Question

what are the coordinates of the point that corresponds to $-\frac{5pi}{4}$ on the unit circle?

$\big(\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\big)$

$\big(-\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\big)$

$\big(-\frac{\sqrt{2}}{2}, \frac{\sqrt{2}}{2}\big)$

$\big(\frac{\sqrt{2}}{2}, -\frac{\sqrt{2}}{2}\big)$

Explanation:

Step1: Find Coterminal Angle

To find the coterminal angle of $-\frac{5\pi}{4}$, we add $2\pi$ (since coterminal angles differ by multiples of $2\pi$).
$$-\frac{5\pi}{4}+2\pi = -\frac{5\pi}{4}+\frac{8\pi}{4}=\frac{3\pi}{4}$$
Wait, no, actually, another way: we can also add $2\pi$ but maybe a better way is to find the reference angle. Wait, $-\frac{5\pi}{4}$ is equivalent to rotating clockwise $\frac{5\pi}{4}$ radians. The reference angle for an angle $\theta$ (in standard position) is the acute angle it makes with the x - axis. For $-\frac{5\pi}{4}$, we can find the positive coterminal angle by adding $2\pi$: $-\frac{5\pi}{4}+2\pi=\frac{3\pi}{4}$? No, wait, $-\frac{5\pi}{4}+2\pi=\frac{3\pi}{4}$? Wait, $2\pi=\frac{8\pi}{4}$, so $-\frac{5\pi}{4}+\frac{8\pi}{4}=\frac{3\pi}{4}$. But the reference angle for $\frac{3\pi}{4}$ is $\pi-\frac{3\pi}{4}=\frac{\pi}{4}$. Wait, no, maybe I made a mistake. Let's think again. The angle $-\frac{5\pi}{4}$: when we rotate clockwise $\frac{5\pi}{4}$, which is the same as rotating counter - clockwise $2\pi-\frac{5\pi}{4}=\frac{3\pi}{4}$. Wait, no, $2\pi-\frac{5\pi}{4}=\frac{8\pi - 5\pi}{4}=\frac{3\pi}{4}$. But the reference angle for an angle in the second quadrant ($\frac{\pi}{2}<\theta<\pi$) is $\pi-\theta$. For $\frac{3\pi}{4}$, reference angle is $\pi-\frac{3\pi}{4}=\frac{\pi}{4}$. But wait, actually, the angle $-\frac{5\pi}{4}$ is in the second quadrant? No, wait, let's plot the angle. A negative angle means clockwise rotation. So starting from the positive x - axis, rotating clockwise $\frac{\pi}{2}$ (to negative y - axis), $\pi$ (to negative x - axis), $\frac{3\pi}{2}$ (to positive y - axis), and $\frac{5\pi}{4}$ is $\pi+\frac{\pi}{4}$, so clockwise $\frac{5\pi}{4}$ is equivalent to counter - clockwise $2\pi-\frac{5\pi}{4}=\frac{3\pi}{4}$. Wait, $\frac{3\pi}{4}$ is in the second quadrant. But the coordinates on the unit circle are given by $(\cos\theta,\sin\theta)$. Wait, no, maybe I messed up the coterminal angle. Let's do it correctly. The angle $-\frac{5\pi}{4}$: let's find the reference angle. The formula for reference angle:

- If $\theta$ is in quadrant I, reference angle is $\theta$.
- If $\theta$ is in quadrant II, reference angle is $\pi - \theta$.
- If $\theta$ is in quadrant III, reference angle is $\theta-\pi$.
- If $\theta$ is in quadrant IV, reference angle is $2\pi-\theta$.

First, find the positive coterminal angle of $-\frac{5\pi}{4}$: $-\frac{5\pi}{4}+2\pi=\frac{3\pi}{4}$. Wait, $\frac{3\pi}{4}$ is in quadrant II. But wait, $-\frac{5\pi}{4}$: let's see, when we rotate clockwise $\frac{5\pi}{4}$, we pass through clockwise $\pi$ (to negative x - axis) and then another $\frac{\pi}{4}$, so we are in the second quadrant? No, that's not right. Wait, clockwise rotation: from positive x - axis, clockwise $\frac{\pi}{2}$ is (0, - 1), clockwise $\pi$ is (- 1,0), clockwise $\frac{3\pi}{2}$ is (0,1), clockwise $2\pi$ is (1,0). So clockwise $\frac{5\pi}{4}$: $\frac{5\pi}{4}=\pi+\frac{\pi}{4}$, so after rotating clockwise $\pi$ (to (- 1,0)), we rotate another $\frac{\pi}{4}$ clockwise, which is towards the second quadrant? Wait, no, clockwise from (- 1,0) rotating $\frac{\pi}{4}$ clockwise would be towards the second quadrant (x negative, y positive). Wait, but the coordinates on the unit circle are $(\cos\theta,\sin\theta)$. Let's calculate $\cos(-\frac{5\pi}{4})$ and $\sin(-\frac{5\pi}{4})$.

We know that $\cos(-\alpha)=\cos\alpha$ and $\sin(-\alpha)=-\sin\alpha$. So:

$\cos(-\frac{5\pi}{4})=\cos(\frac{5\pi}{4})$ and $\sin(-\frac{5\pi}{4})=-\sin(\frac{5\pi}{4})$

Now, $\frac{5\pi}{4}$ is in the third quadrant, where both cosine and sine are negative. The reference angle for $\frac{5\pi}{4}$ is $\frac{5\pi}{4}-\pi=\frac{\pi}{4}$. So $\cos(\frac{5\pi}{4})=-\cos(\frac{\pi}{4})=-\frac{\sqrt{2}}{2}$ and $\sin(\frac{5\pi}{4})=-\sin(\frac{\pi}{4})=-\frac{\sqrt{2}}{2}$

Then $\cos(-\frac{5\pi}{4})=\cos(\frac{5\pi}{4})=-\frac{\sqrt{2}}{2}$ and $\sin(-\frac{5\pi}{4})=-\sin(\frac{5\pi}{4})=-(-\frac{\sqrt{2}}{2})=\frac{\sqrt{2}}{2}$

Wait, that gives $(\cos(-\frac{5\pi}{4}),\sin(-\frac{5\pi}{4}))=(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$? Wait, no, wait:

Wait, $\cos(-\theta)=\cos\theta$, so $\cos(-\frac{5\pi}{4})=\cos(\frac{5\pi}{4})$. $\frac{5\pi}{4}$ is in the third quadrant, so $\cos(\frac{5\pi}{4})=-\cos(\frac{\pi}{4})=-\frac{\sqrt{2}}{2}$

$\sin(-\theta)=-\sin\theta$, so $\sin(-\frac{5\pi}{4})=-\sin(\frac{5\pi}{4})$. $\sin(\frac{5\pi}{4})=-\sin(\frac{\pi}{4})=-\frac{\sqrt{2}}{2}$, so $-\sin(\frac{5\pi}{4})=-(-\frac{\sqrt{2}}{2})=\frac{\sqrt{2}}{2}$

So the coordinates are $(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$? Wait, but let's check again.

Wait, maybe a better approach: The angle $-\frac{5\pi}{4}$: let's find the coterminal angle by adding $2\pi$: $-\frac{5\pi}{4}+2\pi = -\frac{5\pi}{4}+\frac{8\pi}{4}=\frac{3\pi}{4}$. Now, $\frac{3\pi}{4}$ is in the second quadrant, where x is negative and y is positive. The reference angle for $\frac{3\pi}{4}$ is $\pi-\frac{3\pi}{4}=\frac{\pi}{4}$. So $\cos(\frac{3\pi}{4})=-\cos(\frac{\pi}{4})=-\frac{\sqrt{2}}{2}$ and $\sin(\frac{3\pi}{4})=\sin(\frac{\pi}{4})=\frac{\sqrt{2}}{2}$. Since $-\frac{5\pi}{4}$ and $\frac{3\pi}{4}$ are coterminal (they have the same terminal side), their cosine and sine values are the same. So $\cos(-\frac{5\pi}{4})=\cos(\frac{3\pi}{4})=-\frac{\sqrt{2}}{2}$ and $\sin(-\frac{5\pi}{4})=\sin(\frac{3\pi}{4})=\frac{\sqrt{2}}{2}$. So the coordinates are $(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$

Step2: Match with Options

Looking at the options:
- Option 1: $(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$ (both positive, first quadrant) - incorrect.
- Option 2: $(-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})$ (both negative, third quadrant) - incorrect.
- Option 3: $(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$ (x negative, y positive, second quadrant) - correct.
- Option 4: $(\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2})$ (x positive, y negative, fourth quadrant) - incorrect.

Answer:

$(-\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2})$ (the third option)