prove that cos5A = 16cos5A - 20cos3A + 5cosA
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LHS
cos5A
= cos(3A+2A)
= cos3Acos2A- sin3Asin2A
= (4sin³A-3cosA)(2cos²A-1)-(3sinA-4sin³A)2sinAcosA
=8cos^5A-4cos³A-6cosA-6sin²AcosA+8sin⁴AcosA
=8cos^5A-4cos³A-6cos³A+3cosA-6(1-cos²A)cosA+8(1-cos²A)²cosA
=8xos^5A-10cos³A+3cosA-6cos³A+8(1-2cos²A+cos⁴A)cosA
=8cos^5A-4cos³A-3cosA+8cosA-16cos³A+8cos^5A
=16cos^5A-20cos³A+5cosA
=RHS
cos5A
= cos(3A+2A)
= cos3Acos2A- sin3Asin2A
= (4sin³A-3cosA)(2cos²A-1)-(3sinA-4sin³A)2sinAcosA
=8cos^5A-4cos³A-6cosA-6sin²AcosA+8sin⁴AcosA
=8cos^5A-4cos³A-6cos³A+3cosA-6(1-cos²A)cosA+8(1-cos²A)²cosA
=8xos^5A-10cos³A+3cosA-6cos³A+8(1-2cos²A+cos⁴A)cosA
=8cos^5A-4cos³A-3cosA+8cosA-16cos³A+8cos^5A
=16cos^5A-20cos³A+5cosA
=RHS
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