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Establish the identity
$$
1+z+z^{2}+\cdots+z^{n}=\frac{1-z^{n+1}}{1-z} \quad(z \neq 1)
$$
and then use it to derive Lagrange's trigonometric identity:
$$
1+\cos \theta+\cos 2 \theta \cdots+\cos n \theta=\frac{1}{2}+\frac{\sin \frac{(2 n+1) \theta}{2}}{2 \sin \frac{\theta}{2}} \quad(0<\theta<2 \pi)
$$
Hint: As for the first identity, write $S=1+z+z^{2}+\cdots+z^{n}$ and consider the difference $S-z S$. To derive the second identity, write $z=e^{i \theta}$ in the first one.
in Mathematics by Gold Status (10,261 points)
edited by | 19 views

1 Answer

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Best answer
Proof. If $z \neq 1$, then
$$
\begin{aligned}
(1-z)\left(1+z+\cdots+z^{n}\right) &=1+z+\cdots+z^{n}-\left(z+z^{2}+\cdots+z^{n+1}\right) \\
&=1-z^{n+1}
\end{aligned}
$$
Thus
$$
1+z+z^{2}+\cdots+z^{n}=\left\{\begin{array}{ll}
\frac{1-z^{n+1}}{1-z}, & \text { if } z \neq 1 \\
n+1, & \text { if } z=1 .
\end{array}\right.
$$
Taking $z=e^{i \theta}$, where $0<\theta<2 \pi$, then $z \neq 1$. Thus
$$
\begin{aligned}
1+e^{i \theta}+e^{2 i \theta}+\cdots+e^{n i \theta} &=\frac{1-e^{(n+1) \theta}}{1-e^{i \theta}}=\frac{1-e^{(n+1) \theta}}{-e^{i \theta / 2}\left(e^{i \theta / 2}-e^{-i \theta / 2}\right)} \\
&=\frac{-e^{-i \theta / 2}\left(1-e^{(n+1) \theta}\right)}{2 i \sin (\theta / 2)} \\
&=\frac{i\left(e^{-i \theta / 2}-e^{\left(n+\frac{1}{2}\right) i \theta}\right)}{2 \sin (\theta / 2)}
\end{aligned}
$$

$$
=\frac{1}{2}+\frac{\sin \left[\left(n+\frac{1}{2}\right) \theta\right]}{2 \sin (\theta / 2)}+i \frac{\cos (\theta / 2)-\cos \left[\left(n+\frac{1}{2}\right) \theta\right]}{2 \sin (\theta / 2)}
$$
Equating real and imaginary parts, we obtain
$$
1+\cos \theta+\cos 2 \theta \cdots+\cos n \theta=\frac{1}{2}+\frac{\sin \left[\left(n+\frac{1}{2}\right) \theta\right]}{2 \sin (\theta / 2)}
$$
and
$$
\sin \theta+\sin 2 \theta \cdots+\sin n \theta=\frac{\cos (\theta / 2)-\cos \left[\left(n+\frac{1}{2}\right) \theta\right]}{2 \sin (\theta / 2)} .
$$
by Gold Status (10,261 points)

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