Question

Prove this: `(1-sin^4x-cos^4x)/(1-sin^6x-cos^6x)=2/3`?

`(1-sin^4x-cos^4x)/(1-sin^6x-cos^6x)=2/3`

Answer 1

`LHS=(1-sin^4x-cos^4x)/(1-sin^6x-cos^6x)`

`=(1-((sin^2x)^2+(cos^2x)^2))/(1-((sin^2x)^3+(cos^2x)^3))`

`=(1-((sin^2x+cos^2x)^2-2sin^2cos^2x))/(1-((sin^2x+cos^2x)^3-3sin^2xcos^2x(sin^2x+cos^2x))`

`=(1-(sin^2x+cos^2x)^2+2sin^2cos^2x)/(1-(sin^2x+cos^2x)^3+3sin^2xcos^2x(sin^2x+cos^2x))`

`=(1-1^2+2sin^2cos^2x)/(1-1^3+3sin^2xcos^2x)`

`=(2sin^2cos^2x)/(3sin^2xcos^2x)=2/3=RHS`

Proved

In step 3 the following formulae are used

`a^2+b^2=(a+b)^2-2ab`

and

`a^3+b^3=(a+b)^3-3ab(a+b)`

Answer 2

Please see the explanation. I confirmed each step of this proof using www.WolframAlpha.com

Explanation:

Multiply both sides by `3(1-sin^6(x)-cos^6(x))`

`3-3sin^4(x)-3cos^4(x) = 2-2sin^6(x)-2cos^6(x)`

Substitute `-3(1 - cos^2(x))^2" for " -3sin^4(x)`

`3-3(1 - cos^2(x))^2-3cos^4(x) = 2-2sin^6(x)-2cos^6(x)`

Multiply the square:

`3-3(1 - 2cos^2(x)+cos^4(x))-3cos^4(x) = 2-2sin^6(x)-2cos^6(x)`

Distribute the -3:

`3-3 + 6cos^2(x)-3cos^4(x)-3cos^4(x) = 2-2sin^6(x)-2cos^6(x)`

Combine like terms:

`6cos^2(x)-6cos^4(x) = 2-2sin^6(x)-2cos^6(x)`

Divide both sides by 2:

`3cos^2(x)-3cos^4(x) = 1-sin^6(x)-cos^6(x)`

Substitute `-(1 - cos^2(x))^3" for " -sin^6(x)`

`3cos^2(x)-3cos^4(x) = 1-(1 - cos^2(x))^3-cos^6(x)`

Expand the cube:

`3cos^2(x)-3cos^4(x) = 1-(1 - 3cos^2(x)+3cos^4(x)-cos^6(x))-cos^6(x)`

Distribute the -1:

`3cos^2(x)-3cos^4(x) = 1-1 + 3cos^2(x)-3cos^4(x)+cos^6(x)-cos^6(x)`

Combine like terms:

`3cos^2(x)-3cos^4(x) = 3cos^2(x)-3cos^4(x)`

The right is identical to the left. Q.E.D.