# How do you find the terminal point on the unit

How do you find the terminal point on the unit circle determined by $$\displaystyle{t}={5}\frac{\pi}{{12}}$$ ?

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lovagwb
Explanation:
The arc is $$\displaystyle{t}={\frac{{{5}\pi}}{{{12}}}}$$
x-cordinate is : $$\displaystyle{x}={\cos{{\left({\frac{{{5}\pi}}{{{12}}}}\right)}}}$$
y-coordinate: $$\displaystyle{y}={\sin{{\left({\frac{{{5}\pi}}{{{12}}}}\right)}}}$$
Calculator gives $$\displaystyle{x}={\sin{{\left({\frac{{{5}\pi}}{{{12}}}}\right)}}}={\sin{{75}}}={0.97}$$
and $$\displaystyle{y}={\cos{{75}}}={0.26}$$
###### Not exactly what you’re looking for?
jean2098
The terminal point on the unit circle has the cosine as x-coordinated and the sine as y-coordinate.
$$\displaystyle{\left({\cos{\theta}},\ {\sin{\theta}}\right)}$$
In this case $$\displaystyle\theta={\frac{{{5}\pi}}{{{12}}}}$$
$$\displaystyle{\left({\cos{{\frac{{{5}\pi}}{{{12}}}}}},{\sin{{\frac{{{5}\pi}}{{{12}}}}}}\right)}$$
As $$\displaystyle\theta={\frac{{{5}\pi}}{{{12}}}}$$ is not one of the special angles, we will use a calcutator to evaluate the cousine and the sine at $$\displaystyle\theta={\frac{{{5}\pi}}{{{12}}}}$$.
$$\displaystyle{\cos{{\frac{{{5}\pi}}{{{12}}}}}}={{\cos{{75}}}^{\circ}\approx}{0.2588}$$
$$\displaystyle{\sin{{\frac{{{5}\pi}}{{{12}}}}}}={{\sin{{75}}}^{\circ}\approx}{0.9659}$$
Thus the coordinates of the terminal point at $$\displaystyle\theta={\frac{{{5}\pi}}{{{12}}}}$$ are then:
$$\displaystyle{\left({\cos{{\frac{{{5}\pi}}{{{12}}}}}},{\sin{{\frac{{{5}\pi}}{{{12}}}}}}\right)}={\left({0.2588},{0.9659}\right)}$$
nick1337
Maybe this will help me, thanks