Răspuns:
[tex]Ex.3)[/tex] [tex]x[/tex]∈[tex](0,\frac{\pi }{2} ) => sin x >0;cos x > 0[/tex]
[tex]sin^2x+cos^2x=1 => cos^2x=1-sin^2x <=>1-\frac{9}{25} <=>\frac{25-9}{25}[/tex]
[tex]=\frac{16}{25} => cosx=\frac{4}{5}[/tex]
[tex]sin(180-x)=sin(180)*cosx-sin(x)*cos(180)=0*cosx-sinx*(-1)[/tex]
[tex]=sinx=\frac{3}{5}[/tex]
[tex]cos(180-x)=cos(180)*cos(x)+sin(180)*sin(x)=-1*cos(x)+0*sin(x)[/tex]
[tex]=-cos(x)=-\frac{4}{5}[/tex]
[tex]tg(180-x)=-tgx =-\frac{sinx}{cosx}=-\frac{\frac{3}{5} }{\frac{4}{5} } <=>-\frac{3*5}{4*5}=-\frac{3}{4}[/tex]
[tex]ctg(180-x)=-ctgx=-\frac{cosx}{sinx}=-\frac{4}{3}[/tex]
[tex]Ex.4)[/tex] [tex]x[/tex]∈[tex](0,\frac{\pi }{2} ) => sin x >0;cos x > 0[/tex]
[tex]sin^2x+cos^2x=1=>sin^2x=1-cos^2x <=>1-\frac{1}{9}=\frac{9-1}{9}=\frac{8}{9}=>\\sinx=\frac{2\sqrt{2} }{3}[/tex]
[tex]sin(\pi -x)=sin\pi *cosx-sinx*cos\pi =0*cosx-\frac{2\sqrt{2} }{3} *(-1)\\=\frac{2\sqrt{2} }{3}[/tex]
[tex]cos(\pi -x)=cos\pi *cosx+sin\pi *sinx=(-1)*\frac{1}{3}+0*sinx=[/tex]
[tex]=-\frac{1}{3}[/tex]
[tex]tg(\pi -x)=-tgx=-\frac{sinx}{cosx}=-\frac{\frac{2\sqrt{2} }{3} }{\frac{1}{3} }=-\frac{2\sqrt{2}*3 }{1*3} =-2\sqrt{2}[/tex]
[tex]ctg(\pi -x)=-ctgx=-\frac{1}{tgx}=-\frac{1}{2\sqrt{2} }=-\frac{\sqrt{2} }{4}[/tex]
