3a = 4cos {}^{3} a - 3cos \: a \: \\ \\ = > > we \: have \: \\ \\ (3 \: sin \: a \: - 4sin {}^{3} a + sin \: a \: )sin \: a \: + (4cos {}^{3} a - 3cos \: a \: - \: cos \: a \: )cos \: a \: \\ \\ \\ = 4 \: sin {}^{2} a(1 - sin {}^{2} a) - 4cos {}^{2} a(1 - cos {}^{2} a) \\ \\ = 4sin {}^{2} a \: \: cos {}^{2} a - 4 \: cos {}^{2} a \: \: sin {}^{2} a \\ \\ = 0 = rhs$
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