Prove that cos^3Acos(3A)+sin^3A(sin3A)=cos^3 2Acos3Acos(3A)+sin3A(sin3A)=cos32A ?

3 Answers
Oct 25, 2017

Please refer to a Proof given in the Explanation Section.

Explanation:

Recall that,

cos3A=4cos^3A-3cosA, &, sin3A=3sinA-4sin^3A.cos3A=4cos3A3cosA,&,sin3A=3sinA4sin3A.

Sub.ing these, we have,

:." The L.H.S.="cos^3A(4cos^3A-3cosA)+sin^3A(3sinA-4sin^3A),

=4cos^6A-3cos^4A+3sin^4A-4sin^6A,

=4(cos^6A-sin^6A)-3(cos^4A-sin^4A),

=4{(cos^2A)^3-(sin^2A)^3}-3{(cos^2A)^2-(sin^2A)^2},

=4{cos^2A-sin^2A){(cos^2A)^2+cos^2Asin^2A+(sin^2A)^2}

-3(cos^2A-sin^2A)(cos^2A+sin^2A),

=(cos^2A-sin^2A)[4{(cos^2A)^2+cos^2Asin^2A+(sin^2A)^2}-3*1],

=(cos^2A-sin^2A)[4cos^4A+4cos^2Asin^2A+4sin^4A-3(cos^2A+sin^2A)^2],

=(cos^2A-sin^2A)[4cos^4A+4cos^2Asin^2A+4sin^4A-3(cos^4A+2cos^2Asin^2A+sin^4a)],

=(cos^2A-sin^2A)[cos^4A-2cos^2Asin^2A+sin^4A],

=(cos^2A-sin^2A)(cos^2A-sin^2A)^2,

=(cos^2A-sin^2A)^3,

=(cos2A)^3,

=cos^3 2A,

"=The R.H.S."

Enjoy Maths.!

Oct 25, 2017

Kindly refer to a Proof in the Explanation.

Explanation:

Here is a Second Method to solve the Problem.

We know the following Identities :

cos3A=4cos^3A-3cosA, and, sin3A=3sinA-4sin^3A.

:. cos^3A=1/4(cos3A+3cosA).......(1), and,

sin^3A=1/4(3sinA-sin3A).............(2). Using these, we have,

"The L.H.S="1/4{(cos3A+3cosA)cos3A+(3sinA-sin3A)sin3A],

=1/4{cos^2 3A+3cos3AcosA+3sinAsin3A-sin^2 3A},

=1/4{(cos^2 3A-sin^2 3A)+3(cos3AcosA+sin3AsinA)},

=1/4{cos(2xx3A)+3cos(3A-A)},

=1/4{cos(3xx2A)+3cos2A},

=1/4(cos3theta+3costheta), where, theta=2A,

=cos^3 theta..........[because, (1)],

=cos^3 2A,

"=The R.H.S."

Enjoy Maths.!

LHS=cos^3Acos3A+sin^3Asin3A

=1/2(cos^2A*2cosAcos3A+sin^2A*2sinAsin3A)

=1/2[cos^2A(cos4A+cos2A)+sin^2A(cos2A-cos4A)]

=1/2[cos^2A*cos4A+cos^2A*cos2A+sin^2A*cos2A-sin^2A*cos4A]

=1/2[cos^2A*cos4A-sin^2A*cos4A+cos^2A*cos2A+sin^2A*cos2A]

=1/2[(cos^2A-sin^2A)cos4A+(cos^2A+sin^2A)cos2A]

=1/2[cos2A*cos4A+cos2A]

=1/2[cos2A(cos4A+1)]

=1/2*cos2A*2cos^2 2A

=cos^3 2A=RHS