Express $2\cos (n\theta)$ in terms of $z$

Question

How can I show that $$2\cos(n\theta)=z^n + \frac{1}{z^n}$$

if $z=\cos\theta+i\sin \theta$

Can some one help me? thx!

Answer 1

Starting from the Euler's formula $$z=\cos\theta+i\sin \theta$$ and from De-Moivres theorem $$\color{red}{z^n=\cos(n\theta) +i\sin(n\theta)=e^{in\theta}}\tag{1}$$ therefore $$\color{blue}{\frac{1}{z^n}=(\cos (n\theta)+i\sin(n\theta))^{-1}=\cos(n\theta)-i\sin(n\theta)=e^{-in\theta}}\tag{2}$$ since $\cos(-n\theta)=\cos(n\theta)$ and $\sin(-n\theta)=-\sin(n\theta)$

Finally, add equations $(1)$ and $(2)$:

$$\require{enclose}\cos(n\theta)+\enclose{updiagonalstrike}{i\sin(n\theta)}+\cos(n\theta)\enclose{updiagonalstrike}{-i\sin(n\theta)}=2\cos(n\theta)=\frac{1}{z^n}+{z^n}$$

Answer 2

Since $z=e^{i\theta}$, $z^n=e^{in\theta}=\cos(n\theta)+i\sin(n\theta)$. Similarly, $z^{-n}=e^{-in\theta}=\cos(-n\theta)+i\sin(-n\theta)=\cos(n\theta)-i\sin(n\theta)$ since $\cos$ is even and $\sin$ is odd.

Answer 3

setting the value of $z=\cos\theta+i\sin \theta$, one should have $$z^n+\frac{1}{z^n}=(\cos\theta+i\sin\theta)^n+\frac{1}{(\cos\theta+i\sin\theta)^n}$$ $$=(\cos\theta+i\sin\theta)^n+(\cos\theta+i\sin\theta)^{-n}$$ using De-Moivre's theorem, $$=\cos(n\theta)+i\sin(n\theta)+\cos(-n\theta)+i\sin(-n\theta)$$ $$=\cos(n\theta)+i\sin(n\theta)+\cos(n\theta)-i\sin(n\theta)$$ $$=2\cos(n\theta)$$ or $$2\cos(n\theta)=z^n+\frac{1}{z^n}$$