prove that 3(sinA-cosA)^⁴+6(sinA+cosA)^²+(sin^⁶A+cos^⁶A)=14

Asked by gaddamyashwanthr | 21st Sep, 2020, 09:00: PM

Expert Answer:

To prove: 3(sinA-cosA)4 + 6(sinA+cosA)2 + 4(sin6A+cos6A) = 13
 
3 open parentheses sin squared straight A plus cos squared straight A minus 2 sinAcosA close parentheses squared plus 6 open parentheses sin squared straight A plus cos squared straight A plus 2 sinAcosA close parentheses plus open parentheses sin squared straight A plus cos squared straight A close parentheses open parentheses sin to the power of 4 straight A plus cos to the power of 4 straight A minus sin squared Acos squared straight A close parentheses
equals 3 open parentheses 1 minus 2 sinAcosA close parentheses squared plus 6 open parentheses 1 plus 2 sinAcosA close parentheses plus 4 open parentheses open parentheses sin squared straight A plus cos squared straight A close parentheses squared minus 2 sin squared Acos squared straight A minus sin squared Acos squared straight A close parentheses
equals 3 open parentheses 1 minus 2 sinAcosA close parentheses squared plus 6 open parentheses 1 plus 2 sinAcosA close parentheses plus 4 open parentheses 1 minus 3 sin squared Acos squared straight A close parentheses
equals 3 open parentheses 1 plus 4 sin squared Acos squared straight A minus 4 sinAcosA close parentheses plus 6 plus 12 sinAcosA plus 4 minus 12 sin squared Acos squared straight A
equals 3 plus 12 sin squared Acos squared straight A minus 12 sinAcosA plus 6 plus 12 sinAcosA plus 4 minus 12 sin squared Acos squared straight A
equals 13

Answered by Renu Varma | 22nd Sep, 2020, 12:06: PM