2[tex]sin^{2} [/tex]a+[tex] \sqrt{2} [/tex]cosa+tga, ctga=1. Created by МелисСая. algebra-ru

tex]cosa+tga, ctga=1...

August 2022 1 7 Report

2[tex]sin^{2} [/tex]a+[tex] \sqrt{2} [/tex]cosa+tga, ctga=1

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red321 $ctg(a)=1\\a=arcctg1\\a=\frac{\pi}{4}$

$(sina)^2+\sqrt{2}cosa+tga=(sin\frac{\pi}{4})^2+\sqrt{2}*cos\frac{\pi}{4}+tg\frac{\pi}{4}=\\=(\frac{\sqrt{2}}{2})^2+\sqrt{2}*\frac{\sqrt{2}}{2}+1=\frac{1}{2}+1+1=2.5$

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