Show that cos theta plus I sin theta whole cube is equal to cos 3 theta plus I sin 3 theta
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we have to show that,
(cosθ + isinθ)³ = cos3θ + isin3θ
LHS = (cosθ + isinθ)³
= cos³θ + 3icos²θ.sinθ + 3i²cosθ.sin²θ + i³sin³θ
[ we know, i² = -1 and i³ = -i ]
= cos³θ + 3icos²θ.sinθ - 3cosθ.sin²θ - isin³θ
= (cos³θ - 3cosθ.sin²θ) + (3icos²θ.sinθ - isin³θ)
= (4cos³θ - 3cosθ) + i (3sinθ - 4sin³θ)
[ we know, 4cos³x - 3cosx = cos3x and 3sinx - 4sin³x = sin3x ]
= cos3θ + isin3θ = RHS
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