Using just algebra and Trig show that

Question

(cosx+isinx)^3=cos3x+isin3x

Answer 1

Formula: e^ix = cos (x) + i sin(x)

And e^iax = cos(ax) + i sin(ax)

(e^i3x) = [ cos(3x) + i sin (3x) ]

(e^ix)³ = (e^i3x) = [ cos(3x) + i sin (3x) ]

e^ix = cos (x) + i sin(x)

Take cube each side.

(e^ix)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)

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

(e^i3x) = 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)

Now add to

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

There fore

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

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