Question

What is `cos^3theta-6sin^3theta` in terms of non-exponential trigonometric functions?

Answer 1

`cos^3(theta)-6 sin^3(theta)`

`=1/4(cos(3theta)+3cos(theta))+3/2(sin(3theta)-3sin(theta))`

Explanation:

Since this is of degree `3`, the place to start is the expansion of `cos(3 theta)` and `sin(3 theta)`.

By De Moivre's formula:

`cos(3 theta) + i sin(3 theta)`

`=(cos(theta)+i sin(theta))^3`

`=cos^3(theta)+3 cos^2(theta) sin(theta) i - 3 cos(theta)sin^2(theta) - sin^3(theta) i`

`=(cos^3(theta)-3cos(theta)sin^2(theta))-(sin^3(theta)-3sin(theta)cos^2(theta)) i`

`=(4cos^3(theta)-3cos(theta)) + (3sin(theta)-4sin^3(theta))i`

So equating Real and Imaginary parts we find:

`cos(3theta) = 4cos^3(theta)-3cos(theta)`

`sin(3theta) = 3sin(theta)-4sin^3(theta)`

Hence:

`cos^3(theta) = 1/4(cos(3theta)+3cos(theta))`

`sin^3(theta) = 1/4(3sin(theta)-sin(3theta))`

So:

`cos^3(theta)-6 sin^3(theta)`

`=1/4(cos(3theta)+3cos(theta))+3/2(sin(3theta)-3sin(theta))`