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Find the coordinates of a point on the unit circle that corresponds to an angle of $frac{7pi}{4}$

Answer 1

Abigail Nelson

Samuel Scott

To find the coordinates of the point on the unit circle that corresponds to an angle of $\frac{7\pi}{4}$, we use the sine and cosine functions:

$x = \cos\left(\frac{7\pi}{4}\right)$

$y = \sin\left(\frac{7\pi}{4}\right)$

Since $\frac{7\pi}{4}$ is in the fourth quadrant, we have:

$\cos\left(\frac{7\pi}{4}\right) = \frac{1}{\sqrt{2}}$

$\sin\left(\frac{7\pi}{4}\right) = -\frac{1}{\sqrt{2}}$

So, the coordinates are:

$\left(\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$

Answer 2

Alex Thompson

William King

To find the coordinates of the point corresponding to $frac{7pi}{4}$, we use:

$x = cosleft(frac{7pi}{4}
ight)$

$y = sinleft(frac{7pi}{4}
ight)$

In the fourth quadrant:

$cosleft(frac{7pi}{4}
ight) = frac{1}{sqrt{2}}$

$sinleft(frac{7pi}{4}
ight) = -frac{1}{sqrt{2}}$

Coordinates:

$left(frac{1}{sqrt{2}}, -frac{1}{sqrt{2}}
ight)$

Answer 3

Amelia Mitchell

Daniel Carter

For $frac{7pi}{4}$:

$cosleft(frac{7pi}{4}
ight) = frac{1}{sqrt{2}}$

$sinleft(frac{7pi}{4}
ight) = -frac{1}{sqrt{2}}$

Thus,

$left(frac{1}{sqrt{2}}, -frac{1}{sqrt{2}}
ight)$