Find Coordinates Given Angle And Cosecant Value

by Alex Johnson 48 views

Let's dive into a fun problem where we need to find the coordinates of a point P(x,y){P(x, y)} on the terminal ray of an angle θ{\theta}. We know that θ{\theta} lies between π{\pi} and 3π2{\frac{3\pi}{2}} radians, which means it's in the third quadrant. We also have the cosecant of θ{\theta}, csc⁡θ=−52{\csc \theta = -\frac{5}{2}}. Buckle up, it's going to be an exciting ride!

Understanding the Basics

Before we jump into solving this, let's refresh some trigonometric concepts. Remember, in a unit circle, for any angle θ{\theta}, we have:

  • x=rcos⁡θ{x = r \cos \theta}
  • y=rsin⁡θ{y = r \sin \theta}
  • csc⁡θ=ry{\csc \theta = \frac{r}{y}}

Where r{r} is the distance from the origin to the point P(x,y){P(x, y)}. Since csc⁡θ=−52{\csc \theta = -\frac{5}{2}}, we can deduce that ry=−52{\frac{r}{y} = -\frac{5}{2}}. Because r{r} is always positive, y{y} must be negative, which makes sense since we are in the third quadrant where both x{x} and y{y} are negative.

Setting Up the Equations

Given csc⁡θ=−52{\csc \theta = -\frac{5}{2}}, we can write:

ry=−52{\frac{r}{y} = -\frac{5}{2}}

Let's express y{y} in terms of r{r}:

y=−25r{y = -\frac{2}{5}r}

Now, we need to find x{x}. We know that:

x2+y2=r2{x^2 + y^2 = r^2}

Substitute y{y} into this equation:

x2+(−25r)2=r2{x^2 + \left(-\frac{2}{5}r\right)^2 = r^2}

x2+425r2=r2{x^2 + \frac{4}{25}r^2 = r^2}

Now, solve for x2{x^2}:

x2=r2−425r2{x^2 = r^2 - \frac{4}{25}r^2}

x2=2125r2{x^2 = \frac{21}{25}r^2}

Taking the square root of both sides, we get:

x=±215r{x = \pm \frac{\sqrt{21}}{5}r}

Since we are in the third quadrant, x{x} must be negative, so:

x=−215r{x = -\frac{\sqrt{21}}{5}r}

Finding the Coordinates

Now we have both x{x} and y{y} in terms of r{r}:

x=−215r{x = -\frac{\sqrt{21}}{5}r}

y=−25r{y = -\frac{2}{5}r}

We need to find specific values for x{x} and y{y}. Notice that the ratio x:y{x:y} will give us the relationship without needing to know r{r}. Let's assume a simple value for r{r} that helps clear out the fractions. If we let r=5{r = 5}, we get:

x=−215(5)=−21{x = -\frac{\sqrt{21}}{5}(5) = -\sqrt{21}}

y=−25(5)=−2{y = -\frac{2}{5}(5) = -2}

So the coordinates of point P{P} are P(−21,−2){P(-\sqrt{21}, -2)}.

Verification

Let's verify our solution.

x=−21{x = -\sqrt{21}}

y=−2{y = -2}

r=5{r = 5}

csc⁡θ=ry=5−2=−52{\csc \theta = \frac{r}{y} = \frac{5}{-2} = -\frac{5}{2}}

This matches the given information. Also, since both x{x} and y{y} are negative, the point lies in the third quadrant, which is consistent with π<θ<3π2{\pi < \theta < \frac{3\pi}{2}}.

Conclusion

Therefore, the coordinates of point P(x,y){P(x, y)} are P(−21,−2){P(-\sqrt{21}, -2)}. This problem combines trigonometric identities with quadrant analysis to pinpoint the exact location of a point on the coordinate plane. Remember to always consider the quadrant when determining the signs of x{x} and y{y}.

The point P(x,y){P(x, y)} lies on the terminal ray of angle θ{\theta}. If θ{\theta} is between π{\pi} radians and 3π2{\frac{3 \pi}{2}} radians and csc⁡θ=−52{\csc \theta=-\frac{5}{2}}, the coordinates of P(x,y){P (x, y)} are P(−21,−2){P(-\sqrt{21}, -2)}.

Trigonometry can seem complex, but breaking it down into smaller, manageable steps and understanding the underlying principles makes it much easier. Keep practicing, and you'll become a trig wizard in no time!

Visualizing the Problem

To truly grasp this concept, let's visualize the problem. Imagine a unit circle centered at the origin. The angle θ{\theta} starts at the positive x-axis and rotates counter-clockwise. Since θ{\theta} lies between π{\pi} and 3π2{\frac{3\pi}{2}}, the terminal ray of θ{\theta} is in the third quadrant. In this quadrant, both the x and y coordinates are negative.

The cosecant function, csc⁡θ{\csc \theta}, is the reciprocal of the sine function, sin⁡θ{\sin \theta}. Since csc⁡θ=−52{\csc \theta = -\frac{5}{2}}, this means sin⁡θ=−25{\sin \theta = -\frac{2}{5}}. The sine function corresponds to the y-coordinate on the unit circle (or yr{\frac{y}{r}} in a circle of radius r{r}).

So, we are looking for a point in the third quadrant where the y-coordinate is −25{-\frac{2}{5}} of the radius. Using the Pythagorean theorem, we can find the corresponding x-coordinate, keeping in mind that it must be negative in the third quadrant.

Alternative Approach

Another way to approach this problem is using trigonometric identities directly. Since csc⁡θ=−52{\csc \theta = -\frac{5}{2}}, we know that sin⁡θ=−25{\sin \theta = -\frac{2}{5}}. We can use the Pythagorean identity sin⁡2θ+cos⁡2θ=1{\sin^2 \theta + \cos^2 \theta = 1} to find cos⁡θ{\cos \theta}.

(−25)2+cos⁡2θ=1{\left(-\frac{2}{5}\right)^2 + \cos^2 \theta = 1}

425+cos⁡2θ=1{\frac{4}{25} + \cos^2 \theta = 1}

cos⁡2θ=1−425{\cos^2 \theta = 1 - \frac{4}{25}}

cos⁡2θ=2125{\cos^2 \theta = \frac{21}{25}}

cos⁡θ=±215{\cos \theta = \pm \frac{\sqrt{21}}{5}}

Since θ{\theta} is in the third quadrant, cos⁡θ{\cos \theta} must be negative, so:

cos⁡θ=−215{\cos \theta = -\frac{\sqrt{21}}{5}}

Now we have sin⁡θ=−25{\sin \theta = -\frac{2}{5}} and cos⁡θ=−215{\cos \theta = -\frac{\sqrt{21}}{5}}. To find the coordinates (x,y){(x, y)}, we can use:

x=rcos⁡θ{x = r \cos \theta}

y=rsin⁡θ{y = r \sin \theta}

Again, let's assume r=5{r = 5} to simplify the calculations:

x=5⋅(−215)=−21{x = 5 \cdot \left(-\frac{\sqrt{21}}{5}\right) = -\sqrt{21}}

y=5⋅(−25)=−2{y = 5 \cdot \left(-\frac{2}{5}\right) = -2}

This gives us the same coordinates P(−21,−2){P(-\sqrt{21}, -2)} as before.

Real-World Applications

Understanding trigonometry is essential in various real-world applications, including navigation, engineering, physics, and computer graphics. For example, engineers use trigonometric functions to calculate angles and distances in structural designs. Physicists use them to analyze wave motion and projectile trajectories. Computer graphics rely heavily on trigonometry to create realistic 3D models and animations.

Knowing how to find coordinates given trigonometric values is a fundamental skill that extends beyond theoretical math problems. It allows you to solve practical problems in various fields that require precise calculations of angles and distances.

Tips for Mastering Trigonometry

To truly master trigonometry, consider these tips:

  1. Memorize Key Identities: Knowing the basic trigonometric identities, such as sin⁡2θ+cos⁡2θ=1{\sin^2 \theta + \cos^2 \theta = 1}, tan⁡θ=sin⁡θcos⁡θ{\tan \theta = \frac{\sin \theta}{\cos \theta}}, and the reciprocal identities, is crucial.
  2. Practice Regularly: The more you practice, the more comfortable you will become with applying trigonometric concepts and solving problems.
  3. Visualize Problems: Draw diagrams and visualize the unit circle to help understand the relationships between angles and trigonometric functions.
  4. Understand the Quadrants: Pay attention to the signs of trigonometric functions in each quadrant to avoid errors in calculations.
  5. Use Real-World Examples: Relate trigonometric concepts to real-world applications to make the learning process more engaging and meaningful.

By following these tips and consistently practicing, you can build a strong foundation in trigonometry and excel in your math studies.

In conclusion, remember that trigonometry is not just about memorizing formulas, it's about understanding the relationships between angles and sides in triangles and applying these concepts to solve various problems. Keep exploring and practicing, and you'll discover the beauty and power of trigonometry!

For more information about trigonometry, visit Khan Academy's Trigonometry section.