To convert from polar coordinates [tex]\((r, \theta)\)[/tex] to rectangular coordinates [tex]\((x, y)\)[/tex], we use the following formulas:
[tex]\[ x = r \cos(\theta) \][/tex]
[tex]\[ y = r \sin(\theta) \][/tex]
Given the polar coordinates [tex]\(Q = \left(6, \frac{5\pi}{6}\right)\)[/tex]:
1. Calculate [tex]\( x \)[/tex]:
[tex]\[ x = 6 \cos\left(\frac{5\pi}{6}\right) \][/tex]
2. Calculate [tex]\( y \)[/tex]:
[tex]\[ y = 6 \sin\left(\frac{5\pi}{6}\right) \][/tex]
Now, let's determine the values of [tex]\( \cos\left(\frac{5\pi}{6}\right) \)[/tex] and [tex]\( \sin\left(\frac{5\pi}{6}\right) \)[/tex]:
- The angle [tex]\(\frac{5\pi}{6}\)[/tex] is in the second quadrant where cosine is negative and sine is positive.
- [tex]\(\cos\left(\frac{5\pi}{6}\right) = -\cos\left(\frac{\pi}{6}\right)\)[/tex] because [tex]\(\frac{5\pi}{6}\)[/tex] can be written as [tex]\(\pi - \frac{\pi}{6}\)[/tex]. Since [tex]\(\cos(\pi - \alpha) = -\cos(\alpha)\)[/tex]:
[tex]\[ \cos\left(\frac{5\pi}{6}\right) = -\frac{\sqrt{3}}{2} \][/tex]
- [tex]\(\sin\left(\frac{5\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right)\)[/tex] because [tex]\(\frac{5\pi}{6}\)[/tex] can be written as [tex]\(\pi - \frac{\pi}{6}\)[/tex]. Since [tex]\(\sin(\pi - \alpha) = \sin(\alpha)\)[/tex]:
[tex]\[ \sin\left(\frac{5\pi}{6}\right) = \frac{1}{2} \][/tex]
Now substitute these values back into our formulas:
1. Calculate [tex]\( x \)[/tex]:
[tex]\[ x = 6 \cos\left(\frac{5\pi}{6}\right) = 6 \left(-\frac{\sqrt{3}}{2}\right) = -3\sqrt{3} \][/tex]
2. Calculate [tex]\( y \)[/tex]:
[tex]\[ y = 6 \sin\left(\frac{5\pi}{6}\right) = 6 \left(\frac{1}{2}\right) = 3 \][/tex]
Thus, the rectangular coordinates for point [tex]\(Q = \left(6, \frac{5\pi}{6}\right)\)[/tex] are:
[tex]\[ (-3\sqrt{3}, 3) \][/tex]
Among the given options, this corresponds to:
[tex]\[ \boxed{(-3\sqrt{3}, 3)} \][/tex]
