Cos(x-y)(tgx tgy-1)+(1+tgx tgy)cos(x+y) 30 баллов!
Cos(x-y)(tgx tgy-1)+(1+tgx tgy)cos(x+y)
30 баллов!
Ответ(ы) на вопрос:
[latex]cos(x-y)*(tgx*tgy-1)+(1+tgx*tgy)*cos(x+y)=[/latex]
[latex]=cos(x-y)*(\frac{sinx}{cosx}*\frac{siny}{cosy}-1)+(1+\frac{sinx}{cosx}*\frac{siny}{cosy})*cos(x+y)=[/latex]
[latex]=cos(x-y)*\frac{sinx*siny-cosx*cosy}{cosx*cosxy}+cos(x+y)*\frac{cosx*cosy+sinx*siny}{cosx*cosy}=[/latex]
[latex]=\frac{cos(x-y)*(sinx*siny-cosx*cosy)}{cosx*cosy}+\frac{cos(x+y)*(cosx*cosy+sinx*siny)}{cosx*cosy}=[/latex]
[latex]=\frac{cos(x-y)*(sinx*siny-cosx*cosy)+cos(x+y)*(cosx*cosy+sinx*siny)}{cosx*cosy}=[/latex]
[latex]=cos(x-y)(sinxsiny-cosxcosy)+cos(x+y)(cosxcosy+sinxsiny)[/latex]
[latex]=(cos(x)cos(y)+sin(x)sin(y))(sin(x)sin(y)-cos(x)cos(y))+[/latex]
[latex]+(cos(x)cos(y)-sin(x)sin(y))*(cos(x)cos(y)+sin(x)sin(y))=[/latex]
[latex]=sin^2(x)sin^2(y)-cos^2(x)cos^2(y)+cos^2(x)cos^2(y)-sin^2(x)sin^2(y)=[/latex]
[latex]=0[/latex]
Не нашли ответ?
Похожие вопросы