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
a3+B3 ≡ (A+B)-(A2-A B+B2)
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
.