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A Tricky Trigonometric Equation

For each integer \(n\), how many solutions are there with \( 0 \leq \theta \leq 2 \pi \) such that

\[ \cos \theta + 2 \cos 2 \theta + 3 \cos 3 \theta + \ldots + n \cos n \theta = \frac{ n(n+1) } { 2} ?\]

(Your answer might depend on \(n\).)

Note by Calvin Lin
3 years, 3 months ago

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According To me Answer is : \(\theta =\quad 0\quad ,\quad 2\pi \).

Explanation

\(\sum _{ 1 }^{ n }{ r } =\frac { n(n+1) }{ 2 } \\ 1+2+3+\quad .\quad .\quad .\quad .\quad .\quad +n\quad =\frac { n(n+1) }{ 2 } \quad \).

Now using the Fact That :

\(\sum _{ 1 }^{ n }{ r\cos { (r\theta ) } } \quad \le \quad \sum _{ 1 }^{ n }{ r } \quad (\because \quad \cos { (r\theta ) } \quad \le \quad 1\quad )\).

equality only occurs when

\(\cos { (r\theta ) } \quad =\quad 1\quad \quad \forall \quad r\quad \in \quad [1,n]\\ \Rightarrow \quad r\theta \quad =\quad 0\quad or\quad 2m\pi \quad \quad \forall \quad r\quad \in \quad [1,n]\quad \quad (where\quad m\quad \in \quad I)\\ \therefore \quad in\quad \quad \theta \quad \in \quad [0,2\pi ]\quad \).

only possible solution are:

\(\theta =\quad 0\quad ,\quad 2\pi \). which satisfying above equality always

Deepanshu Gupta - 3 years, 3 months ago

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2

Hint: Use the fact that \(\cos \phi \leq 1\)

Jatin Yadav - 3 years, 3 months ago

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it's awesome!! :D

Aswad H. Mangalaeng - 3 years, 3 months ago

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