Question

Buna seara , poate cineva sa ma ajute cu aceste 2 exercitii, dau coroana si funda , multumesc anticipat

Answer 1

Răspuns:

$Ex.3)$ $x$∈$(0,\frac{\pi }{2} ) => sin x >0;cos x > 0$

$sin^2x+cos^2x=1 => cos^2x=1-sin^2x <=>1-\frac{9}{25} <=>\frac{25-9}{25}$

$=\frac{16}{25} => cosx=\frac{4}{5}$

$sin(180-x)=sin(180)*cosx-sin(x)*cos(180)=0*cosx-sinx*(-1)$

$=sinx=\frac{3}{5}$

$cos(180-x)=cos(180)*cos(x)+sin(180)*sin(x)=-1*cos(x)+0*sin(x)$

$=-cos(x)=-\frac{4}{5}$

$tg(180-x)=-tgx =-\frac{sinx}{cosx}=-\frac{\frac{3}{5} }{\frac{4}{5} } <=>-\frac{3*5}{4*5}=-\frac{3}{4}$

$ctg(180-x)=-ctgx=-\frac{cosx}{sinx}=-\frac{4}{3}$

$Ex.4)$ $x$∈$(0,\frac{\pi }{2} ) => sin x >0;cos x > 0$

$sin^2x+cos^2x=1=>sin^2x=1-cos^2x <=>1-\frac{1}{9}=\frac{9-1}{9}=\frac{8}{9}=>\\sinx=\frac{2\sqrt{2} }{3}$

$sin(\pi -x)=sin\pi *cosx-sinx*cos\pi =0*cosx-\frac{2\sqrt{2} }{3} *(-1)\\=\frac{2\sqrt{2} }{3}$

$cos(\pi -x)=cos\pi *cosx+sin\pi *sinx=(-1)*\frac{1}{3}+0*sinx=$

$=-\frac{1}{3}$

$tg(\pi -x)=-tgx=-\frac{sinx}{cosx}=-\frac{\frac{2\sqrt{2} }{3} }{\frac{1}{3} }=-\frac{2\sqrt{2}*3 }{1*3} =-2\sqrt{2}$

$ctg(\pi -x)=-ctgx=-\frac{1}{tgx}=-\frac{1}{2\sqrt{2} }=-\frac{\sqrt{2} }{4}$