Prove -1 ≤ Sin(x) ≤ 1 Using AM-GM Inequality
Let's dive into a fascinating exploration of how to prove that the sine function, denoted as sin(x), is bounded between -1 and 1 for all real numbers x. This is a fundamental concept in trigonometry and calculus, and there are several elegant ways to demonstrate it. Here, we'll focus on a proof approach leveraging the Arithmetic Mean - Geometric Mean (AM-GM) inequality, along with some insightful algebraic manipulations. So, buckle up, math enthusiasts! We're about to embark on a journey to solidify this cornerstone of trigonometric knowledge.
Applying AM-GM Inequality
Okay, guys, let's start with the core of our proof: the AM-GM inequality. For any non-negative real numbers, the Arithmetic Mean is always greater than or equal to the Geometric Mean. Mathematically, for non-negative numbers a and b, this is expressed as:
(a + b) / 2 ≥ √(ab)
This inequality is a powerful tool, and we're going to use it cleverly to box in our sin(x). Now, consider the numbers 1 and sin²(x). Since the square of any real number is non-negative, sin²(x) is always greater than or equal to 0. Thus, both 1 and sin²(x) are non-negative, and we can happily apply the AM-GM inequality. Plugging these into our formula, we get:
(1 + sin²(x)) / 2 ≥ √(1 * sin²(x))
This sets the stage for our proof. It might seem like a simple step, but it’s the foundation upon which we'll build our argument. By carefully choosing 1 and sin²(x), we’ve created an inequality that involves sin(x) in a way that we can manipulate. Next, we'll explore the two cases that arise from this inequality, carefully teasing out the bounds of sin(x). It’s like a mathematical puzzle, and we're slowly fitting the pieces together!
Case 1: 1 + sin²(x) ≥ 2 sin(x)
Let's dig deeper into our first case. Multiplying both sides of the AM-GM inequality by 2, we arrive at:
1 + sin²(x) ≥ 2√(sin²(x))
Since the square root of sin²(x) is the absolute value of sin(x), we can rewrite this as:
1 + sin²(x) ≥ 2 |sin(x)|
However, the original inequality derived directly from AM-GM was:
1 + sin²(x) ≥ 2 sin(x)
This is a crucial point. Now, let’s rearrange this inequality by subtracting 2 sin(x) from both sides. This gives us:
sin²(x) - 2 sin(x) + 1 ≥ 0
Do you notice anything special about the left-hand side? It's a perfect square! We can factor it as:
(sin(x) - 1)² ≥ 0
This is a powerful result. The square of any real number is always non-negative. Therefore, (sin(x) - 1)² is always greater than or equal to 0, regardless of the value of x. This tells us that our inequality holds true, which is excellent news. But what does it actually mean for sin(x)? Well, it implies that sin(x) - 1 must be a real number (which we already knew), and squaring it results in a non-negative value. This reinforces the idea that sin(x) cannot be arbitrarily large. It gives us a hint that sin(x) is bounded from above. We're getting closer to our goal of proving that sin(x) ≤ 1. So, let's keep this in mind as we move on to the next step!
Subtracting 2sin(x) from both sides, we get:
sin²(x) − 2sin(x) + 1 ≥ 0
This can be written as:
(sin(x) − 1)² ≥ 0
Since the square of any real number is non-negative, this inequality holds true. This tells us that sin(x) - 1 is a real number, which means that sin(x) has an upper bound.
Case 2: 1 + sin²(x) ≥ -2 sin(x)
Alright, let's flip the script and consider the second scenario that arises from our AM-GM inequality exploration. We're still working with the same fundamental principle, but we're going to approach it from a slightly different angle. Remember, the absolute value sneaked into our consideration, so now we're looking at the negative possibility. So, this time, let's consider the case where:
1 + sin²(x) ≥ -2 sin(x)
Notice the subtle but significant change in sign. This might seem like a minor tweak, but it opens up a new avenue for us to explore the behavior of sin(x). Just like before, our goal is to massage this inequality into a form that reveals more about the bounds of sin(x). So, what's our next move? Let’s try rearranging the terms, similar to what we did in the first case. We want to isolate the sin(x) terms and see if we can spot any familiar patterns.
Adding 2 sin(x) to both sides of the inequality, we get:
sin²(x) + 2 sin(x) + 1 ≥ 0
Hey, this looks familiar! It's another perfect square trinomial, just like in the previous case. This is a delightful surprise because it means we're on the right track. Perfect squares are our friends in this kind of mathematical puzzle. They often lead to elegant simplifications and reveal hidden relationships. So, let's put on our factoring hats and see what this perfect square has in store for us. Factoring the left-hand side, we have:
(sin(x) + 1)² ≥ 0
Aha! Another square, and squares are always non-negative, as we discussed before. This means that (sin(x) + 1)² is greater than or equal to 0, no matter what value x takes. This is a powerful statement, and it gives us crucial information about the lower bound of sin(x). Just like the previous case hinted at an upper bound, this one is whispering secrets about a lower bound. We're homing in on our target, folks! Now, let's think carefully about what this inequality tells us about the possible values of sin(x).
Adding 2sin(x) to both sides, we have:
sin²(x) + 2sin(x) + 1 ≥ 0
Which can be written as:
(sin(x) + 1)² ≥ 0
Again, since the square of any real number is non-negative, this inequality also holds true. Thus, sin(x) + 1 is a real number, indicating that sin(x) has a lower bound.
Combining the Results
Alright, guys, let's take a step back and look at the big picture. We've explored two cases using the AM-GM inequality, and each case has given us a valuable piece of the puzzle. In the first case, we showed that (sin(x) - 1)² ≥ 0. This tells us something important about the upper bound of sin(x). Specifically, it implies that sin(x) - 1 cannot be a