Question

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

Answer 1

Please refer to a Proof given in the Explanation Section.

Explanation:

Recall that,

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

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.!

Answer 2

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.!

Answer 3

`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`