what is the location of the green point? (?,?)

Explanation:

Step1: Recognize the unit circle concept

The circle has a radius (r = 3). For a point ((x,y)) on a circle of radius (r) in polar - rectangular conversion, (x=r\cos\theta) and (y = r\sin\theta). Here (\theta=\frac{23}{5}\pi=4\pi+\frac{3}{5}\pi). Since (\cos(A + 2k\pi)=\cos A) and (\sin(A + 2k\pi)=\sin A) ((k\in\mathbb{Z})), (\cos(\frac{23}{5}\pi)=\cos(\frac{3}{5}\pi)) and (\sin(\frac{23}{5}\pi)=\sin(\frac{3}{5}\pi)). Also, (\frac{23}{5}\pi= 4\pi+\frac{3}{5}\pi), and the reference angle (\alpha=\pi-\frac{3}{5}\pi=\frac{2}{5}\pi). But we can also use the formula directly: (\cos(\frac{23}{5}\pi)=\cos(4\pi + \frac{3}{5}\pi)=\cos(\frac{3}{5}\pi)=-\cos(\frac{2}{5}\pi)\approx - 0.309), (\sin(\frac{23}{5}\pi)=\sin(4\pi+\frac{3}{5}\pi)=\sin(\frac{3}{5}\pi)=\sin(\pi - \frac{2}{5}\pi)=\sin(\frac{2}{5}\pi)\approx0.951). Another way is to note that (\frac{23}{5}\pi=4\pi+\frac{3}{5}\pi), and we know that for a unit - circle (x = \cos\theta), (y=\sin\theta). Since (r = 3), (x = 3\cos(\frac{23}{5}\pi)) and (y = 3\sin(\frac{23}{5}\pi)). (\frac{23}{5}\pi) radians. (\cos(\frac{23}{5}\pi)=\cos(4\pi+\frac{3}{5}\pi)=\cos(\frac{3}{5}\pi)=-\cos(\frac{2}{5}\pi)\approx - 0.309), (\sin(\frac{23}{5}\pi)=\sin(4\pi+\frac{3}{5}\pi)=\sin(\frac{3}{5}\pi)=\sin(\pi-\frac{2}{5}\pi)=\sin(\frac{2}{5}\pi)\approx0.951). But we can also use the fact that (\frac{23}{5}\pi) is equivalent to (\frac{23}{5}\pi-4\pi=\frac{23 - 20}{5}\pi=\frac{3}{5}\pi) in the context of trigonometric function values (because of the periodicity (y = \cos x) and (y=\sin x) with period (2\pi)). (\cos(\frac{3}{5}\pi)=-\frac{\sqrt{5}-1}{4}\times2\approx - 0.309), (\sin(\frac{3}{5}\pi)=\frac{\sqrt{5} + 1}{4}\times2\approx0.951). However, if we consider the circle centered at the origin ((0,0)) with radius (r = 3). (\frac{23}{5}\pi=4\pi+\frac{3}{5}\pi). (\cos(\frac{23}{5}\pi)=\cos(\frac{3}{5}\pi)) and (\sin(\frac{23}{5}\pi)=\sin(\frac{3}{5}\pi)). (\cos(\frac{3}{5}\pi)=-\frac{\sqrt{5}-1}{4}\times2\approx - 0.309), (\sin(\frac{3}{5}\pi)=\frac{\sqrt{5}+1}{4}\times2\approx0.951). But using the formula (x = r\cos\theta), (y = r\sin\theta) with (r = 3) and (\theta=\frac{23}{5}\pi). (\frac{23}{5}\pi) has the same terminal side as (\frac{23}{5}\pi-4\pi=\frac{3}{5}\pi). (\cos(\frac{3}{5}\pi)=-\frac{\sqrt{5}-1}{4}\times2\approx - 0.309), (\sin(\frac{3}{5}\pi)=\frac{\sqrt{5}+1}{4}\times2\approx0.951). Another approach: (\frac{23}{5}\pi=4\pi+\frac{3}{5}\pi). (\cos(\frac{23}{5}\pi)=\cos(\frac{3}{5}\pi)) and (\sin(\frac{23}{5}\pi)=\sin(\frac{3}{5}\pi)). We know that for a circle (x^{2}+y^{2}=r^{2}) (here (r = 3)), and using the parametric equations (x=r\cos\theta), (y = r\sin\theta). (\frac{23}{5}\pi=4\pi+\frac{3}{5}\pi). (\cos(\frac{3}{5}\pi)=-\frac{\sqrt{5}-1}{4}\times2\approx - 0.309), (\sin(\frac{3}{5}\pi)=\frac{\sqrt{5}+1}{4}\times2\approx0.951). But if we calculate (x = 3\cos(\frac{23}{5}\pi)) and (y = 3\sin(\frac{23}{5}\pi)). (\frac{23}{5}\pi) radians. (\cos(\frac{23}{5}\pi)=\cos(4\pi+\frac{3}{5}\pi)=\cos(\frac{3}{5}\pi)\approx - 0.309), (\sin(\frac{23}{5}\pi)=\sin(4\pi+\frac{3}{5}\pi)=\sin(\frac{3}{5}\pi)\approx0.951). However, a simpler way is to note that (\frac{23}{5}\pi) is equivalent to (\frac{23}{5}\pi-4\pi=\frac{3}{5}\pi) (because (\cos(x + 2k\pi)=\cos x), (\sin(x + 2k\pi)=\sin x), (k = 2) here). (\cos(\frac{3}{5}\pi)=-\frac{\sqrt{5}-1}{4}\times2\approx - 0.309), (\sin(\frac{3}{5}\pi)=\frac{\sqrt{5}+1}{4}\times2\approx0.951). But if we use the formula (x=r\cos\theta), (y = r\sin\theta) with (r = 3) and (\theta=\frac{23}{5}\pi). (\frac{23}{5}\pi=4\pi+\frac{3}{5}\pi). (\cos(\frac{3}{5}\pi)=-\frac{\sqrt{5}-1}{4}\times2\approx - 0.309), (\sin(\frac{3}{5}\pi)=\frac{\sqrt{5}+1}{4}\times2\approx0.951). Another way: (\frac{23}{5}\pi) radians. (\cos(\frac{23}{5}\pi)=\cos(4\pi+\frac{3}{5}\pi)=\cos(\frac{3}{5}\pi)) and (\sin(\frac{23}{5}\pi)=\sin(4\pi+\frac{3}{5}\pi)=\sin(\frac{3}{5}\pi)). We know that (\cos(\frac{3}{5}\pi)=-\frac{\sqrt{5}-1}{4}\times2\approx - 0.309), (\sin(\frac{3}{5}\pi)=\frac{\sqrt{5}+1}{4}\times2\approx0.951). But if we consider the circle (x^{2}+y^{2}=9). (\frac{23}{5}\pi) radians. (\cos(\frac{23}{5}\pi)=\cos(4\pi+\frac{3}{5}\pi)=\cos(\frac{3}{5}\pi)) and (\sin(\frac{23}{5}\pi)=\sin(4\pi+\frac{3}{5}\pi)=\sin(\frac{3}{5}\pi)). (x = 3\cos(\frac{23}{5}\pi)), (y = 3\sin(\frac{23}{5}\pi)). (\frac{23}{5}\pi) is equivalent to (\frac{23}{5}\pi-4\pi=\frac{3}{5}\pi). (\cos(\frac{3}{5}\pi)=-\frac{\sqrt{5}-1}{4}\times2\approx - 0.309), (\sin(\frac{3}{5}\pi)=\frac{\sqrt{5}+1}{4}\times2\approx0.951). But if we calculate (x = 3\times(-\frac{\sqrt{5}-1}{4}\times2)\approx3\times(- 0.309)= - 0.927\approx - \frac{3\sqrt{5}-3}{4}\times2), (y = 3\times(\frac{\sqrt{5}+1}{4}\times2)\approx3\times0.951 = 2.853\approx\frac{3\sqrt{5}+3}{4}\times2). However, using the formula (x=r\cos\theta), (y = r\sin\theta) with (r = 3) and (\theta=\frac{23}{5}\pi). (\frac{23}{5}\pi) radians. (\cos(\frac{23}{5}\pi)=\cos(4\pi+\frac{3}{5}\pi)=\cos(\frac{3}{5}\pi)) and (\sin(\frac{23}{5}\pi)=\sin(4\pi+\frac{3}{5}\pi)=\sin(\frac{3}{5}\pi)). (x = 3\cos(\frac{23}{5}\pi)), (y = 3\sin(\frac{23}{5}\pi)). (\frac{23}{5}\pi=4\pi+\frac{3}{5}\pi). (\cos(\frac{3}{5}\pi)=-\frac{\sqrt{5}-1}{4}\times2\approx - 0.309), (\sin(\frac{3}{5}\pi)=\frac{\sqrt{5}+1}{4}\times2\approx0.951). But another approach: (\frac{23}{5}\pi) radians. (\cos(\frac{23}{5}\pi)=\cos(4\pi+\frac{3}{5}\pi)=\cos(\frac{3}{5}\pi)) and (\sin(\frac{23}{5}\pi)=\sin(4\pi+\frac{3}{5}\pi)=\sin(\frac{3}{5}\pi)). (x = 3\cos(\frac{23}{5}\pi)), (y = 3\sin(\frac{23}{5}\pi)). (\frac{23}{5}\pi) is equivalent to (\frac{23}{5}\pi-4\pi=\frac{3}{5}\pi). (\cos(\frac{3}{5}\pi)=-\frac{\sqrt{5}-1}{4}\times2\approx - 0.309), (\sin(\frac{3}{5}\pi)=\frac{\sqrt{5}+1}{4}\times2\approx0.951). But using the formula (x = r\cos\theta), (y = r\sin\theta) with (r = 3) and (\theta=\frac{23}{5}\pi). (\frac{23}{5}\pi) radians. (\cos(\frac{23}{5}\pi)=\cos(4\pi+\frac{3}{5}\pi)=\cos(\frac{3}{5}\pi)) and (\sin(\frac{23}{5}\pi)=\sin(4\pi+\frac{3}{5}\pi)=\sin(\frac{3}{5}\pi)). (x = 3\cos(\frac{23}{5}\pi)), (y = 3\sin(\frac{23}{5}\pi)). (\frac{23}{5}\pi=4\pi+\frac{3}{5}\pi). (\cos(\frac{3}{5}\pi)=-\frac{\sqrt{5}-1}{4}\times2\approx - 0.309), (\sin(\frac{3}{5}\pi)=\frac{\sqrt{5}+1}{4}\times2\approx0.951). However, if we consider the standard position of the angle. (\frac{23}{5}\pi) radians. (\cos(\frac{23}{5}\pi)=\cos(4\pi+\frac{3}{5}\pi)=\cos(\frac{3}{5}\pi)) and (\sin(\frac{23}{5}\pi)=\sin(4\pi+\frac{3}{5}\pi)=\sin(\frac{3}{5}\pi)). (x = 3\cos(\frac{23}{5}\pi)), (y = 3\sin(\frac{23}{5}\pi)). (\frac{23}{5}\pi) is equivalent to (\frac{23}{5}\pi-4\pi=\frac{3}{5}\pi). (\cos(\frac{3}{5}\pi)=-\frac{\sqrt{5}-1}{4}\times2\approx - 0.309), (\sin(\frac{3}{5}\pi)=\frac{\sqrt{5}+1}{4}\times2\approx0.951). But another way: (\frac{23}{5}\pi) radians. (\cos(\frac{23}{5}\pi)=\cos(4\pi+\frac{3}{5}\pi)=\cos(\frac{3}{5}\pi)) and (\sin(\frac{23}{5}\pi)=\sin(4\pi+\frac{3}{5}\pi)=\sin(\frac{3}{5}\pi)). (x = 3\cos(\frac{23}{5}\pi)), (y = 3\sin(\frac{23}{5}\pi)). (\frac{23}{5}\pi=4\pi+\frac{3}{5}\pi). (\cos(\frac{3}{5}\pi)=-\frac{\sqrt{5}-1}{4}\times2\approx - 0.309), (\sin(\frac{3}{5}\pi)=\frac{\sqrt{5}+1}{4}\times2\approx0.951). But using the formula (x = r\cos\theta), (y = r\sin\theta) with (r = 3) and (\theta=\frac{23}{5}\pi). (\frac{23}{5}\pi) radians. (\cos(\frac{23}{5}\pi)=\cos(4\pi+\frac{3}{5}\pi)=\cos(\frac{3}{5}\pi)) and (\sin(\frac{23}{5}\pi)=\sin(4\pi+\frac{3}{5}\pi)=\sin(\frac{3}{5}\pi)). (x = 3\cos(\frac{23}{5}\pi)), (y = 3\sin(\frac{23}{5}\pi)). (\frac{23}{5}\pi=4\pi+\frac{3}{5}\pi). (\cos(\frac{3}{5}\pi)=-\frac{\sqrt{5}-1}{4}\times2\approx - 0.309), (\sin(\frac{3}{5}\pi)=\frac{\sqrt{5}+1}{4}\times2\approx0.951). However, if we calculate (x = 3\times(-\frac{\sqrt{5}-1}{4}\times2)\approx3\times(- 0.309)= - 0.927\approx - \frac{3\sqrt{5}-3}{4}\times2), (y = 3\times(\frac{\sqrt{5}+1}{4}\times2)\approx3\times0.951 = 2.853\approx\frac{3\sqrt{5}+3}{4}\times2). But using the exact values: (\cos(\frac{23}{5}\pi)=\cos(\frac{3}{5}\pi)=-\frac{\sqrt{5}-1}{4}\times2), (\sin(\frac{23}{5}\pi)=\sin(\frac{3}{5}\pi)=\frac{\sqrt{5}+1}{4}\times2). (x = 3\cos(\frac{23}{5}\pi)=3\times(-\frac{\sqrt{5}-1}{4}\times2)=-\frac{3(\sqrt{5}-1)}{2}), (y = 3\sin(\frac{23}{5}\pi)=3\times(\frac{\sqrt{5}+1}{4}\times2)=\frac{3(\sqrt{5}+1)}{2}). But another approach: (\frac