Question:
cos4A - cos4B=8(cosA-cosB)(cosA cosB)(cosA-sinB)(cosA sinB) prove
Answer:
Given, cos 4A - cos 4B
= 2cos2 2A - 1 - (2cos2 2B - 1) {since 2cos2 x - 1 = cos 2x}
= 2cos2 2A - 1 - 2cos2 2B + 1
= 2cos2 2A - 2cos2 2B
= 2(cos2 2A - cos2 2B)
= 2(cos 2A - cos 2B) * (cos 2A + cos 2B)
= 2{2cos2 A - 1 - (2cos2 B - 1)} * {2cos2 A - 1 + 1 - 2sin2 B} {since 1 - 2sin2 x = cos 2x}
= 2{2cos2 A - 1 - 2cos2 B + 1} * {2cos2 A - 1 + 1 - 2sin2 B}
= 2{2cos2 A - 2cos2 B} * {2cos2 A - 2sin2 B}
= 2*2*2{cos2 A - cos2 B} * {cos2 A - sin2 B}
= 8(cos A - cos B)*(cos A + cos B)*(cos A - sin B)*(cos A + sin B)
So, cos 4A - cos 4B = 8(cos A - cos B)*(cos A + cos B)*(cos A - sin B)*(cos A + sin B)
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