# Using just algebra and Trig show that

(cosx+isinx)^3=cos3x+isin3x

+1 vote

Formula: eix = cos (x) + i sin(x)

And eiax = cos(ax) + i sin(ax)

(ei3x) = [ cos(3x) + i sin (3x) ]

(eix)3 = (ei3x) = [ cos(3x) + i sin (3x) ]

eix = cos (x) + i sin(x)

Take cube each side.

(eix)3= [ cos(x) + i sin (x) ]3

Recall: (a+b)3 = a3 + 3a2b + 3ab2 +b3

[ cos(x) + i sin (x) ]3 = cos3(x) + 3[cos(x)]2sin(x) + 3cos(x)[isin(x)]2 + [isin(x)]3

= cos3(x) + 3[cos(x)]2[isin(x)] + 3cos(x)[isin(x)]2 + [isin(x)]3

= cos3(x) + 3i[cos(x)]2sin(x) + 3cos(x)(i)2[sin(x)]2 + (i)3[sin(x)]3

Recall:  i2 = -1and i3 = -i

= cos3(x) + 3i[cos(x)]2sin(x) + 3cos(x)(-1)[sin(x)]2 + (-i)[sin(x)]3

= cos3(x) + 3i cos2(x)sin(x) - 3cos(x)sin2(x) - i sin3(x)

[ cos(x) + i sin (x) ]3 = cos3(x) + 3i cos2(x)sin(x) - 3cos(x)sin2(x) - i sin3(x)

[ cos(x) + i sin (x) ]3 = cos3(x)  - 3cos(x)sin2(x) + 3i cos2(x)sin(x) - i sin3(x)

(eix)3 = cos3(x)  - 3cos(x)sin2(x) + 3i cos2(x)sin(x) - i sin3(x)

(ei3x) = cos3(x)  - 3cos(x)sin2(x) + 3i cos2(x)sin(x) - i sin3(x)

[cos(3x) + i sin(3x)] = cos3(x)  - 3cos(x)sin2(x) + 3i cos2(x)sin(x) - i sin3(x)

Now compare real and imaginary terms.

cos (3x) = cos3(x) - 3cos(x)sin2(x)

sin(3x) = 3 cos2(x)sin(x) - sin3(x)