Срочноооо! Пожалуйста умоляю Вариант Б1. Created by marinayarosh1984. algebra-ru

Срочноооо! Пожалуйста умоляю Вариант Б1...

marinayarosh1984 @marinayarosh1984

July 2022 1 11 Report

Срочноооо! Пожалуйста умоляю
Вариант Б1

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$a) \: \: \: \: \frac{ \cos70 ° \cos10° +\sin70° \sin10°}{\sin50°\cos10° +\cos50 ° \sin10°} = \\ = \frac{ \cos(70° - 10°) }{ \sin(50° + 10°) } = \\ = \frac{ \cos60°}{ \sin60°} = \ctg60° = \frac{ \sqrt{3} }{3}$

$б) \cos105 ° = \cos(45 °+ 60 °) = \\ = cos45°cos60° - sin45°sin60° = \\ = \frac{ \sqrt{2} }{2} \times \frac{ 1 }{2} - \frac{ \sqrt{2} }{2} \times \frac{ \sqrt{3} }{2} = \frac{ \sqrt{2} - \sqrt{6} }{4}$

$a) \: \: \: \: \: \: \frac{ \cos( \alpha - \beta ) }{sin \alpha \: cos\beta } = ctg\alpha + \: tg\beta \\ \frac{cos \alpha cos \beta + sin \alpha sin \beta }{sin \alpha \: cos\beta } = ctg\alpha + \: tg\beta \\ \frac{cos \alpha cos \beta}{sin \alpha \: cos\beta} + \frac{sin \alpha sin \beta}{sin \alpha \: cos\beta} = ctg\alpha + \: tg\beta \\ ctg\alpha + \: tg\beta =ctg\alpha + \: tg\beta \\$

$б) \: \: \: \sin( \frac{\pi}{6} + \alpha ) + \sin( \frac{\pi}{6} - \alpha ) = \cos \alpha \\ sin \frac{\pi}{6} cos \alpha + cos \frac{\pi}{6} sin \alpha + sin \frac{\pi}{6} cos \alpha - cos \frac{\pi}{6} sin \alpha = cos \alpha \\ \frac{1}{2} cos \alpha + \frac{1}{2} cos \alpha = cos \alpha \\ cos \alpha = cos \alpha$

$tg \alpha = 2 \\ tg \beta = 1 \\ \tg( \alpha + \beta ) = \frac{tg \alpha + tg \beta }{1 - tg \alpha \: tg \beta } = \\ = \frac{2 + 1}{1 - 2 \times 1} = \frac{3}{ - 1} = - 3$

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