Finding Cosine Given Sine And Quadrant Location A Step-by-Step Guide

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Hey everyone! Today, we're diving into a trigonometric problem where we need to find the value of $\cos \theta$ given that $\sin \theta = -\frac{3\sqrt{13}}{13}$ and $ heta$ lies in the fourth quadrant. Trigonometry can seem daunting at first, but don't worry, we'll break it down step by step so it's super easy to follow. We'll use our knowledge of trigonometric identities, the unit circle, and quadrant rules to nail this problem. So, let's get started and make sure we understand how to find cosine when sine and the quadrant are given.

Understanding the Basics of Trigonometry

Before we jump into solving the problem, let's quickly review some fundamental trigonometric concepts. This will help ensure we're all on the same page and make the solution process much smoother. Understanding these basics is crucial for tackling any trig problem, and it's like having the right tools in your toolbox – you'll be ready for anything!

The Unit Circle

The unit circle is a circle with a radius of 1 centered at the origin (0, 0) in the Cartesian coordinate system. It’s a super important tool in trigonometry because it helps us visualize trigonometric functions. Any point on the unit circle can be represented as $(\cos \theta, \sin \theta)$, where $ heta$ is the angle formed between the positive x-axis and the line connecting the origin to that point. Visualizing angles on the unit circle is key to understanding the signs of trigonometric functions in different quadrants. For example, in the first quadrant (0° to 90°), both cosine and sine are positive. In the second quadrant (90° to 180°), sine is positive, but cosine is negative. This pattern continues around the circle, helping us keep track of the signs.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that are true for all values of the variables for which the functions are defined. They're like the formulas in math – we can use them to simplify expressions and solve equations. The most famous and crucial identity is the Pythagorean identity:

sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1

This identity is derived from the Pythagorean theorem and is the cornerstone of many trigonometric proofs and solutions. We'll be using this identity today to solve our problem. Other important identities include reciprocal identities (like $\csc \theta = \frac{1}{\sin \theta}$), quotient identities (like $\tan \theta = \frac{\sin \theta}{\cos \theta}$), and various angle sum and difference formulas. But for our current problem, the Pythagorean identity is the star of the show!

Quadrant Rules

The coordinate plane is divided into four quadrants, and each quadrant has its own set of rules regarding the signs of trigonometric functions. Knowing these rules is essential for determining whether sine, cosine, and tangent are positive or negative in a given quadrant. Here’s a quick rundown:

  • Quadrant I (0° to 90°): All trigonometric functions are positive.
  • Quadrant II (90° to 180°): Sine (and its reciprocal, cosecant) is positive; cosine and tangent are negative.
  • Quadrant III (180° to 270°): Tangent (and its reciprocal, cotangent) is positive; sine and cosine are negative.
  • Quadrant IV (270° to 360°): Cosine (and its reciprocal, secant) is positive; sine and tangent are negative.

A helpful mnemonic to remember this is “All Students Take Calculus” (ASTC), which stands for:

  • All in Quadrant I
  • Sine in Quadrant II
  • Tangent in Quadrant III
  • Cosine in Quadrant IV

In our problem, we're told that $ heta$ is in the fourth quadrant, which means cosine will be positive, and sine (as given) is negative. This quadrant rule will be super important when we determine the sign of our final answer.

Solving for Cosine

Now that we've reviewed the basics, let's tackle the problem head-on! We're given that $\sin \theta = -\frac{3\sqrt{13}}{13}$ and that $\theta$ is in the fourth quadrant. Our mission is to find the value of $\cos \theta$. We'll use the Pythagorean identity to get us there, and remember, quadrant rules will help us decide on the sign of the answer. So, let's break it down step by step.

Applying the Pythagorean Identity

The Pythagorean identity, as we mentioned earlier, is:

sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1

This identity relates sine and cosine, making it perfect for our problem. We know the value of $\sin \theta$, so we can plug it into the equation and solve for $\cos \theta$. Let's do it! First, we substitute the given value of $\sin \theta$ into the identity:

(−31313)2+cos⁡2θ=1\left(-\frac{3\sqrt{13}}{13}\right)^2 + \cos^2 \theta = 1

Now, we need to square the sine value. Remember, when you square a fraction, you square both the numerator and the denominator. So, let's square $-\frac{3\sqrt{13}}{13}$:

(−31313)2=(−3)2⋅(13)2132=9⋅13169=117169\left(-\frac{3\sqrt{13}}{13}\right)^2 = \frac{(-3)^2 \cdot (\sqrt{13})^2}{13^2} = \frac{9 \cdot 13}{169} = \frac{117}{169}

We can simplify this fraction by dividing both the numerator and the denominator by 13:

117169=913\frac{117}{169} = \frac{9}{13}

So, our equation now looks like this:

913+cos⁡2θ=1\frac{9}{13} + \cos^2 \theta = 1

Isolating Cosine

Next, we need to isolate $\cos^2 \theta$ on one side of the equation. To do this, we'll subtract $\frac{9}{13}$ from both sides:

cos⁡2θ=1−913\cos^2 \theta = 1 - \frac{9}{13}

To subtract the fraction from 1, we need to express 1 as a fraction with the same denominator, which is $\frac{13}{13}$:

cos⁡2θ=1313−913\cos^2 \theta = \frac{13}{13} - \frac{9}{13}

Now, we can subtract the numerators:

cos⁡2θ=13−913=413\cos^2 \theta = \frac{13 - 9}{13} = \frac{4}{13}

Finding Cosine

We've found $\cos^2 \theta$, but we want $\cos \theta$. To find it, we need to take the square root of both sides of the equation:

cos⁡θ=±413\cos \theta = \pm \sqrt{\frac{4}{13}}

When we take the square root of a fraction, we take the square root of both the numerator and the denominator:

cos⁡θ=±413=±213\cos \theta = \pm \frac{\sqrt{4}}{\sqrt{13}} = \pm \frac{2}{\sqrt{13}}

Rationalizing the Denominator

It's standard practice to rationalize the denominator, which means we want to get rid of the square root in the denominator. To do this, we multiply both the numerator and the denominator by $\sqrt{13}$:

cos⁡θ=±213⋅1313=±21313\cos \theta = \pm \frac{2}{\sqrt{13}} \cdot \frac{\sqrt{13}}{\sqrt{13}} = \pm \frac{2\sqrt{13}}{13}

Determining the Sign

Now, we have two possible values for $\cos \theta$: $\frac{2\sqrt{13}}{13}$ and $-\frac{2\sqrt{13}}{13}$. This is where our quadrant rules come into play! We were given that $\theta$ is in the fourth quadrant. In the fourth quadrant, cosine is positive (remember “All Students Take Calculus”?). Therefore, we choose the positive value:

cos⁡θ=21313\cos \theta = \frac{2\sqrt{13}}{13}

Conclusion: The Value of Cosine

So, after all that awesome math, we've found that if $\sin \theta = -\frac{3\sqrt{13}}{13}$ and $\theta$ is in the fourth quadrant, then $\cos \theta = \frac{2\sqrt{13}}{13}$. Wasn't that a fun journey through trigonometry? We used the Pythagorean identity, simplified fractions, rationalized denominators, and applied quadrant rules – a true mathematical adventure! By understanding these steps, you can confidently tackle similar problems and become a trigonometry whiz.

Remember, guys, the key to mastering trigonometry is practice, practice, practice! Work through more examples, and soon you'll be solving these problems in your sleep. And most importantly, don't be afraid to ask questions and seek help when you need it. We're all in this together, and the more we learn, the better we become. Keep up the great work, and happy calculating!