What are the coordinates for the point $p (\frac{5 π}{6})$ $p((5pi)/6)$ where $p (θ) = (x, y)$ $p(theta)=(x,y)$ is the point where the terminal arm of an angle $θ$ $theta$ intersects the unit circl?

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What are the coordinates for the point $p (\frac{5 π}{6})$ $p((5pi)/6)$ where $p (θ) = (x, y)$ $p(theta)=(x,y)$ is the point where the terminal arm of an angle $θ$ $theta$ intersects the unit circl?

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Answer 1 · 2017-12-31T12:13:08

$p (\frac{5 π}{6}) = (- \frac{\sqrt{3}}{2}, \frac{1}{2})$ $p((5pi)/6)=(-sqrt(3)/2,1/2)$

Explanation:

$\frac{5 π}{6}$ $(5pi)/6$ has reference angle $\frac{π}{6}$ $pi/6$ because $π - \frac{5 π}{6} = \frac{π}{6}$ $pi-(5pi)/6 = pi/6$ .

So we know that:
$∣ ∣ ∣ cos (\frac{5 π}{6}) ∣ ∣ ∣ = cos (\frac{π}{6}) = \frac{\sqrt{3}}{2}$ $|cos((5pi)/6)| = cos(pi/6) = sqrt(3)/2$
$∣ ∣ ∣ sin (\frac{5 π}{6}) ∣ ∣ ∣ = sin (\frac{π}{6}) = \frac{1}{2}$ $|sin((5pi)/6)| = sin(pi/6) = 1/2$

Since $\frac{5 π}{6} \in$ $(5pi)/6 in$ QII, we know that $cos (\frac{5 π}{6}) < 0$ $cos((5pi)/6)<0$ and $sin (\frac{5 π}{6}) > 0$ $sin((5pi)/6)>0$ .

Combining what we know, we have that $p (\frac{5 π}{6}) = (- \frac{\sqrt{3}}{2}, \frac{1}{2})$ $p((5pi)/6)=(-sqrt(3)/2,1/2)$ .

I recommend memorizing the entire Unit Circle as it's not that difficult and is extremely useful.

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What are the coordinates for the point $p (\frac{5 π}{6})$ $p((5pi)/6)$ where $p (θ) = (x, y)$ $p(theta)=(x,y)$ is the point where the terminal arm of an angle $θ$ $theta$ intersects the unit circl?

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What are the coordinates for the point p(5π6)p((5pi)/6) where p(θ)=(x,y)p(theta)=(x,y) is the point where the terminal arm of an angle θtheta intersects the unit circl?

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Explanation:

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What are the coordinates for the point $p (\frac{5 π}{6})$ $p((5pi)/6)$ where $p (θ) = (x, y)$ $p(theta)=(x,y)$ is the point where the terminal arm of an angle $θ$ $theta$ intersects the unit circl?