cos(x-y)(tgx tgy-1)+(1+tgx tgy)cos(x+y) 30 баллов!. Created by Ismailova00. algebra-ru

cos(x-y)(tgx tgy-1)+(1+tgx tgy)cos(x+y) 30 баллов!...

July 2022 1 10 Report

cos(x-y)(tgx tgy-1)+(1+tgx tgy)cos(x+y)
30 баллов!

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o1l7eg17 $cos(x-y)*(tgx*tgy-1)+(1+tgx*tgy)*cos(x+y)=$

$=cos(x-y)*(\frac{sinx}{cosx}*\frac{siny}{cosy}-1)+(1+\frac{sinx}{cosx}*\frac{siny}{cosy})*cos(x+y)=$

$=cos(x-y)*\frac{sinx*siny-cosx*cosy}{cosx*cosxy}+cos(x+y)*\frac{cosx*cosy+sinx*siny}{cosx*cosy}=$

$=\frac{cos(x-y)*(sinx*siny-cosx*cosy)}{cosx*cosy}+\frac{cos(x+y)*(cosx*cosy+sinx*siny)}{cosx*cosy}=$

$=\frac{cos(x-y)*(sinx*siny-cosx*cosy)+cos(x+y)*(cosx*cosy+sinx*siny)}{cosx*cosy}=$

$=cos(x-y)(sinxsiny-cosxcosy)+cos(x+y)(cosxcosy+sinxsiny)$

$=(cos(x)cos(y)+sin(x)sin(y))(sin(x)sin(y)-cos(x)cos(y))+$
$+(cos(x)cos(y)-sin(x)sin(y))*(cos(x)cos(y)+sin(x)sin(y))=$

$=sin^2(x)sin^2(y)-cos^2(x)cos^2(y)+cos^2(x)cos^2(y)-sin^2(x)sin^2(y)=$
$=0$

1 votes Thanks 8

Ismailova00 Спасибо большое)

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Ismailova00 August 2022 | 0 Ответы

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