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 $n$ .)

#Geometry

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`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
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Comments

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 - 6 years, 8 months ago

Hint: Use the fact that $\cos \phi \leq 1$

jatin yadav - 6 years, 8 months ago

it's awesome!! :D

Aswad Hariri Mangalaeng - 6 years, 8 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$