You can
put this solution on YOUR website!
It's immediate by DeMoivre's theorem, but I suppose they want
you to prove it from scratch without using DeMoivre's theorem.
OK, here's how I approach it:
[cos(x) + i·sin(x)]³ =
[cos(x) + i·sin(x)]²[cos(x) + i·sin(x)]
We'll work on the left bracketed expression only for awhile,
and just keep bringing down the right bracket expression
until we get the left bracket simplified:
[cos²(x) + 2i·cos(x)sin(x)+ i²·sin²(x)][cos(x) + i·sin(x)]
[cos²(x) + 2i·cos(x)sin(x)+ (-1)·sin²(x)][cos(x) + i·sin(x)]
[cos²(x) + 2i·cos(x)sin(x) - sin²(x)][cos(x) + i·sin(x)]
[{cos²(x) - sin²(x)} + i·{2cos(x)sin(x)}][cos(x) + i·sin(x)]
[{cos²(x) - sin²(x)} + i·{2sin(x)cos(x)}][cos(x) + i·sin(x)]
Use the identities:
cos²(A) - sin²(A) = cos(2A)
2sin(A)cos(A) = sin(2A)
[cos(2x) + i·sin(2x)][cos(x) + i·sin(x)] =
Now we finally use the right bracket and multiply out:
cos(2x)cos(x) + i·cos(2x)sin(x) + i·sin(2x)cos(x) + i²·sin(2x)cos(x) =
cos(2x)cos(x) + i·[cos(2x)sin(x) + sin(2x)cos(x)] + (-1)·sin(2x)cos(x) =
cos(2x)cos(x) + i·[cos(2x)sin(x) + sin(2x)cos(x)] - sin(2x)cos(x) =
cos(2x)cos(x) - sin(2x)sin(x) + i·[cos(2x)sin(x) + sin(2x)cos(x)] =
cos(2x)cos(x) - sin(2x)sin(x) + i·[sin(2x)cos(x) + cos(2x)sin(x)] =
Use the identities:
cos(A)cos(B) - sin(A)sin(B) = cos(A+B)
sin(A)cos(B) + cos(A)sin(B) = sin(A+B)
cos(2x+x) + i·sin(2x+x)
cos(3x) + i·sin(3x)
You can probably leave out some of those steps where I just regrouped
terms.
Edwin
