prove that cosA.cos(π/3+A).cos(π/3-A)=1/4 cos3A
Answers
Answer:
os(π/3 - A) = cos(π/3)cosA + sin(π/3)sinA = (1/2)cosA + (√3/2)sinA
Similarly, cos(π/3 + A) = (1/2)cosA - (√3/2)sinA
So, cosAcos(π/3-A)cos(π/3+A) = cosA[(1/4)cos2A - (3/4)sin2A]
= cosA[(1/4)cos2A - (3/4)(1-cos2A)]
= cosA[cos2A - 3/4]
= (1/4)(4cos3A - 3cosA) = (1/4)cos(3A)
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NOTE: cos(3A) = cos(2A+A) = cos(2A)cosA - sin(2A)sinA
= (2cos2A-1)cosA - 2sinAcosAsinA
= 2cos3A-cosA - 2sin2AcosA
= 2cos3A - cosA - 2(1-cos2A)cosA
= 4cos3A - 3cosA