Помогите решить два задания ! (Подробное решение). Created by DomnikaMaksimova. algebra-ru

Помогите решить два задания ! (Подробное решение)...

DomnikaMaksimova @DomnikaMaksimova

August 2022 1 56 Report

Помогите решить два задания ! (Подробное решение)

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$sin 2x=?$

1) $x= \frac{ \pi }{4}$

$Sin 2( \frac{ \pi }{4})=Sin ( \frac{ \pi }{2})=1$

2) $x= \frac{ \pi }{3}$

$Sin 2( \frac{ \pi }{3})=Sin \frac{2 \pi }{3}= \frac{ \sqrt{3} }{2}$

3) $x= \frac{-2 \pi }{3}$

$Sin 2( \frac{-2 \pi }{3})=Sin\frac{-4 \pi }{3}=-Sin \frac{4 \pi }{3}=-Sin( \pi + \frac{ \pi }{3})=Sin \frac{ \pi }{3}= \frac{ \sqrt{3} }{2}$

4) $x= \frac{-5 \pi }{6}$

$Sin2(- \frac{5 \pi }{6})=-Sin \frac{5 \pi }{3}=-Sin( \pi + \frac{2 \pi }{3})=Sin \frac{2 \pi }{3}= \frac{ \sqrt{3} }{2}$

$y= \frac{1}{Sin^2x}= \frac{1}{ \frac{1-Cos2x}{2}}= \frac{2}{1-Cos2x}$

1) $x=- \frac{ \pi }{2}$

$\frac{2}{1-Cos(- \pi )} = \frac{1}{1-Cos \pi }= \frac{2}{1+1}=1$

2) $x= \frac{3 \pi }{4}$

$\frac{2}{1-Cos( \frac{3 \pi }{2})}= \frac{2}{1-0}=2$

3) $x=- \frac{10 \pi }{3}$

$\frac{2}{1-Cos(-20 \pi /3)}= \frac{2}{1-Cos( 6\pi + \frac{2 \pi }{3})}= \frac{2}{1-Cos \frac{2 \pi }{3}}= \frac{2}{1+ \frac{1}{2}}= \frac{4}{3}$

4) $x=- \frac{19 \pi}{4}$

$\frac{2}{1-Cos(- \frac{19 \pi }{2})}= \frac{2}{1-Cos(10 \pi- \frac{ \pi }{2})}= \frac{2}{1-Cos \frac{ \pi }{2}}= \frac{2}{1-0}=2$

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DomnikaMaksimova спасибо огромное !!!

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