Show that (cos x +i sin x )³=cos 3 x +i sin 3 x
Answers
Step-by-step explanation:
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)+ i�·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) + i�·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.