# What is the Cartesian form of #(-7,(11pi)/8)#?

Cartesian form of coordinates are

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The Cartesian form of the point ((-7, \frac{11\pi}{8})) can be found by converting the polar coordinates to rectangular coordinates using the following formulas:

[ x = r \cdot \cos(\theta) ] [ y = r \cdot \sin(\theta) ]

where ( r ) is the radius or distance from the origin, and ( \theta ) is the angle measured counterclockwise from the positive x-axis.

Given that ( r = -7 ) and ( \theta = \frac{11\pi}{8} ), we can plug these values into the formulas to find the rectangular coordinates:

[ x = -7 \cdot \cos\left(\frac{11\pi}{8}\right) ] [ y = -7 \cdot \sin\left(\frac{11\pi}{8}\right) ]

Computing the values:

[ x = -7 \cdot \cos\left(\frac{11\pi}{8}\right) \approx -7 \cdot \cos\left(\frac{3\pi}{8}\right) ] [ x \approx -7 \cdot \left(-\frac{\sqrt{2+\sqrt{2}}}{2}\right) ] [ x \approx -7 \cdot \left(-\frac{\sqrt{2}+\sqrt{2}}{2}\right) ] [ x \approx -7 \cdot \left(-\frac{2\sqrt{2}}{2}\right) ] [ x \approx 7\sqrt{2} ]

[ y = -7 \cdot \sin\left(\frac{11\pi}{8}\right) \approx -7 \cdot \sin\left(\frac{3\pi}{8}\right) ] [ y \approx -7 \cdot \left(\frac{\sqrt{2-\sqrt{2}}}{2}\right) ] [ y \approx -7 \cdot \left(\frac{\sqrt{2}-\sqrt{2}}{2}\right) ] [ y \approx -7\sqrt{2} ]

So, the Cartesian form of the point ((-7, \frac{11\pi}{8})) is ((7\sqrt{2}, -7\sqrt{2})).

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