[tex]\mathsf{Given : Terminal\;point - \left(\dfrac{\sqrt{3}}{2},\dfrac{1}{2} \right)}[/tex]
[tex]\mathsf{\bigstar\;\;x - coordinate : \dfrac{\sqrt{3}}{2}}[/tex]
[tex]\mathsf{\bigstar\;\;y - coordinate : \dfrac{1}{2}}[/tex]
Now consider each option :
[tex]\mathsf{\bigstar\;\;Option\;(A) : cos\left(\dfrac{\pi}{6}\right) = cos30^{\circ} = \dfrac{\sqrt{3}}{2}}[/tex]
[tex]\mathsf{\bigstar\;\;Option\;(B) : sin\left(\dfrac{\pi}{6}\right) = sin30^{\circ} = \dfrac{1}{2}}[/tex]
[tex]\mathsf{\bigstar\;\;Option\;(C) : cos\left(\dfrac{\pi}{3}\right) = cos60^{\circ} = \dfrac{1}{2}}[/tex]
[tex]\mathsf{\bigstar\;\;Option\;(D) : sin\left(\dfrac{\pi}{3}\right) = sin60^{\circ} = \dfrac{\sqrt{3}}{2}}[/tex]
From the Above, We can notice that :
[tex]\mathsf{\bigstar\;\;sin\left(\dfrac{\pi}{6}\right)\;and\;cos\left(\dfrac{\pi}{3}\right)\;are\;equivalent\;to\;\dfrac{1}{2}}[/tex]
Answers : Option (B) and Option (C)
