срочно помогите пожалуйста срочно!!! . Created by akylai42. algebra-ru

Объяснение:1) 45° = (π×45°)/180° =π/4120° = (π×120°)/180° = 2π/3540° = (π×540°)/180° = 3π2)π/6 = (π×180°)/(6×π) = 180°/6 = 30°7π/2 = (7π×180°)/(2×π) = 7×90° = 630°3π = (3π×180°)/π = 540° 3)Найти: sin(a)-?, tg(a)-?, ctg(a)-?4)5)

срочно помогите пожалуйста срочно!!! ...

akylai42 @akylai42

November 2021 1 7 Report

срочно помогите пожалуйста срочно!!!

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VlAdYsLaV15

Объяснение:

45° = (π×45°)/180° =π/4

120° = (π×120°)/180° = 2π/3

540° = (π×540°)/180° = 3π

π/6 = (π×180°)/(6×π) = 180°/6 = 30°

7π/2 = (7π×180°)/(2×π) = 7×90° = 630°

3π = (3π×180°)/π = 540°

$\cos( \alpha ) = - \frac{12}{13} \\ \pi < \alpha < \frac{3\pi}{2}$

Найти: sin(a)-?, tg(a)-?, ctg(a)-?

$\sin^{2} ( \alpha ) + \cos^{2} ( \alpha ) = 1 \\ \sin ^{2} ( \alpha ) = 1 - \cos^{2} ( \alpha ) \\ \sin {}^{2} ( \alpha ) = 1 - ( - \frac{12}{13} )^{2} \\ \sin {}^{2} ( \alpha ) = \frac{25}{169} \\ \sin( \alpha ) = - \frac{5}{13} \\ \tan ( \alpha ) = \frac{ \sin( \alpha ) }{ \cos( \alpha ) } = \frac{ - \frac{5}{13} }{ - \frac{12}{13} } = \frac{5}{12} \\ \cot( \alpha ) = \frac{ \cos( \alpha ) }{ \sin( \alpha ) } = \frac{12}{5}$

$a) \cos(180) + 5 \tan(45) = - 1 + 5 \times 1 = 4 \\ b) \: \frac{\pi}{60} \times \cos( \frac{\pi}{9} ) - 2 \sin( \frac{\pi}{6} ) = \frac{\pi \times \cos( \frac{\pi}{9} ) }{60} - 2 \times \frac{1}{2} = \frac{\pi - \cos( \frac{\pi}{9} ) }{60} - 1$

$a)\frac{ - \sin( \frac{\pi}{2} + \alpha ) + \sin( \alpha ) }{ 2\cos( \alpha ) \times \sin( - \alpha ) + 1 } = \frac{ - \cos( \alpha ) + \sin( \alpha ) }{ - \sin(2 \alpha ) + 1} \\ b)(1 - \cos( \alpha ) )(1 - \cos( \alpha ) ) = 1 - 2 \cos( \alpha ) + \cos ^{2} ( \alpha ) \\ c) \frac{ \cos( \alpha ) }{1 - \sin( \alpha ) } - \frac{ \cos( \alpha ) }{1 + \sin( \alpha ) } = \frac{ \cos( \alpha ) + \sin( \alpha ) \cos( \alpha ) - \cos( \alpha ) + \sin( \alpha ) \cos( \alpha ) }{1 - \sin^{2} ( \alpha ) } = \\ \frac{2 \sin( \alpha ) \cos( \alpha ) }{ \cos^{2} ( \alpha ) } = \frac{2 \sin( \alpha ) }{ \cos( \alpha ) } = 2 \tan( \alpha )$

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