Problem 14 - Precalculus: A Concise Course

Question

Find the point \((x, y)\) on the unit circle that corresponds to the real number \(t\). $$ t=\frac{3 \pi}{4} $$

Step-by-Step Solution

Answer

The point on the unit circle that corresponds to \(t = \frac{3\pi}{4}\) is \(\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\).

Step 1: Understand the problem

The problem asks to find the point on the unit circle that corresponds to the real number \(t = \frac{3\pi}{4}\). It should be remembered that the unit circle is a circle of radius 1 centered at the origin of a coordinate plane. Each point \((x, y)\) on the unit circle is determined by falling at an angle \(t\) counter-clockwise from the positive x-axis. The x-coordinate of the point is \(\cos(t)\) and the y-coordinate is \(\sin(t)\).

Step 2: Find the x and y coordinates

Given that \(t = \frac{3\pi}{4}\), the x-coordinate of the point is \(\cos\left(\frac{3\pi}{4}\right)\) and the y-coordinate is \(\sin\left(\frac{3\pi}{4}\right)\). By evaluating these expressions, we find that the x-coordinate is \(-\frac{1}{\sqrt{2}}\) and the y-coordinate is \(\frac{1}{\sqrt{2}}\).

Step 3: Provide the final solution

Therefore, the point on the unit circle for \(t = \frac{3\pi}{4}\) is \(\left(-\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}\right)\).