Question

Find the terminal point for $\frac{\pi }{6}$, suppose the terminal point determined by$t=\frac{\pi }{6}$ is$P\left(x,y\right)$and the points role="math" localid="1648559189794" $Q$and width="14" height="19" role="math">R are as shown in the figure.

Why are the distance $PQ$and $PR$the same? Use this fact, together with the Distance Formula, to show that the coordinates of $P$satisfy the equation $2y=\sqrt{x^2+{\left(y-1\right)}^2}$. Simplify this equation using the fact that $x^2+y^2=1$.

Solve the simplified equation to find $P\left(x,y\right)$.

Short answer

The terminal point is $\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$.

Step-by-step solution

Given information.

The given value $t$ is $\frac{\pi }{6}$.

Calculation.

Now, find the terminal point.

$$\begin{gathered} P\left(x,y\right)=\left(\sin \theta ,\cos \theta \right) \\ =\left(\sin \frac{\pi }{6},\cos \frac{\pi }{6}\right) \\ =\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right) \end{gathered}$$

If two equal chords of a circle intersect inside the circle, then the line joining the point of intersection to the center makes equal angles with the chords.

Now, substitute the value of terminal point in the given equation.

$$\begin{gathered} 2y=\sqrt{x^2+{\left(y-1\right)}^2} \\ 4y^2=x^2+y^2+1-2y \\ 4{\left(\frac{\sqrt{3}}{2}\right)}^2={\left(\frac{1}{2}\right)}^2+{\left(\frac{\sqrt{3}}{2}\right)}^2-2\left(\frac{\sqrt{3}}{2}\right) \\ 4\left(\frac{3}{4}\right)=\left(\frac{1}{4}+\frac{3}{4}\right)-\sqrt{3} \\ 3=1-\sqrt{3} \\ 2=-1.732 \end{gathered}$$

Since, the both sides are not equation. Thus, the given equation is not satisfied the terminal point $\left(\frac{1}{2},\frac{\sqrt{3}}{2}\right)$.

Now, simplify the equation.

localid="1648560574363" $$\begin{gathered} 2y=\sqrt{x^2+{\left(y-1\right)}^2} \\ 4y^2=x^2+y^2+1-2y\left[x^2+y^2=1\right] \\ 4y^2+2y=2+1 \\ 4y^2+2y-3=0 \end{gathered}$$

Now, solve the equation.

$$\begin{gathered} y=\frac{-b\pm \sqrt{b^2-4ac}}{2a} \\ =\frac{-2\pm \sqrt{4+48}}{8} \\ =\frac{-1+\sqrt{13}}{4},-\frac{1+\sqrt{13}}{4} \\ =\left(0.65138,-1.15138\right) \end{gathered}$$