נתון:
צ"ל:
הוכחה:
∠ β = 90 ∘ ∠ α = 30 ∘ ↓ ∠ A = 90 ∘ − 30 ∘ = 60 ∘ ↓ A B = 1 2 O A ( △ ) A O = 1 c m = r ( U n i t c i r c l e s ) ↓ sin ( α ∘ ) = A B = 1 2 ↓ ( 1 2 ) 2 + O B 2 = 1 O B 2 = 1 − 1 4 = 3 4 O B = 3 4 = 3 2 {\displaystyle {\begin{aligned}&\angle \beta =90^{\circ }\\&\angle \alpha =30^{\circ }\\&\downarrow \\&\angle A=90^{\circ }-30^{\circ }=60^{\circ }\\&\downarrow \\&AB={\frac {1}{2}}OA({\color {Gray}\triangle })\\&AO=1_{cm}=r(Unitcircles)\\&\downarrow \\&{\color {blue}\sin(\alpha ^{\circ })=AB={\frac {1}{2}}}\\&\downarrow \\&({\frac {1}{2}})^{2}+OB^{2}=1\\&OB^{2}=1-{\frac {1}{4}}={\frac {3}{4}}\\&{\color {blue}OB={\sqrt {\frac {3}{4}}}={\frac {\sqrt {3}}{2}}}\\\end{aligned}}}
ע"פ הנחה זו נוכל למצוא את tan ( 30 ∘ ) {\displaystyle \tan(30^{\circ })} :
t a n ( 30 ∘ ) = sin ( 30 ∘ ) cos ( 30 ∘ ) = 1 2 3 2 = 1 3 = 3 3 {\displaystyle tan(30^{\circ })={\frac {\sin(30^{\circ })}{\cos(30^{\circ })}}={\frac {\frac {1}{2}}{\frac {\sqrt {3}}{2}}}={\frac {1}{\sqrt {3}}}={\frac {\sqrt {3}}{3}}}