Question

If acos^3A+3acosAsin^2A=m , asin^3A+3acos^2BsinA=n , prove that (m+n)^2/3+(m-n)^2/3=2a^2/3

Answer 1

acos³A+3acosAsin²A=m
asin³A+3acos²AsinA=n
∴, m+n
=acos³A+3acosAsin²A+asin³A+3acos²AsinA
=a(sin³A+3sin²AcosA+3sinAcos²A+cos³A)
=a(sinA+cos)³
m-n
=acos³A+3acosAsin²A-asin³A-3acos²AsinA
=a(cos³A-3cos²AsinA+3cosAsin²A-sin³A)
=a(cosA-sinA)³
∴, (m+n)²/³+(m-n)²/³
={a(sinA+cosA)³}²/³+{a(cosA-sinA)³}²/³
=a²/³(sinA+cosA)²+a²/³(cosA-sinA)²
=a²/³(sin²A+2sinAcosA+cos²A+cos²A-2sinAcosA+sin²A)
=a²/³(1+1) [∵, sin²A+cos²A=1]
=2a²/³ (Proved)

Answer 2

Step-by-step explanation:

This is the correct explanation of the sum

hope it's help ful and plzz mark me as brainlist