Розв'язання Ctg Α = 8/√10 Знаходження Значення Виразу

Hey guys, ever stumbled upon a math problem that seems like a riddle wrapped in an enigma? Today, we're going to unravel one such puzzle, focusing on finding the value of trigonometric expressions, particularly when we're given the cotangent of an angle, specifically, ctg α = 8/√10. Now, before your eyes glaze over with mathematical jargon, let's break this down in a way that's not only easy to grasp but also kinda fun. So, grab your thinking caps, and let's dive into the fascinating world of trigonometry!

Delving into Trigonometric Identities and Relationships

First things first, let's chat about the trigonometric landscape. Think of trigonometry as a family of functions – sine, cosine, tangent, cotangent, secant, and cosecant – all interlinked and playing off each other. Understanding their relationships is key to cracking problems like ours. Our main player today is cotangent (ctg), which, if you recall, is the reciprocal of the tangent function. In mathematical terms, ctg α = 1 / tan α. This is our golden ticket, folks! But wait, there's more to this family drama. Tangent, in turn, is the ratio of sine to cosine (tan α = sin α / cos α), making cotangent the flip side – ctg α = cos α / sin α. See how everything's connected? Now, let's circle back to our mission: finding the value of expressions when ctg α = 8/√10. This piece of information is like a treasure map leading us to the solution. We're not just given a random number; we're handed a crucial relationship between the cosine and sine of angle alpha. This is where the magic happens – the point where we can start substituting and simplifying. But before we jump into calculations, let's talk strategy. Knowing ctg α directly doesn't give us sine or cosine in isolation, but it does give us their ratio. To find individual values, we need another equation involving sine and cosine. Enter the Pythagorean identity, our trusty sidekick in this trigonometric quest.

Unpacking the Pythagorean Identity: A Trigonometric Cornerstone

The Pythagorean identity, a cornerstone in the world of trigonometry, states that sin² α + cos² α = 1. This equation is like the ultimate relationship status update for sine and cosine – they're always connected, always summing up to one (when their squares are added, that is). This is huge for us because it provides the extra piece of the puzzle we need. We have one equation (ctg α = cos α / sin α) and we need another to solve for two unknowns (sin α and cos α). The Pythagorean identity swoops in as our second equation, setting the stage for a system of equations that we can solve. Now, the fun part begins. We have two equations:

1. ctg α = cos α / sin α = 8/√10
2. sin² α + cos² α = 1

Our mission is to massage these equations, mix and match them, until we isolate sin α and cos α. Think of it as a trigonometric tango, a dance of algebra and identities leading us to the grand reveal. The strategy here is to use the first equation to express either cos α in terms of sin α (or vice versa) and then substitute that expression into the Pythagorean identity. This transforms the Pythagorean identity into a single-variable equation, which we can then solve using good ol' algebraic techniques. Once we've cracked the code for sin α (or cos α), we can easily find the other using our initial equations. And remember, folks, in the land of trigonometry, angles can live in different quadrants, each with its own personality – a.k.a., sign conventions for sine, cosine, and tangent. So, we need to keep an eye out for those pesky plus or minus signs, as they can significantly impact our final answer. It's like baking a cake; you need to get the ingredients and the measurements just right, or you might end up with a flat disaster. Let's start by expressing cos α in terms of sin α using the cotangent relationship. This will pave the way for our substitution and simplification adventure.

The Algebraic Tango: Solving for Sine and Cosine

Alright, let's get our hands dirty with some algebra! We know from ctg α = 8/√10 that cos α / sin α = 8/√10. Our first move is to isolate cos α, so we'll multiply both sides of the equation by sin α. This gives us:

cos α = (8/√10) * sin α

See how we've elegantly expressed cosine in terms of sine? This is our secret sauce for the next step: substitution. We'll take this expression for cos α and plug it into our trusty Pythagorean identity, sin² α + cos² α = 1. When we do this, the identity transforms into a single equation with just one variable, sin α, which is exactly what we want. Brace yourselves; here comes the substitution:

sin² α + ((8/√10) * sin α)² = 1

Now, let's simplify this bad boy. We need to square the term inside the parentheses, which means squaring both the constant and the sin α part. Remember, squaring a fraction means squaring both the numerator and the denominator. So, (8/√10)² becomes 64/10. Our equation now looks like this:

sin² α + (64/10) * sin² α = 1

We're almost there, guys! Notice that we have two terms, both involving sin² α. This means we can combine them by factoring out the sin² α. It's like having two slices of the same pizza; you can combine them into one bigger slice. Factoring out sin² α gives us:

sin² α * (1 + 64/10) = 1

Now, let's simplify the term inside the parentheses. We need to add 1 to 64/10. To do this, we'll rewrite 1 as 10/10 so that we have a common denominator. This gives us:

sin² α * (10/10 + 64/10) = 1

sin² α * (74/10) = 1

We're on the home stretch! To isolate sin² α, we'll divide both sides of the equation by 74/10. Remember, dividing by a fraction is the same as multiplying by its reciprocal. So, we'll multiply both sides by 10/74:

sin² α = 1 * (10/74)

sin² α = 10/74

We can simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 2. This gives us:

sin² α = 5/37

Now, to find sin α, we need to take the square root of both sides. But remember, taking the square root can give us two possible solutions: a positive one and a negative one. This is where the quadrant of α comes into play, which we'll discuss later. For now, let's just write down both possibilities:

sin α = ±√(5/37)

We have successfully found the possible values for sin α! Now, we can use these values and our earlier expression for cos α in terms of sin α to find the corresponding values for cos α. This is like finding the matching socks in your drawer; once you have one, you can easily find the other. Let's proceed to calculate cos α using the values we've found for sin α.

Cracking the Cosine Code: Calculating cos α

Okay, we've nailed down sin α, which is a huge win! We found that sin α = ±√(5/37). Now, it's time to uncover the value of cos α. Remember our little equation from earlier, cos α = (8/√10) * sin α? This is our golden key to unlock the cosine value. We're going to plug in our two possible values for sin α and see what cos α reveals. Let's start with the positive root:

If sin α = √(5/37), then

cos α = (8/√10) * √(5/37)

Now, this looks a bit messy, but fear not! We can simplify this expression using some algebraic wizardry. First, let's rewrite √10 as √(2 * 5) so we can potentially cancel out some terms. Our equation becomes:

cos α = (8/√(2 * 5)) * √(5/37)

We can rewrite this as:

cos α = (8 * √5) / (√(2) * √(5) * √37)

Notice that we have a √5 in both the numerator and the denominator, which we can cancel out. This simplifies our expression to:

cos α = 8 / (√(2) * √37)

We can combine the square roots in the denominator:

cos α = 8 / √(2 * 37)

cos α = 8 / √74

Now, let's rationalize the denominator by multiplying both the numerator and the denominator by √74:

cos α = (8 * √74) / (√74 * √74)

cos α = (8 * √74) / 74

We can simplify this fraction by dividing both the numerator and the denominator by their greatest common divisor, which is 2:

cos α = (4 * √74) / 37

So, when sin α = √(5/37), we find that cos α = (4 * √74) / 37. Awesome! Now, let's tackle the case where sin α is negative. If sin α = -√(5/37), then

cos α = (8/√10) * (-√(5/37))

This is the same calculation as before, but with a negative sign. So, we can simply take our previous result and slap a negative sign on it:

cos α = -(4 * √74) / 37

So, when sin α = -√(5/37), we have cos α = -(4 * √74) / 37. We've now found the two possible values for cos α that correspond to our two possible values for sin α. We're on fire, guys! But hold on, we're not quite at the finish line yet. We need to consider the implications of these two solutions and think about which quadrant α might live in. This will help us determine the correct signs for sin α and cos α and give us the most accurate answer. Let's dive into the world of quadrants and signs to complete our trigonometric quest.

Navigating the Quadrants: Sign Conventions and Final Answers

Alright, we've arrived at the final leg of our journey: understanding how the quadrant of angle alpha (α) affects the signs of sine and cosine. This is like the final twist in a detective novel, where we piece together the clues to reveal the whole picture. Remember, the trigonometric world is divided into four quadrants, each with its own personality. In the first quadrant (0° to 90°), both sine and cosine are positive. It's like the land of sunshine and happiness for trigonometric functions. In the second quadrant (90° to 180°), sine is positive, but cosine is negative. Think of it as the sine having a party, while the cosine is feeling a bit gloomy. The third quadrant (180° to 270°) is where both sine and cosine are negative, a rather melancholic place for our functions. Finally, in the fourth quadrant (270° to 360°), cosine is positive, but sine is negative. It's like a reverse of the second quadrant, where the cosine gets to shine while the sine is in the shadows. Now, how does this relate to our problem? We know that ctg α = 8/√10, which is a positive number. Cotangent, remember, is the ratio of cosine to sine (ctg α = cos α / sin α). For this ratio to be positive, either both cosine and sine must be positive, or both must be negative. This narrows down our possibilities to two quadrants: the first quadrant (where both are positive) and the third quadrant (where both are negative). Let's revisit our calculated values for sine and cosine:

• sin α = ±√(5/37)
• cos α = ±(4 * √74) / 37

We have two scenarios to consider:

1. Scenario 1: α in the First Quadrant

   In this case, both sin α and cos α are positive. So, we take the positive roots:

   • sin α = √(5/37)
   • cos α = (4 * √74) / 37

2. Scenario 2: α in the Third Quadrant

   Here, both sin α and cos α are negative. We take the negative roots:

   • sin α = -√(5/37)
   • cos α = -(4 * √74) / 37

And there you have it, folks! We've successfully navigated the trigonometric terrain, conquered the algebraic challenges, and considered the quadrant quandaries. We've found two possible sets of values for sin α and cos α that satisfy the condition ctg α = 8/√10. Depending on the specific context or any additional information about the angle α (which quadrant it lies in), we can choose the appropriate set of solutions. This journey through trigonometry has shown us the power of relationships, the elegance of identities, and the importance of considering all possibilities. So, the next time you encounter a trigonometric puzzle, remember the tools and strategies we've explored today, and you'll be well-equipped to crack the code! Now, go forth and conquer those angles!