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July 2022 1 10 Report

Найти значение sinx и cosx если tgx=-3 и 3п/2 <х<2п

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$1+tg ^{2} x= \frac{1}{Cos ^{2}x }\\\\Cos ^{2} x= \frac{1}{1+tg ^{2}x }= \frac{1}{1+(-3) ^{2} } = \frac{1}{1+9} = \frac{1}{10}$
x - угол четвёртой четверти, значит Cosx > 0
$Cosx= \sqrt{ \frac{1}{10} }= \frac{1}{ \sqrt{10} } \\\\Sinx= -\sqrt{1-Cos ^{2} x}= -\sqrt{1- \frac{1}{10} }= -\sqrt{ \frac{9}{10} } = -\frac{3}{ \sqrt{10} }$

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v share na rasstoyanii 12 sm vid jogo centra provedeno sechenie ploshad kotorogo 64

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vikamolchanova1 July 2022 | 0 Ответы

ochen srochno ploshad pryamougolnoj trapecii ravna 32 kornya iz 3 a ostryj ugol 6

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ya treugolnike avs av8 sm vs10 sm as13 smm seredina avk seredina vsnajt

koren iz cos2x cosx0

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v share na rasstoyanii 12 sm vid jogo centra provedeno sechenie ploshad kotorogo 64

ochen srochno ploshad pryamougolnoj trapecii ravna 32 kornya iz 3 a ostryj ugol 6

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