Solving -sin²(x) = Cos(2x) In The Interval [-π, Π] A Trigonometric Exploration

Introduction

Hey guys! Today, we're diving deep into the fascinating world of trigonometry to tackle the equation $-\sin^2(x) = \cos(2x)$ within the interval $[-\pi, \pi]$. This isn't just about crunching numbers; it's about understanding the relationship between trigonometric functions and how they dance together on the unit circle. Think of it as uncovering a secret language of waves and angles. We will meticulously explore the equation, leveraging trigonometric identities and graphical analysis to pinpoint the solutions. By the end of this journey, you'll not only have the answers but also a stronger intuition for how these functions behave. So, grab your thinking caps, and let's embark on this mathematical adventure!

Understanding the Core Concepts

Before we jump into solving, let's solidify our understanding of the key players: sine, cosine, and their identities. Sine, denoted as $\sin(x)$, represents the y-coordinate of a point on the unit circle corresponding to an angle x. Cosine, $\cos(x)$, mirrors this but for the x-coordinate. These functions are like two sides of the same coin, intricately linked and periodic, meaning their values repeat after a certain interval (2$\pi$ for both). Now, the star of our show is the double-angle identity for cosine: $\cos(2x) = \cos^2(x) - \sin^2(x)$. This identity is a game-changer, allowing us to rewrite our equation in a more manageable form. It's like having a translator that can convert a complex phrase into simple words. But we can go even further! Remember the Pythagorean identity, $\sin^2(x) + \cos^2(x) = 1$? This gem lets us express $\cos^2(x)$ as $1 - \sin^2(x)$, which will be crucial in simplifying our equation and bringing it closer to a solution. Without these fundamental concepts and identities, we'd be navigating in the dark. They provide the light and the map we need to reach our destination: the solutions to the equation.

Transforming the Equation

Okay, let's get our hands dirty and transform our equation. We start with $-\sin^2(x) = \cos(2x)$. The first move is to apply the double-angle identity, replacing $\cos(2x)$ with $\cos^2(x) - \sin^2(x)$. This gives us $-\sin^2(x) = \cos^2(x) - \sin^2(x)$. See how things are starting to simplify? Now, let’s use the Pythagorean identity to replace $\cos^2(x)$ with $1 - \sin^2(x)$. Our equation now looks like this: $-\sin^2(x) = 1 - \sin^2(x) - \sin^2(x)$. We're making progress! Notice that $-\sin^2(x)$ appears on both sides. We can add $\sin^2(x)$ to both sides, which neatly eliminates it from the left side, leaving us with $0 = 1 - 2\sin^2(x)$. This is fantastic! We've successfully transformed the equation into a much simpler form, isolating the $\sin^2(x)$ term. It's like peeling away the layers of an onion to reveal the core. Now, we're just one step away from isolating $\sin^2(x)$ completely and unlocking the solutions. By adding $2\sin^2(x)$ to both sides and then dividing by 2, we arrive at the elegant equation $\sin^2(x) = \frac{1}{2}$. This is a pivotal moment! We've transformed a seemingly complex trigonometric equation into a straightforward algebraic one, setting the stage for finding the values of x that make it true.

Solving for $\sin(x)$

Now comes the fun part: actually solving for $\sin(x)$. We've arrived at the equation $\sin^2(x) = \frac{1}{2}$. To get $\sin(x)$ by itself, we need to take the square root of both sides. But hold on! Remember that taking the square root introduces both positive and negative solutions. So, we get $\sin(x) = \pm\sqrt{\frac{1}{2}}$. Let's simplify this a bit. The square root of $\frac{1}{2}$ is the same as $\frac{\sqrt{1}}{\sqrt{2}}$, which is simply $\frac{1}{\sqrt{2}}$. To make things even neater, we can rationalize the denominator by multiplying both the numerator and denominator by $\sqrt{2}$, giving us $\frac{\sqrt{2}}{2}$. Therefore, we have two possibilities: $\sin(x) = \frac{\sqrt{2}}{2}$ or $\sin(x) = -\frac{\sqrt{2}}{2}$. These are our target values for $\sin(x)$. It's like having two targets to hit on a dartboard. Now, we need to figure out which angles x in our interval $[-\pi, \pi]$ produce these sine values. This involves thinking about the unit circle and where sine is positive and negative. Remember, sine corresponds to the y-coordinate on the unit circle, so we're looking for points where the y-coordinate is either $\frac{\sqrt{2}}{2}$ or $-\frac{\sqrt{2}}{2}$. This is where our understanding of the unit circle and reference angles becomes crucial in finding the specific solutions for x.

Finding the Solutions for $x$ within $[-\pi, \pi]$

Alright, let's hunt down the values of x within the interval $[-\pi, \pi]$ that satisfy $\sin(x) = \frac{\sqrt{2}}{2}$ or $\sin(x) = -\frac{\sqrt{2}}{2}$. Visualizing the unit circle is our superpower here. Think of the unit circle as a clock, with 0 radians at the 3 o'clock position and increasing counterclockwise. The sine function, remember, corresponds to the y-coordinate. First, let's tackle $\sin(x) = \frac{\sqrt{2}}{2}$. This value is positive, meaning we're looking for angles in the first and second quadrants. The reference angle whose sine is $\frac{\sqrt{2}}{2}$ is $\frac{\pi}{4}$ (45 degrees). So, in the first quadrant, one solution is simply $x = \frac{\pi}{4}$. In the second quadrant, the angle with the same reference angle is $\pi - \frac{\pi}{4} = \frac{3\pi}{4}$. Great! We've found two solutions. Now, let's consider $\sin(x) = -\frac{\sqrt{2}}{2}$. This value is negative, so we're looking for angles in the third and fourth quadrants. The reference angle is still $\frac{\pi}{4}$. In the third quadrant, the angle is $\pi + \frac{\pi}{4} = \frac{5\pi}{4}$. However, this is outside our interval of $[-\pi, \pi]$. To bring it within the interval, we can subtract $2\pi$, but that would take us further away. Instead, we can express this angle as its negative counterpart in the given interval: $\frac{5\pi}{4} - 2\pi = -\frac{3\pi}{4}$. Finally, in the fourth quadrant, the angle is $2\pi - \frac{\pi}{4} = \frac{7\pi}{4}$, which is also outside our interval. Again, we find the equivalent angle within the interval by subtracting $2\pi$, giving us $\frac{7\pi}{4} - 2\pi = -\frac{\pi}{4}$. So, our complete set of solutions within the interval $[-\pi, \pi]$ is $x = \frac{\pi}{4}, \frac{3\pi}{4}, -\frac{3\pi}{4}, -\frac{\pi}{4}$. We've successfully navigated the unit circle and pinpointed all the angles that satisfy our original equation! It’s like completing a treasure hunt, where the solutions are the hidden gems.

Verification of Solutions

Before we declare victory, let's verify our solutions. It's like double-checking our work to make sure we haven't made any mistakes. We found four potential solutions: $x = \frac{\pi}{4}, \frac{3\pi}{4}, -\frac{\pi}{4}, -\frac{3\pi}{4}$. To verify, we'll plug each of these values back into our original equation, $-\sin^2(x) = \cos(2x)$, and see if both sides are equal. Let's start with $x = \frac{\pi}{4}$. We have $-\sin^2(\frac{\pi}{4}) = -(\frac{\sqrt{2}}{2})^2 = -\frac{1}{2}$. And $\cos(2 * \frac{\pi}{4}) = \cos(\frac{\pi}{2}) = 0$. Uh oh! It seems like there's a mistake, because $-\frac{1}{2}$ is not equal to $0$. This highlights the crucial importance of verification. We need to go back and carefully review our steps to identify where we might have gone astray. It's like being a detective and retracing our steps at the scene of the crime. Upon reviewing, we realize that while we correctly transformed the equation to $\sin^2(x) = \frac{1}{2}$, we made an error in the initial simplification process. Let’s revisit the transformation:

$-\sin^2(x) = \cos(2x)$

$-\sin^2(x) = \cos^2(x) - \sin^2(x)$

Using $\cos^2(x) = 1 - \sin^2(x)$:

$-\sin^2(x) = 1 - \sin^2(x) - \sin^2(x)$

$-\sin^2(x) = 1 - 2\sin^2(x)$

Adding $2\sin^2(x)$ to both sides:

$\sin^2(x) = 1$

Taking the square root:

$\sin(x) = \pm 1$

Now, this gives us a different set of values to consider. Let's correct our solutions based on this new finding.

Corrected Solutions and Verification

Okay, guys, we've caught our mistake and now have the correct equation to solve: $\sin(x) = \pm 1$. This means we're looking for angles where the y-coordinate on the unit circle is either 1 or -1. Within the interval $[-\pi, \pi]$, these angles are $x = \frac{\pi}{2}$ and $x = -\frac{\pi}{2}$. These are the angles where the sine function reaches its maximum and minimum values. Now, let's verify these solutions using the original equation, $-\sin^2(x) = \cos(2x)$.

For $x = \frac{\pi}{2}$:

$-\sin^2(\frac{\pi}{2}) = -(1)^2 = -1$

$\cos(2 * \frac{\pi}{2}) = \cos(\pi) = -1$

So, the equation holds true for $x = \frac{\pi}{2}$.

For $x = -\frac{\pi}{2}$:

$-\sin^2(-\frac{\pi}{2}) = -(-1)^2 = -1$

$\cos(2 * -\frac{\pi}{2}) = \cos(-\pi) = -1$

And the equation also holds true for $x = -\frac{\pi}{2}$.

Therefore, our corrected and verified solutions are $x = \frac{\pi}{2}$ and $x = -\frac{\pi}{2}$. This process highlights the importance of careful verification in mathematics. Even a small error in the beginning can lead to incorrect solutions. But by checking our work, we can catch these mistakes and ensure the accuracy of our results.

Graphical Verification

To further solidify our understanding and verify our solutions, let's take a visual approach using graphs. We'll plot the functions $y = -\sin^2(x)$ and $y = \cos(2x)$ on the same coordinate plane within the interval $[-\pi, \pi]$. The points where these two graphs intersect represent the solutions to our equation, $-\sin^2(x) = \cos(2x)$. When we plot these graphs, we'll observe that they intersect at two points within our interval. These points correspond to the solutions we found algebraically: $x = \frac{\pi}{2}$ and $x = -\frac{\pi}{2}$. At $x = \frac{\pi}{2}$, both functions have a value of -1. The sine function reaches its maximum value of 1 at $\frac{\pi}{2}$, so $-\sin^2(x)$ is -1. The cosine function, $\cos(2x)$, completes a full cycle in $\pi$ radians, so $\cos(2 * \frac{\pi}{2}) = \cos(\pi) = -1$. Similarly, at $x = -\frac{\pi}{2}$, both functions also have a value of -1. The sine function reaches its minimum value of -1 at $-\frac{\pi}{2}$, so $-\sin^2(x)$ is again -1. And $\cos(2 * -\frac{\pi}{2}) = \cos(-\pi) = -1$. The graphical verification provides a visual confirmation that our algebraic solutions are correct. It's like seeing the puzzle pieces fit together perfectly. The intersections of the graphs visually represent the points where the equation holds true, reinforcing our understanding of the relationship between the sine and cosine functions. This multi-faceted approach – algebraic and graphical – strengthens our confidence in the accuracy of our solutions.

Conclusion

So, there you have it, guys! We've successfully navigated the trigonometric equation $-\sin^2(x) = \cos(2x)$ within the interval $[-\pi, \pi]$. We started by understanding the fundamental trigonometric concepts and identities, like the double-angle identity for cosine and the Pythagorean identity. We then skillfully transformed the equation, simplifying it step-by-step until we isolated the sine function. We initially made a mistake but rectified it through careful review and verification, a crucial part of the mathematical process. This journey highlighted the importance of accuracy and diligence in problem-solving. We discovered that the solutions are $x = \frac{\pi}{2}$ and $x = -\frac{\pi}{2}$. We not only found these solutions algebraically but also verified them graphically, providing a visual confirmation of our results. This combined approach reinforces our understanding and confidence in our answers. More than just finding the solutions, we've deepened our understanding of the interplay between trigonometric functions and the unit circle. We've seen how identities act as powerful tools for simplifying equations and how graphical analysis can provide valuable insights. This exploration has not only equipped us with the ability to solve this specific equation but has also strengthened our overall mathematical toolkit. Keep practicing, keep exploring, and keep enjoying the beauty of mathematics!