Problem 1
Question
\(1-6=\) Show that the point is on the unit circle. $$ \left(\frac{4}{5},-\frac{3}{5}\right) $$
Step-by-Step Solution
Verified Answer
The point \( \left( \frac{4}{5}, -\frac{3}{5} \right) \) is on the unit circle since it satisfies \( x^2 + y^2 = 1 \).
1Step 1: Recall the Equation of Unit Circle
The unit circle in the coordinate plane is defined by the equation: \[ x^2 + y^2 = 1 \] This means that any point \((x, y)\) on the unit circle will satisfy this equation.
2Step 2: Identify Coordinates
Identify the given coordinates. In this problem, we have the point:\[ \left( \frac{4}{5}, -\frac{3}{5} \right) \]Here, \( x = \frac{4}{5} \) and \( y = -\frac{3}{5} \).
3Step 3: Substitute and Calculate
Substitute the coordinates into the unit circle equation:\[ \left( \frac{4}{5} \right)^2 + \left( -\frac{3}{5} \right)^2 = ? \]Calculate separately:\[ \left( \frac{4}{5} \right)^2 = \frac{16}{25} \]\[ \left( -\frac{3}{5} \right)^2 = \frac{9}{25} \]
4Step 4: Sum the Values
Add the calculated squares together:\[ \frac{16}{25} + \frac{9}{25} = \frac{25}{25} = 1 \]
5Step 5: Conclusion
Since the sum of the squares equals 1, the point \( \left( \frac{4}{5}, -\frac{3}{5} \right) \) satisfies the unit circle equation. Therefore, this point is on the unit circle.
Key Concepts
Coordinate PlaneUnit Circle EquationCoordinates SubstitutionPythagorean Identity
Coordinate Plane
The coordinate plane is a two-dimensional surface where we can place points using coordinates. Each point on this plane is defined by a pair of numbers \((x, y)\), which tell us the position along the horizontal and vertical axes.
- The horizontal axis is known as the x-axis. It runs left to right.
- The vertical axis is called the y-axis. It runs top to bottom.
Think of this as a map for finding any point using these coordinates.
This plane is essential in geometry and algebra to visualize points, lines, and curves.
In the case of the unit circle, understanding the coordinate plane helps us locate each point precisely to see if it falls on the circle.
- The horizontal axis is known as the x-axis. It runs left to right.
- The vertical axis is called the y-axis. It runs top to bottom.
Think of this as a map for finding any point using these coordinates.
This plane is essential in geometry and algebra to visualize points, lines, and curves.
In the case of the unit circle, understanding the coordinate plane helps us locate each point precisely to see if it falls on the circle.
Unit Circle Equation
The unit circle is a special circle on the coordinate plane. Its center is at the origin \((0, 0)\) and has a radius of 1.
To know if a point is on this circle, we use the unit circle equation: \[ x^2 + y^2 = 1 \]This equation tells us that the sum of the squares of the x and y coordinates of any point on the circle will always equal 1.
- If a point satisfies this equation, it lies precisely on the circle.
- Points not satisfying the equation are outside or inside the circle, depending on their squared sum.
To know if a point is on this circle, we use the unit circle equation: \[ x^2 + y^2 = 1 \]This equation tells us that the sum of the squares of the x and y coordinates of any point on the circle will always equal 1.
- If a point satisfies this equation, it lies precisely on the circle.
- Points not satisfying the equation are outside or inside the circle, depending on their squared sum.
Coordinates Substitution
Coordinates substitution involves replacing the variables in an equation with actual numeric values of a point.
This step is crucial to test whether a point belongs to a particular curve or line, like the unit circle in this exercise.
For example, given the point \((\frac{4}{5}, -\frac{3}{5})\):
This step is crucial to test whether a point belongs to a particular curve or line, like the unit circle in this exercise.
For example, given the point \((\frac{4}{5}, -\frac{3}{5})\):
- Replace \(x\) with \(\frac{4}{5}\) and \(y\) with \(-\frac{3}{5}\) in the equation \(x^2 + y^2 = 1\).
- Calculate this to check if the left side matches the right side (1).
Pythagorean Identity
The Pythagorean identity is an extension of the Pythagorean theorem, related closely to trigonometry and geometry. In the context of the unit circle, it explains why \(x^2 + y^2 = 1\) for all points on the unit circle.
For any point \((x, y)\) on the unit circle:- The distance from the origin (center) to the point is the radius (1 in the unit circle).- This formula originates from the Pythagorean theorem: in a right triangle, \(a^2 + b^2 = c^2\), where \(c\) is the hypotenuse.- Here, \(x\) and \(y\) are like the legs of a right triangle, and the radius (1) is the hypotenuse.Thus, the Pythagorean identity underlines the inherent geometric truth of a circle's structure on the plane.
For any point \((x, y)\) on the unit circle:- The distance from the origin (center) to the point is the radius (1 in the unit circle).- This formula originates from the Pythagorean theorem: in a right triangle, \(a^2 + b^2 = c^2\), where \(c\) is the hypotenuse.- Here, \(x\) and \(y\) are like the legs of a right triangle, and the radius (1) is the hypotenuse.Thus, the Pythagorean identity underlines the inherent geometric truth of a circle's structure on the plane.
Other exercises in this chapter
Problem 1
The given function models the displacement of an object moving in simple harmonic motion. (a) Find the amplitude, period, and frequency of the motion. (b) Sketc
View solution Problem 1
Graph the function. $$ f(x)=1+\cos x $$
View solution Problem 2
The given function models the displacement of an object moving in simple harmonic motion. (a) Find the amplitude, period, and frequency of the motion. (b) Sketc
View solution Problem 2
Graph the function. $$ f(x)=3+\sin x $$
View solution