Find Trig Functions Given Secant: A Step-by-Step Guide

by Pedro Alvarez 55 views

Hey guys! Today, we're diving into the world of trigonometry, and we've got a fun problem to solve. We're given that $\sec \theta = \frac{\sqrt{6}}{2}$, and our mission is to find the values of all the other trigonometric functions. Don't worry, it's not as scary as it sounds! We'll break it down step by step and make sure we understand everything along the way. So, grab your calculators and let's get started!

Understanding the Basics

Before we jump into the problem, let's quickly recap the basic trigonometric functions and their relationships. Remember SOH CAH TOA? This handy mnemonic helps us remember the definitions of sine, cosine, and tangent:

  • Sine (sin θ) = Opposite / Hypotenuse
  • Cosine (cos θ) = Adjacent / Hypotenuse
  • Tangent (tan θ) = Opposite / Adjacent

Now, let's talk about the reciprocal functions:

  • Cosecant (csc θ) is the reciprocal of sine: csc θ = 1 / sin θ = Hypotenuse / Opposite
  • Secant (sec θ) is the reciprocal of cosine: sec θ = 1 / cos θ = Hypotenuse / Adjacent
  • Cotangent (cot θ) is the reciprocal of tangent: cot θ = 1 / tan θ = Adjacent / Opposite

Key Concept: The reciprocal trigonometric functions are simply the flipped versions of the primary trigonometric functions. This relationship is crucial for solving trigonometric problems, as knowing one function's value immediately gives us the value of its reciprocal.

Also, remember the Pythagorean Identity: $\sin^2 \theta + \cos^2 \theta = 1$. This identity is a cornerstone of trigonometry and helps us relate sine and cosine. We can also derive other useful identities from it by dividing both sides by $\cos^2 \theta$ or $\sin^2 \theta$. These identities are powerful tools in our trigonometric toolbox. This identity, along with the definitions of the trigonometric functions, allows us to solve a wide variety of problems, even when we only have limited information. Trigonometry is all about relationships, and these identities help us understand and utilize those relationships effectively.

Understanding these fundamental relationships is key to tackling any trigonometry problem. With these definitions and identities in our arsenal, we're well-equipped to find the remaining trigonometric function values when given the secant.

Finding Cosine (cos θ)

Okay, so we know that $\sec \theta = \frac{\sqrt{6}}{2}$. Remember that secant is the reciprocal of cosine. This means that:

cos⁡θ=1sec⁡θ\cos \theta = \frac{1}{\sec \theta}

To find $\cos \theta$, we simply take the reciprocal of $ rac{\sqrt{6}}{2}$:

cos⁡θ=162=26\cos \theta = \frac{1}{\frac{\sqrt{6}}{2}} = \frac{2}{\sqrt{6}}

Now, we usually don't like having radicals in the denominator, so let's rationalize it. We multiply both the numerator and denominator by $\sqrt{6}$:

cos⁡θ=26⋅66=266\cos \theta = \frac{2}{\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{2\sqrt{6}}{6}

We can simplify this fraction by dividing both the numerator and denominator by 2:

cos⁡θ=63\cos \theta = \frac{\sqrt{6}}{3}

Therefore, we've found that $\cos \theta = \frac{\sqrt{6}}{3}$. This is a significant step forward. By understanding the reciprocal relationship between secant and cosine, we were able to quickly determine the value of cosine. This illustrates the power of knowing your definitions and being able to apply them effectively. The process of rationalizing the denominator is also a crucial skill in mathematics, ensuring that we present our answers in the most simplified and conventional form. Remember, mathematical elegance often lies in simplicity and clarity.

Finding Sine (sin θ)

Now that we know $\cos \theta$, we can use the Pythagorean Identity to find $\sin \theta$. The identity is:

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

Let's plug in the value we found for $\cos \theta$:

sin⁡2θ+(63)2=1\sin^2 \theta + \left(\frac{\sqrt{6}}{3}\right)^2 = 1

Now, let's simplify:

sin⁡2θ+69=1\sin^2 \theta + \frac{6}{9} = 1

sin⁡2θ+23=1\sin^2 \theta + \frac{2}{3} = 1

Subtract $\frac{2}{3}$ from both sides:

sin⁡2θ=1−23\sin^2 \theta = 1 - \frac{2}{3}

sin⁡2θ=13\sin^2 \theta = \frac{1}{3}

To find $\sin \theta$, we take the square root of both sides:

sin⁡θ=±13\sin \theta = \pm\sqrt{\frac{1}{3}}

sin⁡θ=±13\sin \theta = \pm\frac{1}{\sqrt{3}}

Again, let's rationalize the denominator by multiplying both numerator and denominator by $\sqrt{3}$:

sin⁡θ=±13⋅33=±33\sin \theta = \pm\frac{1}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \pm\frac{\sqrt{3}}{3}

So, we have two possible values for $\sin \theta$: $\frac{\sqrt{3}}{3}$ and $-\frac{\sqrt{3}}{3}$. To determine the correct sign, we need more information about the angle $ heta$. For now, let's assume $\theta$ is in the first quadrant, where sine is positive. Therefore:

sin⁡θ=33\sin \theta = \frac{\sqrt{3}}{3}

This step highlights the importance of the Pythagorean Identity in linking sine and cosine. By strategically using this identity, we were able to navigate from the known value of cosine to the value of sine. The appearance of the $\pm$ sign when taking the square root reminds us that trigonometric functions can be positive or negative depending on the quadrant of the angle. To definitively determine the correct sign, we would need additional information about the angle's location. This underscores the importance of considering the context of the problem and paying attention to details like quadrant restrictions.

Finding Cosecant (csc θ)

Cosecant is the reciprocal of sine. We found that $\sin \theta = \frac{\sqrt{3}}{3}$, so:

csc⁡θ=1sin⁡θ=133\csc \theta = \frac{1}{\sin \theta} = \frac{1}{\frac{\sqrt{3}}{3}}

csc⁡θ=33\csc \theta = \frac{3}{\sqrt{3}}

Let's rationalize the denominator:

csc⁡θ=33⋅33=333\csc \theta = \frac{3}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{3}}{3}

csc⁡θ=3\csc \theta = \sqrt{3}

Therefore, $\csc \theta = \sqrt{3}$. Once we've determined the sine of an angle, finding the cosecant is a straightforward application of the reciprocal relationship. This further emphasizes the interconnectedness of the trigonometric functions. Each function's value provides a direct pathway to finding its reciprocal, streamlining the problem-solving process. The simplicity of this step underscores the importance of mastering the fundamental definitions of the trigonometric functions and their relationships.

Finding Tangent (tan θ)

Tangent is defined as $\tan \theta = \frac{\sin \theta}{\cos \theta}$. We know that $\sin \theta = \frac{\sqrt{3}}{3}$ and $\cos \theta = \frac{\sqrt{6}}{3}$, so:

tan⁡θ=3363\tan \theta = \frac{\frac{\sqrt{3}}{3}}{\frac{\sqrt{6}}{3}}

We can simplify this by multiplying the numerator by the reciprocal of the denominator:

tan⁡θ=33⋅36=36\tan \theta = \frac{\sqrt{3}}{3} \cdot \frac{3}{\sqrt{6}} = \frac{\sqrt{3}}{\sqrt{6}}

Let's rationalize the denominator:

tan⁡θ=36⋅66=186\tan \theta = \frac{\sqrt{3}}{\sqrt{6}} \cdot \frac{\sqrt{6}}{\sqrt{6}} = \frac{\sqrt{18}}{6}

We can simplify $\sqrt{18}$ as $\sqrt{9 \cdot 2} = 3\sqrt{2}$, so:

tan⁡θ=326\tan \theta = \frac{3\sqrt{2}}{6}

tan⁡θ=22\tan \theta = \frac{\sqrt{2}}{2}

Therefore, $\tan \theta = \frac{\sqrt{2}}{2}$. This step demonstrates how the tangent function acts as a bridge between sine and cosine. By understanding the ratio definition of tangent, we were able to leverage our previously calculated values of sine and cosine to determine the tangent. The simplification of radicals and fractions is a recurring theme in trigonometry, and mastering these skills allows for a more streamlined and accurate problem-solving process. This step reinforces the idea that each trigonometric function provides a unique perspective on the relationships within a right triangle, and by combining these perspectives, we can unlock a more complete understanding.

Finding Cotangent (cot θ)

Cotangent is the reciprocal of tangent. We found that $\tan \theta = \frac{\sqrt{2}}{2}$, so:

cot⁡θ=1tan⁡θ=122\cot \theta = \frac{1}{\tan \theta} = \frac{1}{\frac{\sqrt{2}}{2}}

cot⁡θ=22\cot \theta = \frac{2}{\sqrt{2}}

Let's rationalize the denominator:

cot⁡θ=22⋅22=222\cot \theta = \frac{2}{\sqrt{2}} \cdot \frac{\sqrt{2}}{\sqrt{2}} = \frac{2\sqrt{2}}{2}

cot⁡θ=2\cot \theta = \sqrt{2}

Therefore, $\cot \theta = \sqrt{2}$. Finding the cotangent is another direct application of the reciprocal relationship, this time between tangent and cotangent. This final calculation underscores the elegant symmetry within the trigonometric functions. Once we've determined the value of a function, its reciprocal can be found with minimal effort. This reinforces the importance of recognizing and utilizing these fundamental relationships to simplify the problem-solving process. With this final step, we've successfully navigated the problem and found all the remaining trigonometric function values.

Final Answer

Okay, guys, we did it! We successfully found all the remaining trigonometric function values when given $\sec \theta = \frac{\sqrt{6}}{2}$. Here's a summary of our results:

sin⁡θ=33\sin \theta = \frac{\sqrt{3}}{3}

csc⁡θ=3\csc \theta = \sqrt{3}

cos⁡θ=63\cos \theta = \frac{\sqrt{6}}{3}

sec⁡θ=62\sec \theta = \frac{\sqrt{6}}{2}

tan⁡θ=22\tan \theta = \frac{\sqrt{2}}{2}

cot⁡θ=2\cot \theta = \sqrt{2}

Remember, the key to solving these types of problems is understanding the definitions of the trigonometric functions, their reciprocal relationships, and the Pythagorean Identity. Keep practicing, and you'll become a trigonometry whiz in no time! This exercise highlights the power of combining fundamental definitions, reciprocal relationships, and key identities like the Pythagorean Identity. By systematically applying these tools, we were able to navigate from a single piece of information (the value of secant) to a complete understanding of all the trigonometric function values for the given angle. This demonstrates the interconnected nature of trigonometry and the beauty of how these functions relate to one another. Keep exploring and practicing, and you'll find that trigonometry becomes less daunting and more like a fascinating puzzle to be solved!