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What is the equation for the pattern sinx, cosx, -sinx, -cosx, sinx, cosx?
The pattern should look like any other equation relating to n. An example like that problem is four derivatives of e^-3x, which give you a pattern of... [(-1)^n times 3^n times e^-3x].. after taking four derivatives the pattern should look like the one in the brackets.
1 Answer
- MathPhDLv 61 decade agoFavorite Answer
These are the derivatives (starting at the 0-th) of sin(x).
If you want Asin(x) + Bcos(x) and you want A and B to go through the patterns
1,0
0,1
-1,0
0,-1
and repeat, then you need the pattern 1, 0, -1, 0...Any pattern of length n can be given using sines and cosines of frequencies up to n, or equivalently, the n-th roots of 1. Here, I will use the 4th roots of 1, whicha re i and -i.
powers of i and -i
1........1
i........-i
-1.....-1
-i.......i
experience allows me to see that [ i^n + (-i)^n ]/2 works! So putting this all together (cosine pattern slightly late) gives
[ [ i^n + (-i)^n ]sin(x) + [ i^(n+1 + (-i)^(n+1) ]cos(x) ]/2
