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Chapter 4: Problem 1

Find all numbers \(t\) such that \(\left(\frac{1}{3}, t\right)\) is a point on the unit circle.

Short Answer

Expert verified
The numbers \(t\) that satisfy the condition are \(t = \pm \frac{\sqrt{8}}{3}\), leading to the two points on the unit circle being \(\left(\frac{1}{3}, \frac{\sqrt{8}}{3}\right)\) and \(\left(\frac{1}{3}, -\frac{\sqrt{8}}{3}\right)\).

Step by step solution

01

Substitute the given x-coordinate into the equation of the unit circle

In the equation of the unit circle, replace x with \(\frac{1}{3}\). This will give us: \[\left(\frac{1}{3}\right)^2 + t^2 = 1\]
02

Simplify the equation

Calculate the square of \(\frac{1}{3}\) which is \(\frac{1}{9}\), and rewrite the equation as: \[\frac{1}{9} + t^2 = 1\]
03

Solve for t^2

To find the possible values of t, first solve for t^2. Subtract \(\frac{1}{9}\) from 1: \[t^2 = 1 - \frac{1}{9}\] To simplify, represent 1 as \(\frac{9}{9}\) to have the same denominator as the fraction we are subtracting: \[t^2 = \frac{9}{9} - \frac{1}{9}\]
04

Simplify the right-hand side

Perform the subtraction on the right-hand side: \[t^2 = \frac{8}{9}\]
05

Solve for t

To get the values of t, we need to find the square root of \(\frac{8}{9}\). We can have two possible values, positive and negative: \[t = \pm \sqrt{\frac{8}{9}}\] To further simplify, we can find the square roots of the numerator (8) and the denominator (9): \[t = \pm \frac{\sqrt{8}}{\sqrt{9}}\] Since \(\sqrt{9} = 3\), we have: \[t = \pm \frac{\sqrt{8}}{3}\]
06

List the points on the unit circle

Finally, we have the coordinates of the two points on the unit circle, which can be represented as: \[\left(\frac{1}{3}, \frac{\sqrt{8}}{3}\right) \text{ and } \left(\frac{1}{3}, -\frac{\sqrt{8}}{3}\right)\] These are the two points on the unit circle where the x-coordinate is \(\frac{1}{3}\) and the y-coordinate is t.

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