Notam triunghiul: ΔABC m(∡B)=x ⇒m(∡C)= 90°-x. Conform definitiilor finctiilor trigonometrice in tiunghiul dreptunghic:sinB=sinx=[tex] \frac{cateta,opusa}{ipotenuza}= \frac{AC}{BC},dar:cosC=cos(90-x)= \frac{cateta,alaturata}{ipotenuza}= \frac{AC}{BC}=sinx [/tex]. La fel se demonstreaza toate:
cosx=cosB=AB/BC=sinC=sin(90-x); tgx=tgB=AC/AB=ctgC=ctg(90-x);
ctgx=ctgB=AB/AC=tgC=tg(90-x)