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Chapter 4: Problem 4

Find all numbers \(t\) such that \(\left(t,-\frac{3}{7}\right)\) is a point on the unit circle.

Short Answer

Expert verified
The possible values of \(t\) such that \(\left(t,-\frac{3}{7}\right)\) is a point on the unit circle are \(t = \pm \frac{\sqrt{40}}{7}\).

Step by step solution

01

Write down the unit circle equation

The equation for the unit circle is given by \(x^2 + y^2 = 1\).
02

Substitute the coordinates

We are given the point \(\left(t,-\frac{3}{7}\right)\) and asked to find all possible values of \(t\). Substitute the coordinates of the point into the unit circle equation: \[t^2 + \left(-\frac{3}{7}\right)^2 = 1\]
03

Solve for t

Begin by squaring \(-\frac{3}{7}\) and simplifying the equation: \[\begin{aligned} t^2 + \left(\frac{3}{7}\right)^2 &= 1 \\ t^2 + \frac{9}{49} &= 1 \end{aligned}\] Now, subtract \(\frac{9}{49}\) from both sides of the equation: \[\begin{aligned} t^2 &= 1 - \frac{9}{49} \\ t^2 &= \frac{49 - 9}{49} \\ t^2 &= \frac{40}{49} \end{aligned}\] Since we are looking for all possible values of \(t\), remember that both the positive and negative square roots of a number squared will yield the same result. Thus, we need to find the positive and negative square roots of \(\frac{40}{49}\): \[\begin{aligned} t &= \pm \sqrt{\frac{40}{49}} \\ t &= \pm \frac{\sqrt{40}}{7} \end{aligned}\]
04

Write the final answer

All the possible values of \(t\) such that \(\left(t,-\frac{3}{7}\right)\) is a point on the unit circle are given by: \[t = \pm \frac{\sqrt{40}}{7}\]

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