Finding Trigonometric Function Values Given Sine And Quadrant

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Hey guys! Let's dive into a super common trigonometry problem: figuring out the exact values of all the trig functions when you only know the sine and the quadrant. It might sound tricky, but trust me, we'll break it down step by step. We'll take the given information, use some handy trigonometric identities, and a little bit of geometry to nail this. So, grab your calculators (or not, because we're finding exact values!), and let's get started!

Problem Statement

Okay, so here's the deal. We're given that sin⁡θ=513\sin \theta = \frac{5}{13}, and we know that θ\theta is chilling in Quadrant I. Our mission, should we choose to accept it (and we do!), is to find the exact values of all the other trigonometric functions: cosine (cos⁡θ\cos \theta), tangent (tan⁡θ\tan \theta), cosecant (csc⁡θ\csc \theta), secant (sec⁡θ\sec \theta), and cotangent (cot⁡θ\cot \theta). Sounds like a party, right? Let's get this trigonometry party started!

Understanding the Basics

Before we jump into the calculations, let's quickly refresh some key concepts. This will make the whole process smoother, like butter on a hot skillet. First, remember the definitions of the trigonometric functions. Sine (sin⁡θ\sin \theta) is the ratio of the opposite side to the hypotenuse in a right triangle, cosine (cos⁡θ\cos \theta) is the ratio of the adjacent side to the hypotenuse, and tangent (tan⁡θ\tan \theta) is the ratio of the opposite side to the adjacent side. A handy mnemonic to remember this is SOH CAH TOA:

  • Sine = Opposite / Hypotenuse
  • Cosine = Adjacent / Hypotenuse
  • Tangent = Opposite / Adjacent

The reciprocal functions are just the flipped versions of these: cosecant (csc⁡θ\csc \theta) is the reciprocal of sine, secant (sec⁡θ\sec \theta) is the reciprocal of cosine, and cotangent (cot⁡θ\cot \theta) is the reciprocal of tangent. Also, let's not forget the Pythagorean Identity: sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1. This little gem is going to be our best friend in solving this problem. Finally, let’s keep in mind the quadrant rules. In Quadrant I, all trigonometric functions are positive. This is crucial because it tells us that all the values we calculate should be positive.

Understanding these basics will not only help you solve this specific problem but will also give you a solid foundation for tackling more complex trig problems in the future. It’s like having the right tools in your toolbox – you’re ready for anything!

Using the Pythagorean Identity

Alright, let’s get our hands dirty with some actual calculations! We know that sin⁡θ=513\sin \theta = \frac{5}{13}, and we want to find cos⁡θ\cos \theta. This is where the Pythagorean Identity, sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1, comes to the rescue. Plug in the value of sin⁡θ\sin \theta into the identity:

(513)2+cos⁡2θ=1\left(\frac{5}{13}\right)^2 + \cos^2 \theta = 1

Squaring 513\frac{5}{13} gives us 25169\frac{25}{169}, so the equation becomes:

25169+cos⁡2θ=1\frac{25}{169} + \cos^2 \theta = 1

Now, we need to isolate cos⁡2θ\cos^2 \theta. Subtract 25169\frac{25}{169} from both sides:

cos⁡2θ=1−25169\cos^2 \theta = 1 - \frac{25}{169}

To subtract these, we need a common denominator. Rewrite 1 as 169169\frac{169}{169}:

cos⁡2θ=169169−25169\cos^2 \theta = \frac{169}{169} - \frac{25}{169}

Now we can subtract the numerators:

cos⁡2θ=144169\cos^2 \theta = \frac{144}{169}

To find cos⁡θ\cos \theta, we take the square root of both sides:

cos⁡θ=±144169\cos \theta = \pm\sqrt{\frac{144}{169}}

The square root of 144169\frac{144}{169} is 1213\frac{12}{13}, but we need to consider both positive and negative roots. Since θ\theta is in Quadrant I, where cosine is positive, we choose the positive root:

cos⁡θ=1213\cos \theta = \frac{12}{13}

See? We're already making progress! By using the Pythagorean Identity, we've successfully found the exact value of cos⁡θ\cos \theta. This is a crucial step because now we have two of the main trigonometric functions, and the rest will follow more easily.

Calculating Tangent

With both sin⁡θ\sin \theta and cos⁡θ\cos \theta in our arsenal, finding the tangent (tan⁡θ\tan \theta) is a piece of cake. Remember, tangent is defined as the ratio of sine to cosine:

tan⁡θ=sin⁡θcos⁡θ\tan \theta = \frac{\sin \theta}{\cos \theta}

We know that sin⁡θ=513\sin \theta = \frac{5}{13} and cos⁡θ=1213\cos \theta = \frac{12}{13}, so we can plug these values into the formula:

tan⁡θ=5131213\tan \theta = \frac{\frac{5}{13}}{\frac{12}{13}}

To divide fractions, we multiply by the reciprocal of the denominator:

tan⁡θ=513⋅1312\tan \theta = \frac{5}{13} \cdot \frac{13}{12}

The 13s cancel out, leaving us with:

tan⁡θ=512\tan \theta = \frac{5}{12}

And there you have it! We've found the exact value of tan⁡θ\tan \theta. It's amazing how one piece of information leads to another, like dominoes falling. Knowing sine and cosine makes finding tangent straightforward, and it shows how interconnected these trigonometric functions really are.

Finding Reciprocal Functions

Now that we've got the main trio – sine, cosine, and tangent – let’s tackle their reciprocal buddies: cosecant (csc⁡θ\csc \theta), secant (sec⁡θ\sec \theta), and cotangent (cot⁡θ\cot \theta). Remember, reciprocal functions are just the flipped versions of the originals. This makes finding them super easy!

Cosecant (csc⁡θ\csc \theta)

Cosecant is the reciprocal of sine, so:

csc⁡θ=1sin⁡θ\csc \theta = \frac{1}{\sin \theta}

We know sin⁡θ=513\sin \theta = \frac{5}{13}, so:

csc⁡θ=1513\csc \theta = \frac{1}{\frac{5}{13}}

To divide by a fraction, we multiply by its reciprocal:

csc⁡θ=135\csc \theta = \frac{13}{5}

Secant (sec⁡θ\sec \theta)

Secant is the reciprocal of cosine, so:

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

We know cos⁡θ=1213\cos \theta = \frac{12}{13}, so:

sec⁡θ=11213\sec \theta = \frac{1}{\frac{12}{13}}

Again, multiply by the reciprocal:

sec⁡θ=1312\sec \theta = \frac{13}{12}

Cotangent (cot⁡θ\cot \theta)

Cotangent is the reciprocal of tangent, so:

cot⁡θ=1tan⁡θ\cot \theta = \frac{1}{\tan \theta}

We know tan⁡θ=512\tan \theta = \frac{5}{12}, so:

cot⁡θ=1512\cot \theta = \frac{1}{\frac{5}{12}}

Multiply by the reciprocal:

cot⁡θ=125\cot \theta = \frac{12}{5}

Boom! We've successfully found the exact values of all three reciprocal trigonometric functions. This part is really just about knowing the definitions and doing a quick flip. It's like the victory lap after a good run – satisfying and relatively easy.

Summarizing the Results

Alright, let’s take a moment to bask in our trigonometric glory and summarize what we've found. We started with sin⁡θ=513\sin \theta = \frac{5}{13} and the knowledge that θ\theta is in Quadrant I. Through a combination of the Pythagorean Identity and the definitions of trigonometric functions, we've managed to find the exact values of all the remaining trig functions. Here's a quick recap:

  • sin⁡θ=513\sin \theta = \frac{5}{13} (Given)
  • cos⁡θ=1213\cos \theta = \frac{12}{13}
  • tan⁡θ=512\tan \theta = \frac{5}{12}
  • csc⁡θ=135\csc \theta = \frac{13}{5}
  • sec⁡θ=1312\sec \theta = \frac{13}{12}
  • cot⁡θ=125\cot \theta = \frac{12}{5}

We've gone from knowing just one trig function to knowing them all! This is a testament to the power of trigonometric identities and the relationships between these functions. Being able to find these values is a fundamental skill in trigonometry and is essential for more advanced topics.

Conclusion

So there you have it, guys! We've successfully navigated the world of trigonometric functions and found the exact values of all of them, given just the sine and the quadrant. We used the Pythagorean Identity, the definitions of trigonometric functions, and a little bit of algebraic manipulation to get the job done. Remember, practice makes perfect, so keep tackling these types of problems. The more you do, the more comfortable and confident you'll become. Understanding these foundational concepts is crucial for success in trigonometry and beyond. Keep up the great work, and happy trig-ing!