Răspuns:
[tex]6) sin^2x+cos^2x=1[/tex]
[tex]sinx=\frac{cateta// opusa}{ipotenuza}[/tex]
[tex]cosx=\frac{cateta// alaturata}{ipotenuza}[/tex]
[tex]sinx=\frac{AB}{AC}, cosx=\frac{BC}{AC}[/tex]
[tex]sin^2x+cos^2x=\frac{AB^2}{AC^2}+\frac{BC^2}{AC^2}<=>\frac{AB^2+BC^2}{AC^2} = \frac{AC^2}{AC^2}=1[/tex]
Explicatie: [tex]AB^2+BC^2=AC^2[/tex] - din teorema lui Pitagora
[tex]tgx=\frac{cateta//opusa}{cateta//alaturata} \\ctg=\frac{cateta//alaturata}{cateta//opusa}[/tex]
[tex]tgx=\frac{AB}{BC}, ctgx=\frac{BC}{AB}[/tex]
[tex]tgx*ctgx=\frac{AB}{BC}*\frac{BC}{AB}=1[/tex]
[tex]\frac{sinx}{cosx} =tgx <=>\frac{\frac{AB}{AC} }{\frac{BC}{AC} }=\frac{AB}{BC}=tgx[/tex] "Adevarat"
[tex]\frac{cosx}{sinx}=ctgx<=>\frac{\frac{BC}{AC} }{\frac{AB}{AC} }=\frac{BC}{AB}=ctgx[/tex] "Adevarat"
[tex]7) sinx=\frac{1}{4} = \frac{AB}{AC}=>AB=1k, AC=4k[/tex]
[tex]BC^2=AC^2-AB^2<=>(4k)^2-(k)^2 = 16k^2-k^2=>BC^2=15k^2=>\\=>BC=k\sqrt{15}[/tex]
[tex]cosx=\frac{BC}{AC}=\frac{k\sqrt{15} }{4k} =\frac{\sqrt{15} }{4}[/tex]
[tex]tgx=\frac{AB}{BC}=\frac{k}{k\sqrt{15} } =\frac{\sqrt{15} }{15}[/tex]
[tex]ctgx=\frac{BC}{AB}=\frac{k\sqrt{15} }{k}=\sqrt{15}[/tex]
Unghiul x = m(<BCA), am lucrat pe triunghiul din imagine:
