sin3A+ sin2A -sinA =
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Answer:
sin4a is the correct answer to your question
Answered by
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Step-by-step explanation:
3SinA-4Sin³+2SinACosA-SinA
3SinA-4Sin³A+SinA(2CosA-1)
SinA(3-4Sin²A)+SinA(2CosA-1)
SinA(3-4Sin²A+2CosA-1)
SinA[2{CosA-2Sin²A)}
SinA{2(CosA-2+2Cos²A)}
SinA[2{CosA(1-2+2CosA)}]
SinA[2{CosA(2CosA-1)}]
SinA(4Cos²A-2CosA)
4Cos²A-2SinACosA
2CosA(2CosA-SinA)
2CosA(2√1-Sin²A -√1-Cos²A)
2CosA[2{√-(Cos²A-Sin²A)}]
2CosA[2√{-Cos2A}]
4CosA-√2Cos2A
√2CosA{(√2)³-√2CosA-1)
solve that further
4CosA-2Cos²A-1
2CosA(2-Cos²A-1/2)
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