91Ó°ÊÓ

In our given polar coordinates \(\left(12, \frac{11 \pi}{4}\right)\), the value of \(r\) is 12 and the value of \(\theta\) is \(\frac{11 \pi}{4}\).

Now we will apply the formulas \(x=r \cos (\theta)\) and \(y=r \sin (\theta)\) using the values of \(r\) and \(\theta\) from step 1. \(x = 12 \cos\left(\frac{11 \pi}{4}\right)\) Since \(\frac{11 \pi}{4}\) is equivalent to \(\frac{3 \pi}{4}\) (subtracting \(2 \pi\) or any other multiple of \(2 \pi\) does not affect the cosine value), we can write: \(x = 12 \cos\left(\frac{3 \pi}{4}\right)\) Now, we know that \(\cos\left(\frac{3 \pi}{4}\right) = -\frac{\sqrt{2}}{2}\), so: \(x = 12\left(-\frac{\sqrt{2}}{2}\right)\) Therefore, \(x = -6\sqrt{2}\). Now, let's calculate the y-coordinate: \(y = 12 \sin\left(\frac{11 \pi}{4}\right)\) Similarly, we can write: \(y = 12 \sin\left(\frac{3 \pi}{4}\right)\) We know that \(\sin\left(\frac{3 \pi}{4}\right) = \frac{\sqrt{2}}{2}\), so: \(y = 12\left(\frac{\sqrt{2}}{2}\right)\) Therefore, \(y = 6\sqrt{2}\).

We have calculated the x and y coordinates as \(x = -6\sqrt{2}\) and \(y = 6\sqrt{2}\). Therefore, the rectangular coordinates for the given polar coordinates are \((-6\sqrt{2}, 6\sqrt{2})\).