Given that a point is marked on the unit circle say it is [tex]\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)[/tex]
That meanns base of the triangle
[tex]AB=\frac{1}{2}[/tex]
and hight of the right angle triangle
[tex]BC=\frac{\sqrt{3}}{2}[/tex]
Since it is unit circle so automatically AC=1
or you can use Pythagorean theorem to find side AC.
Now we can use ratios of sine, cos and tan to find their values as follows:
[tex]\sin\left(A\right)=\frac{Perpendicular}{Hypotenuse}=\frac{BC}{AC}=\frac{\frac{\sqrt{3}}{2}}{1}=\frac{\sqrt{3}}{2}[/tex]
[tex]Cos\left(A\right)=\frac{Base}{Hypotenuse}=\frac{AB}{AC}=\frac{\frac{1}{2}}{1}=\frac{1}{2}[/tex]
[tex]\tan\left(A\right)=\frac{Perpendicular}{Base}=\frac{BC}{AB}=\frac{\frac{\sqrt{3}}{2}}{\frac{1}{2}}=\sqrt{3}[/tex]
Same way you can find values of sin, cos, tan etc for any given point on the circle.
