Chapter 5: Problem 18
The point \(P\) is on the unit circle. Find \(P(x, y)\) from the given information. The \(x\) -coordinate of \(P\) is positive, and the \(y\) -coordinate of \(P\) is \(-\sqrt{5} / 5\).
Short Answer
Expert verified
The coordinates of point \(P\) are \(\left(\frac{2\sqrt{5}}{5}, -\frac{\sqrt{5}}{5}\right)\)."
Step by step solution
01
Write the Equation of the Unit Circle
Recall that the unit circle is defined by the equation \(x^2 + y^2 = 1\). This equation represents all points \((x, y)\) that are exactly one unit away from the origin \((0, 0)\). We will use this equation to find the \(x\)-coordinate.
02
Substitute the Given y-coordinate
We are given that the \(y\)-coordinate of the point is \(-\frac{\sqrt{5}}{5}\). Substitute this value into the unit circle equation: \[x^2 + \left(-\frac{\sqrt{5}}{5}\right)^2 = 1\].
03
Simplify the Equation
Simplify the equation by calculating \(\left(-\frac{\sqrt{5}}{5}\right)^2\): \[\left(-\frac{\sqrt{5}}{5}\right)^2 = \frac{5}{25} = \frac{1}{5}.\]Substitute this into the unit circle equation: \[x^2 + \frac{1}{5} = 1.\]
04
Solve for x²
To find \(x^2\), subtract \(\frac{1}{5}\) from both sides of the equation:\[x^2 = 1 - \frac{1}{5} = \frac{5}{5} - \frac{1}{5} = \frac{4}{5}.\]
05
Find the x-coordinates
Take the square root of both sides to solve for \(x\):\[x = \pm \sqrt{\frac{4}{5}} = \pm \frac{2}{\sqrt{5}}.\]Rationalize the denominator by multiplying the numerator and the denominator by \(\sqrt{5}\):\[x = \pm \frac{2}{\sqrt{5}} \cdot \frac{\sqrt{5}}{\sqrt{5}} = \pm \frac{2\sqrt{5}}{5}.\]
06
Determine the Correct Sign for x
Since we are told that the \(x\)-coordinate is positive, choose the positive solution:\[x = \frac{2\sqrt{5}}{5}.\]
07
Write the Coordinates of Point P
Therefore, the coordinates of point \(P\) are \(\left(\frac{2\sqrt{5}}{5}, -\frac{\sqrt{5}}{5}\right)\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Equation of the Unit Circle
The equation of a unit circle is a fundamental concept in trigonometry. It describes a perfect circle with a radius of one unit, centered at the origin of a coordinate system, \(0, 0\). The equation is represented as \(x^2 + y^2 = 1\). This set equation signifies that any point \(x, y\) lying on this circle is exactly one unit away from the origin.
- The unit circle helps to visualize and solve trigonometric problems.
- Its equation is derived from the Pythagorean Theorem, which states that in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.
X-Coordinate
In the given problem, we want to find the x-coordinate of the point on the unit circle. We were given that the y-coordinate is \(-\frac{\sqrt{5}}{5}\). This information needs to be plugged into the equation of the unit circle to solve for the x-coordinate. Here's how you manage that:
- Substitute the y-coordinate into the unit circle equation: \(x^2 + \left(-\frac{\sqrt{5}}{5}\right)^2 = 1\).
- Simplify \(y^2\) to get \(\frac{1}{5}\).
- Then solve the equation \(x^2 + \frac{1}{5} = 1\) to find \(x^2 = \frac{4}{5}\).
- Finally, take the square root to solve for x, noting that \(x\) could be positive or negative.
Y-Coordinate
The y-coordinate in the context of the unit circle can provide insight into the specific position of a point on the circle. Given the problem, the y-coordinate is specified as \(-\frac{\sqrt{5}}{5}\), which influences where the point lies: the lower half of the circle.
- Knowing the y-coordinate allows you to find compatible x-coordinates that fulfill the unit circle's equation.
- The position of the y-coordinate on the negative axis indicates that, for a unit circle, the point is below the x-axis.