Question

Prove that cos 5A = 16 cos^5 A - 20 cos^3 A + 5cos A

Answer 1

LHS
=cos5A
=cos(3A+2A)
=cos3Acos2A-sin3Asin2A [∵, cose(A+B)=cosAcosB-sinAsinB]
=(4cos³A-3cosA)(2cos²A-1)-(3sinA-4sin³A)(2sinAcosA)
=(8cos⁵A-6cos³A-4cos³A+3cosA)-(6sin²AcosA-8sin⁴AcosA)
=8cos⁵A-10cos³A+3cosA-sin²A(6cosA-8sin²AcosA)
=8cos⁵A-10cos³A+3cosA-(1-cos²A){6cosA-(1-cos²A)8cosA}
[sin²A+cos²A=1]
=8cos⁵A-10cos³A+3cosA-(1-cos²A)(6cosA-8cosA+8cos³A)
=8cos⁵A-10cos³A+3cosA-(1-cos²A)(8cos³A-2cosA)
=8cos⁵A-10cos³A+3cosA-(8cos³A-8cos⁵A-2cosA+2cos³A)
=8cos⁵A-10cos³A+3cosA-10cos³A+8cos⁵A+2cosA
=16cos⁵A-20cos³A+5cosA
=RHS (Proved)

Answer 2

Answer:

Step-by-step explanation:

LHS

=cos5A

=cos(3A+2A)

=cos3Acos2A-sin3Asin2A [∵, cose(A+B)=cosAcosB-sinAsinB]

=(4cos³A-3cosA)(2cos²A-1)-(3sinA-4sin³A)(2sinAcosA)

=(8cos⁵A-6cos³A-4cos³A+3cosA)-(6sin²AcosA-8sin⁴AcosA)

=8cos⁵A-10cos³A+3cosA-sin²A(6cosA-8sin²AcosA)

=8cos⁵A-10cos³A+3cosA-(1-cos²A){6cosA-(1-cos²A)8cosA}

[sin²A+cos²A=1]

=8cos⁵A-10cos³A+3cosA-(1-cos²A)(6cosA-8cosA+8cos³A)

=8cos⁵A-10cos³A+3cosA-(1-cos²A)(8cos³A-2cosA)

=8cos⁵A-10cos³A+3cosA-(8cos³A-8cos⁵A-2cosA+2cos³A)

=8cos⁵A-10cos³A+3cosA-10cos³A+8cos⁵A+2cosA

=16cos⁵A-20cos³A+5cosA

=RHS (Proved)