Complex Phobia 4

Geometry Level 4

If α \alpha and β \beta are the roots of the equation z 2 sin ⁡ 2 x − z sin ⁡ x + 1 = 0 z^2\sin^2{x} - z\sin{x} + 1 = 0

Then find α n . β n \large \alpha^n.\beta^n .

tan ⁡ 2 n x \tan^{2n}{x} sin ⁡ 2 n x \sin^{2n}{x} cos ⁡ 2 n x \cos^{2n}{x} cot ⁡ 2 n x \cot^{2n}{x} csc ⁡ 2 n x \csc^{2n}{x}

Krishna Shankar by Krishna Shankar , 23, India

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2 solutions

Prakhar Bindal
Dec 15, 2016

Level 4? seriously??

by vieta's product of roots = cosec^2(x)

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Tom Engelsman
Jan 16, 2017

Let us find the above roots via the Quadratic Formula:

z s i n ( x ) = 1 ± ( − 1 ) 2 − 4 ( 1 ) ( 1 ) 2 = 1 ± i ⋅ 3 2 = e i ⋅ π 3 , e − i ⋅ π 3 ; z sin(x) = \frac{1 \pm \sqrt{(-1)^{2} - 4(1)(1)}}{2} = \frac{1 \pm i \cdot \sqrt{3}}{2} = e^{i \cdot \frac{\pi}{3}}, e^{-i \cdot \frac{\pi}{3}};

which ultimately yields: z = e i ⋅ π 3 ⋅ c s c ( x ) ; e − i ⋅ π 3 ⋅ c s c ( x ) . z = e^{i \cdot \frac{\pi}{3}} \cdot csc(x); e^{-i \cdot \frac{\pi}{3}} \cdot csc(x). Thus the desired root product becomes: [ e i ⋅ π 3 ⋅ c s c ( x ) ] n ⋅ [ e − i ⋅ π 3 ⋅ c s c ( x ) ] n = ( c s c 2 ( x ) ) n = c s c 2 n ( x ) . [e^{i \cdot \frac{\pi}{3}} \cdot csc(x)]^{n} \cdot [e^{-i \cdot \frac{\pi}{3}} \cdot csc(x)]^{n} = (csc^{2}(x))^{n} = \boxed{csc^{2n}(x)}.

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