Find Coordinates On Terminal Ray Given Trig Functions

Hey Plastik Magazine readers! Let's dive into a cool math problem today. We're going to figure out how to find the coordinates of a point on the terminal ray of an angle when we're given some trigonometric functions. Sounds a bit complex, but we'll break it down step by step. Get ready to put on your thinking caps, guys!

Unpacking the Trigonometric Functions

So, here's the deal: we're given three trigonometric functions: ${\csc \theta=\frac{13}{12}}$, ${\sec \theta=-\frac{13}{5}}$, and ${\cot \theta=-\frac{5}{12}}$. What do these mean? Well, let's refresh our memories. These are reciprocal trigonometric functions. Remember the primary ones: sine, cosine, and tangent? These are just their inverses. To understand how these functions help us pinpoint the coordinates ${(x, y)}$, we must first grasp their definitions in relation to the unit circle and the sides of a right triangle formed by the terminal ray of the angle ${\theta}$, the x-axis, and a perpendicular line from the point ${(x, y)}$ to the x-axis.

• Cosecant (${\csc \theta}$): This is the reciprocal of sine, which means ${\csc \theta = \frac{1}{\sin \theta}}$. In terms of the coordinates ${(x, y)}$ and the radius r (the distance from the origin to the point), ${\csc \theta = \frac{r}{y}}$. In our case, ${\csc \theta = \frac{13}{12}}$, which tells us that the ratio of the radius to the y-coordinate is 13/12. This is crucial because it gives us a direct relationship between r and y, allowing us to infer that for some constant k, r = 13k and y = 12k. Understanding this relationship is key to unlocking the problem. We're essentially dealing with ratios here, and this sets the stage for finding the actual coordinates.
• Secant (${\sec \theta}$): This is the reciprocal of cosine, so ${\sec \theta = \frac{1}{\cos \theta}}$. In terms of coordinates, ${\sec \theta = \frac{r}{x}}$. We're given ${\sec \theta = -\frac{13}{5}}$. Notice the negative sign here! This tells us something important about the x-coordinate. It must be negative because the radius r is always positive. This piece of information is vital for determining the quadrant in which the terminal ray lies. We now know that the ratio of the radius to the x-coordinate is -13/5, implying that r = 13k (same k as before, since it's the same point) and x = -5k for the same constant k. The negative sign is a game-changer because it restricts our possibilities to quadrants where cosine (and thus secant) is negative.
• Cotangent (${\cot \theta}$): This is the reciprocal of tangent, meaning ${\cot \theta = \frac{1}{\tan \theta}}$. In terms of coordinates, ${\cot \theta = \frac{x}{y}}$. We're given ${\cot \theta = -\frac{5}{12}}$. Again, we have a negative cotangent, which tells us something about the signs of x and y. For the cotangent to be negative, x and y must have opposite signs. We already know from the secant that x is negative. This confirms that our angle is in the second quadrant, where x is negative and y is positive. The cotangent value reinforces the relationship between the x and y coordinates, aligning perfectly with our findings from the secant and cosecant. This interlocking information ensures we're on the right track.

Putting the Pieces Together

Okay, so we've dissected the trigonometric functions and figured out their relationships with the coordinates. Now, how do we actually find those coordinates? The key is to use these ratios and the Pythagorean theorem.

Cracking the Coordinate Code

Let's recap what we've gathered so far. We know:

• ${r = 13k}$
• ${y = 12k}$
• ${x = -5k}$

Where k is a constant we need to determine. To find k, we can use the Pythagorean theorem, which relates x, y, and r: x² + y² = r². This fundamental theorem is the bridge between the ratios we've extracted and the actual coordinate values. It provides a concrete equation that we can solve for our unknown constant k.

Substituting our values, we get:

${(-5k)^2 + (12k)^2 = (13k)^2}$

Let's simplify this. Squaring each term, we get:

${25k^2 + 144k^2 = 169k^2}$

Combining the terms on the left side:

${169k^2 = 169k^2}$

Whoa! This might look a bit weird at first, but hold on. It tells us that this equation holds true for any value of k. However, in the context of coordinates, we're usually looking for the simplest solution, and that's when k = 1. Why? Because if k is any other value, we'd just be scaling the triangle up or down without changing the underlying relationships between the sides and the angle. Choosing k = 1 gives us the smallest possible integer values for the coordinates, making our lives easier.

Now that we know k = 1, we can find our coordinates:

• ${x = -5(1) = -5}$
• ${y = 12(1) = 12}$

So, the coordinates of the point on the terminal ray are ${(-5, 12)}$. We've successfully navigated through the trigonometric ratios, used the Pythagorean theorem, and arrived at our final coordinates! Amazing work, guys!

Double-Checking Our Work

It's always a good idea to double-check our answer. Let's plug our coordinates back into the original trigonometric functions to see if they hold true. This is a critical step in any mathematical problem-solving process. It ensures that our solution is consistent with the given information and helps us catch any potential errors.

• ${\csc \theta = \frac{r}{y} = \frac{13}{12}}$ (Checks out!)
• ${\sec \theta = \frac{r}{x} = \frac{13}{-5} = -\frac{13}{5}}$ (Checks out!)
• ${\cot \theta = \frac{x}{y} = \frac{-5}{12} = -\frac{5}{12}}$ (Checks out!)

Everything lines up perfectly! Our coordinates ${(-5, 12)}$ satisfy all the given trigonometric functions. This gives us confidence in our solution and confirms that we've correctly applied the trigonometric principles.

Wrapping It Up

So, there you have it! We successfully found the coordinates of a point on the terminal ray of an angle using trigonometric functions. It might have seemed daunting at first, but by breaking it down step by step and understanding the relationships between trigonometric functions and coordinates, we cracked the code. Remember, guys, math is all about understanding the relationships and applying the right tools. Keep practicing, and you'll be math whizzes in no time! Remember, every mathematical problem is a puzzle waiting to be solved, and with the right approach, you can conquer any challenge!