082-Engineering-Mathematics-Anthony-Croft-Robert-Davison-Martin-Hargreaves-James-Flint-Edisi-5-2017

9.9 De Moivre’s theorem 349
Now, writing sin 2 θ = 1 −cos 2 θ in Equation (9.10) and cos 2 θ = 1 −sin 2 θ in Equation
(9.11),wefind
cos3θ =cos 3 θ −3cosθ(1−cos 2 θ)
=cos 3 θ+3cos 3 θ−3cosθ
=4cos 3 θ−3cosθ
sin3θ = 3(1 −sin 2 θ)sinθ −sin 3 θ
as required.
=3sinθ−4sin 3 θ
This technique allows trigonometric functions of multiples of angles to be expressed
in terms of powers. Sometimes we want to carry out the reverse process and express a
power interms ofmultiple angles. Consider Example 9.21.
Example9.21 Ifz = cos θ +jsin θ show that
z + 1 z =2cosθ
z−1 z =2jsinθ
and find similarexpressions forz n + 1 z n andzn − 1 z n.
Solution Consider the complex number
z=cosθ+jsinθ
UsingDe Moivre’s theorem,
1
z =z−1 = (cosθ +jsinθ) −1 = cos(−θ) +jsin(−θ)
Butcos(−θ) = cosθ andsin(−θ) = −sinθ,sothatifz = cosθ +jsinθ
1
z =cosθ−jsinθ
Consequently,
z + 1 z =2cosθ and z−1 z =2jsinθ
Moreover,
so that
z n = cosnθ +jsinnθ and z −n = cosnθ −jsinnθ
z n + 1 z n =2cosnθ and zn − 1 z n = 2jsinnθ
z n + 1 z n =2cosnθ and zn − 1 z n = 2jsinnθ