I can’t find the full formula anywhere!

## Answer

Dear Julie,

I suspect you are referring to the identities that convert cubes into trigonometric functions of the triple of the argument:

4 Sin[x]^{3}≡3 Sin[x]-Sin[3x]

4 Cos[x]^{3}≡3·cos[x]+Cos[3x]

Both identities together then give you an opportunity to convert the power increases in your expression into multiples of the argument.

Your question illustrates another problem:

Without further specification, multiple answers are even possible. Depending on the intended application, one answer may already be “better” than the other.

This way you can use the identities

a^{3}+B^{3} ≡ (A+B)-(A^{2}-A B+B^{2})

and

cos[x]^{2}+Sin[x]^{2} 1

Convert the sum of cubes also to

cos[x]^{3}+Sin[x]^{3} ≡ (Cos[x]+Sin[x])·(Cos[x]^{2}-Cos[x]·Sin[x]+Sin[x]^{2}) ≡ (Cos[x]+Sin[x])·(1-Cos[x]·Sin[x])

For certain applications, this conversion of a cubic polynomial in terms of Cos[x] and Sin[x] to an equivalent second degree polynomial are also interesting.

Philippe J. Roussel

imec

## Answered by

#### eng. Philippe Roussel

Microelectronics Reliability

Kapeldreef 75 3001 Leuven

http://www.imec-int.com

.