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$$\cos(1)+\cos(2)+\ldots+\cos(n-1)= \cos(n)-\cos(n-1)/(2\cos(1)-2 ) -1/2$$

Here is my attempt:

Let $P(n)$ be this statement.

$P(1)$ is true since $0=\cos(1)-\cos(1-1)/(2\cos(1)-2) -1/2$

Suppose $P(k)$ is true for some integer $k$. Then I have to prove $P(k+1)$ is also true. That is :

$$\cos(1)+\cos(2)+\ldots+\cos(k-1)+\cos(k)=\cos(k+1)-\cos(k)/(2\cos(1)-2) -1/2.$$

By inductive hypothesis, we have $\cos(k)-\cos(k-1)/(2\cos(1)-2 ) -1/2 +\cos(k)$. But how does it equal to the right hand side? Can someone help me with this question please, thank you!

Zeus579
  • 133

1 Answers1

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By De Moivre, $e^{ik}=\cos k+i\sin k$. Then

$\begin{eqnarray} \sum_{k=1}^{n-1}\cos k&=&\sum_{k=1}^{n-1}Re(e^{ik})\\ &=&Re\left(\sum_{k=1}^{n-1}e^{ik}\right)\\ &=&Re\left(\frac{e^{in}-1}{e^i-1}-1\right)\\ &=&Re\left(\frac{\cos n+i\sin n}{\cos1+i\sin1-1}\right)-1\\ &=&Re\left(\frac{\cos n+i\sin n}{\cos1+i\sin1-1}\cdot\frac{\cos1-i\sin1-1}{\cos1-i\sin1-1}\right)-1\\ &=&Re\left(\frac{\cos n\cos1-i\cos n\sin 1-\cos n+i\sin n\cos1+\sin n\sin 1-i\sin n}{2-2\cos1}\right)-1\\ &=&\frac{\cos n\cos 1-\cos n+\sin n\sin 1}{2-2\cos 1}-1\\ &=&\frac{\cos(n-1)-\cos n}{2-2\cos1}-1 \end{eqnarray}$

sinbadh
  • 7,521