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Question
Prove that sin 4A = 4sinA cos3A – 4 cosA sin3A
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Solution
sin4A = sin(2A + 2A)
We know that,
sin(A + B) = sinA cosB + cosA sinB
Therefore, sin4A = sin2A cos2A + cos2A sin2A
⇒ sin4A = 2sin2A cos2A
From T-ratios of multiple angles,
We get,
sin2A = 2sinA cosA and cos2A = cos2A – sin2A
⇒ sin4A = 2(2sinA cosA)(cos2A – sin2A)
⇒ sin4A = 4sinA cos3A – 4cosA sin3A
Hence, sin4A = 4sin A cos3A – 4cosA sin3A
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