Problem 16 - Algebra and Trigonometry

Question

15-20 \(\mathbf{m}\) The point \(P\) is on the unit circle. Find \(P(x, y)\) from the given information. The \(y\) -coordinate of \(P\) is \(-\frac{1}{3},\) and the \(x\) -coordinate is positive.

Step-by-Step Solution

Answer

The coordinates of point P are \(\left(\frac{2\sqrt{2}}{3}, -\frac{1}{3}\right)\).

Step 1: Understanding the Unit Circle Equation

The unit circle is defined by the equation \(x^2 + y^2 = 1\). We know that any point \((x, y)\) on the unit circle satisfies this equation. Given the information about \(y\), we need to find the corresponding \(x\) value.

Step 2: Substitute Known y-value into the Unit Circle Equation

We are given that the \(y\)-coordinate is \(-\frac{1}{3}\). Substitute this value into the unit circle equation: \(x^2 + \left(-\frac{1}{3}\right)^2 = 1\). Simplifying gives: \(x^2 + \frac{1}{9} = 1\).

Step 3: Solve for x-coordinate

Rearrange the equation to find \(x^2\): \(x^2 = 1 - \frac{1}{9}\). Simplify the right side: \(x^2 = \frac{9}{9} - \frac{1}{9} = \frac{8}{9}\). Take the square root of both sides: \(x = \sqrt{\frac{8}{9}}\) or \(x = -\sqrt{\frac{8}{9}}\).

Step 4: Choose the Positive x-coordinate

Since we are told that the \(x\)-coordinate is positive, choose the positive root: \(x = \sqrt{\frac{8}{9}} = \frac{\sqrt{8}}{3}\). Simplifying gives \(x = \frac{2\sqrt{2}}{3}\).

Step 5: Write the Coordinates of Point P

With the calculated values of \(x\) and the given \(y\), the coordinates of point \(P\) are \(\left(\frac{2\sqrt{2}}{3}, -\frac{1}{3}\right)\).

Key Concepts

Coordinate Geometry

In the realm of mathematics, coordinate geometry is a powerful tool that combines algebra and geometry to solve problems involving shapes and their coordinates. It gives us a method to represent and analyze geometric figures using a coordinate plane. The unit circle, with its center at the origin \((0, 0)\) and a radius of 1, is a perfect example of how coordinate geometry can be applied.
Understanding a point on the unit circle means recognizing that each point \(P(x, y)\) must satisfy the unit circle equation \(x^2 + y^2 = 1\). Here, both the position and nature of the point are dictated using basic algebraic principles.
By knowing one coordinate, you can find the other using this equation, revealing the precise location of the point on this fundamental geometric shape.

Solving Equations

Solving equations is a quintessential skill in mathematics that allows us to find unknown values and make sense of various mathematical relationships. In our unit circle problem, we use this skill to find the missing \(x\)-coordinate given a known \(y\)-coordinate.
The initial step involves substituting the known value into the unit circle equation \(x^2 + y^2 = 1\). Once you substitute \(y = -\frac{1}{3}\), the equation becomes \(x^2 + \left(-\frac{1}{3}\right)^2 = 1\), simplifying to \(x^2 + \frac{1}{9} = 1\).

• Subtract \(\frac{1}{9}\) from 1 to get \(x^2 = \frac{8}{9}\).
• Take the square root to solve for \(x\), leading to \(x = \pm \sqrt{\frac{8}{9}}\).

Choosing the positive value as specified, you arrive at \(x = \frac{2\sqrt{2}}{3}\), hence solving the equation to find the coordinates of the point.

Trigonometry

Trigonometry beautifully complements the concept of the unit circle and is often considered its heart. It helps us understand the relationship between angles and the coordinates of points on the circle. The unit circle is key to trigonometric functions such as sine, cosine, and tangent.
Each point on the unit circle can be associated with an angle from the positive x-axis, providing a direct linkage between angular measures and trigonometric values:

• The \(x\)-coordinate of a point represents the cosine of the angle.
• The \(y\)-coordinate represents the sine of the angle.

Applying this to our exercise, should you need the angle that corresponds to the point \( \left( \frac{2\sqrt{2}}{3}, -\frac{1}{3} \right) \), you can use inverse trigonometric functions to find it.

Pythagorean Identity

At the core of understanding trigonometry and the unit circle lies the Pythagorean Identity, which is derived from the Pythagorean Theorem. This identity states that for any angle \(\theta\), the equation \(\cos^2(\theta) + \sin^2(\theta) = 1\) holds true.
In the context of the unit circle, every point \(P(x, y)\) satisfies this identity because they are forms of cosine and sine values.

• For \(x = \cos(\theta)\) and \(y = \sin(\theta)\), the combination equals to 1, representing a direct application of the Pythagorean Identity.

This formula is instrumental in verifying the location of points and solving various problems related to angles and distances in a unit circle, reinforcing its universal applicability in trigonometry.