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Siberianwild @Siberianwild

July 2022 1 5 Report

(-2cosx)/(3sinx+cosx+2)
выразить через z=tg(x/2)

Answers & Comments

zЛена Tg(x/2)=(sinx-2cosx)/(sinx+2cosx)

выразим всё через tg(x/2)
sinx=2tg(x/2)/(1+(tg(x/2))^2)
cosx=(1-(tg(x/2))^2)/(1+(tg(x/2))^2)

sinx-2cosx=2tg(x/2)/(1+(tg(x/2))^2)-2*(1-(tg(x/2))^2)/(1+(tg(x/2))^2)=(2tg(x/2)-2+2*(tg(x/2))^2)/(1+(tg(x/2))^2)
sinx+2cosx=2tg(x/2)/(1+(tg(x/2))^2)+2*(1-(tg(x/2))^2)/(1+(tg(x/2))^2)=(2tg(x/2)+2-2*(tg(x/2))^2)/(1+(tg(x/2))^2)

tg(x/2)=(2tg(x/2)-2+2*(tg(x/2))^2)/(2tg(x/2)+2-2*(tg(x/2))^2)
tg(x/2)=(tg(x/2)-1+(tg(x/2))^2)/(tg(x/2)+1-(tg(x/2))^2)
tg(x/2)-1+(tg(x/2))^2=tg(x/2)*(tg(x/2)+1-(tg(x/2))^2)
tg(x/2)-1+(tg(x/2))^2=(tg(x/2))^2+tg(x/2)-(tg(x/2))^3
-1=-(tg(x/2))^3
(tg(x/2))^3=1
tg(x/2)=1
x/2=пи/4+пи*k
x=пи/2+2*пи*k, k - целые числа

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