Understanding When Sin(3x)cos(3x) Increases Or Decreases

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What's up, math enthusiasts! Today, we're diving deep into the wild and wonderful world of trigonometric functions, specifically focusing on the behavior of $\sin(3x)\cos(3x)$. You know, those oscillating waves that pop up everywhere from physics to engineering? Well, understanding when a function is increasing or decreasing is a super crucial concept in calculus. It tells us about the function's trajectory – is it going up, or is it heading down? For $\sin(3x)\cos(3x)$, this analysis is a bit more involved than your average $f(x)=x^2$, but that's what makes it fun, right? We're going to break down exactly how to figure out those intervals of increase and decrease, using our trusty calculus tools. So, grab your calculators, maybe a coffee, and let's get this party started! We'll be using derivatives, the product-to-sum identity, and a little bit of trigonometric magic to uncover the secrets of $\sin(3x)\cos(3x)$'s movement.

The Power of the Derivative

Alright guys, let's talk about the absolute rockstar of determining whether a function is increasing or decreasing: the derivative. You've probably heard of it, maybe even wrestled with it in your calculus class. The fundamental idea is this: if the derivative of a function, let's call it $f'(x)$, is positive over a certain interval, then the original function $f(x)$ is increasing on that interval. Conversely, if $f'(x)$ is negative, $f(x)$ is decreasing. It's like the derivative is the function's speedometer, telling us its instantaneous rate of change. A positive speed means we're moving forward (increasing), and a negative speed means we're going backward (decreasing). So, our first mission, should we choose to accept it, is to find the derivative of our function, $f(x) = \sin(3x)\cos(3x)$.

Now, how do we tackle this beast? We've got a product of two functions, $\sin(3x)$ and $\cos(3x)$. This screams product rule! Remember the product rule? If $f(x) = u(x)v(x)$, then $f'(x) = u'(x)v(x) + u(x)v'(x)$. Let's set $u(x) = \sin(3x)$ and $v(x) = \cos(3x)$.

To find $u'(x)$, we need the chain rule. The derivative of $\sin(u)$ is $\cos(u)$, and the derivative of the inner function, $3x$, is just 3. So, $u'(x) = \cos(3x) \cdot 3 = 3\cos(3x)$.

Similarly, for $v(x) = \cos(3x)$, the derivative of $\cos(u)$ is $-\sin(u)$, and the derivative of $3x$ is 3. So, $v'(x) = -\sin(3x) \cdot 3 = -3\sin(3x)$.

Now, let's plug these into the product rule:

$f'(x) = (3\cos(3x))(\cos(3x)) + (\sin(3x))(-3\sin(3x))$

$f'(x) = 3\cos^2(3x) - 3\sin^2(3x)$

There we have it! The derivative of $\sin(3x)\cos(3x)$ is $3\cos^2(3x) - 3\sin^2(3x)$. But wait, there's a more elegant way to write this. Remember those powerful trigonometric identities? This expression looks suspiciously like something we can simplify using the double angle identity for cosine: $\cos(2\theta) = \cos^2(\theta) - \sin^2(\theta)$.

If we factor out a 3 from our derivative, we get $f'(x) = 3(\cos^2(3x) - \sin^2(3x))$. Now, let $\theta = 3x$. Then the expression inside the parentheses is exactly $\cos(2\theta)$, which is $\cos(2 \cdot 3x) = \cos(6x)$.

So, our simplified derivative is: $f'(x) = 3\cos(6x)$.

This is so much cleaner to work with! And now, our task is to find where $f'(x) > 0$ (for increasing intervals) and where $f'(x) < 0$ (for decreasing intervals). Let's go!

Simplifying the Function First: A Sneaky Trick

Before we go too far down the derivative road, sometimes it's a huge help to simplify the original function itself. For $f(x) = \sin(3x)\cos(3x)$, I'm seeing a pattern that reminds me of the double angle identity for sine: $\sin(2\theta) = 2\sin(\theta)\cos(\theta)$.

Our function looks almost like half of that identity. If we let $\theta = 3x$, then $2\sin(3x)\cos(3x) = \sin(2 \cdot 3x) = \sin(6x)$.

Our function is $\sin(3x)\cos(3x)$. If we multiply and divide by 2, we get:

$f(x) = \frac{1}{2} (2\sin(3x)\cos(3x))$

$f(x) = \frac{1}{2} \sin(6x)$

Boom! How awesome is that? Our original, seemingly complex function $\sin(3x)\cos(3x)$ is actually just a scaled and shifted version of the basic sine function, $\sin(6x)$. This simplification is going to make finding the derivative and analyzing its behavior so much easier. Let's re-calculate the derivative using this new form, just to confirm our previous work and to see how elegant it becomes.

Using the simplified form $f(x) = \frac{1}{2} \sin(6x)$:

To find $f'(x)$, we use the constant multiple rule and the chain rule. The derivative of $\sin(u)$ is $\cos(u)$, and the derivative of the inner function $6x$ is 6.

$f'(x) = \frac{1}{2} \cdot \frac{d}{dx}(\sin(6x))$

$f'(x) = \frac{1}{2} \cdot (\cos(6x) \cdot 6)$

$f'(x) = \frac{1}{2} \cdot 6 \cos(6x)$

$f'(x) = 3\cos(6x)$

See? We arrived at the exact same derivative, $3\cos(6x)$, but with significantly less effort and a much clearer path. This highlights the importance of looking for trigonometric identities to simplify expressions before diving into calculus. It's like taking a shortcut through a beautiful meadow instead of climbing a rocky mountain. So, from here on out, we'll be working with $f'(x) = 3\cos(6x)$ to find our intervals of increase and decrease. This simplified form makes the whole process much more manageable and less prone to silly errors. It’s a prime example of how a little bit of algebraic and trigonometric foresight can save you a ton of headache later on.

Finding Intervals of Increase and Decrease

Okay, team, we've got our simplified function $f(x) = \frac{1}{2} \sin(6x)$ and its derivative $f'(x) = 3\cos(6x)$. Now comes the crucial part: figuring out where $f(x)$ is going uphill (increasing) and where it's going downhill (decreasing). As we established, this depends on the sign of the derivative, $f'(x)$.

1. Intervals of Increase:

Our function $f(x)$ is increasing when $f'(x) > 0$. So, we need to solve the inequality:

$3\cos(6x) > 0$

This simplifies to:

$\\cos(6x) > 0$

We know that the cosine function is positive in the first and fourth quadrants. The general solutions for $\cos(\theta) > 0$ are $-\frac{\pi}{2} + 2n\pi < \theta < \frac{\pi}{2} + 2n\pi$, where $n$ is an integer.

In our case, $\theta = 6x$. So, we have:

$-\frac{\pi}{2} + 2n\pi < 6x < \frac{\pi}{2} + 2n\pi$

To find the intervals for $x$, we divide the entire inequality by 6:

$-\frac{\pi}{12} + \frac{2n\pi}{6} < x < \frac{\pi}{12} + \frac{2n\pi}{6}$

$-\frac{\pi}{12} + \frac{n\pi}{3} < x < \frac{\pi}{12} + \frac{n\pi}{3}$

These are the intervals where our function $f(x) = \sin(3x)\cos(3x)$ is increasing.

2. Intervals of Decrease:

Our function $f(x)$ is decreasing when $f'(x) < 0$. So, we need to solve the inequality:

$3\cos(6x) < 0$

This simplifies to:

$\\cos(6x) < 0$

We know that the cosine function is negative in the second and third quadrants. The general solutions for $\cos(\theta) < 0$ are $\frac{\pi}{2} + 2n\pi < \theta < \frac{3\pi}{2} + 2n\pi$, where $n$ is an integer.

Again, let $\theta = 6x$:

$\\frac{\pi}{2} + 2n\pi < 6x < \\frac{3\pi}{2} + 2n\pi$

Divide by 6 to solve for $x$:

$\\frac{\pi}{12} + \frac{2n\pi}{6} < x < \\frac{3\pi}{12} + \frac{2n\pi}{6}$

$\\frac{\pi}{12} + \frac{n\pi}{3} < x < \\frac{\pi}{4} + \frac{n\pi}{3}$

These are the intervals where our function $f(x) = \sin(3x)\cos(3x)$ is decreasing.

3. Critical Points:

Critical points occur where $f'(x) = 0$ or where $f'(x)$ is undefined. Since $f'(x) = 3\cos(6x)$ is defined for all $x$, we only need to find where $f'(x) = 0$.

$3\cos(6x) = 0$

$\\cos(6x) = 0$

The cosine function is zero at odd multiples of $\frac{\pi}{2}$. So:

$6x = \frac{\pi}{2} + n\pi$

Divide by 6:

$x = \frac{\pi}{12} + \frac{n\pi}{6}$

These critical points are the boundaries between the intervals of increase and decrease. For example, if we take $n=0$, we get $x = \frac{\pi}{12}$. If we take $n=1$, we get $x = \frac{\pi}{12} + \frac{\pi}{6} = \frac{\pi}{12} + \frac{2\pi}{12} = \frac{3\pi}{12} = \frac{\pi}{4}$. These are indeed the endpoints of our intervals.

So, to recap, for any integer $n$:

• Increasing Intervals: $\left(-\frac{\pi}{12} + \frac{n\pi}{3}, \frac{\pi}{12} + \frac{n\pi}{3}\right)$
• Decreasing Intervals: $\left(\frac{\pi}{12} + \frac{n\pi}{3}, \frac{\pi}{4} + \frac{n\pi}{3}\right)$

These intervals repeat infinitely, which is characteristic of periodic trigonometric functions. It's like a roller coaster that goes up and down, up and down, over and over again!

Visualizing the Behavior

Understanding these intervals is one thing, but seeing it can really make it click, right? Let's visualize what's happening with $f(x) = \frac{1}{2} \sin(6x)$. This function is a sine wave with an amplitude of $\frac{1}{2}$ and a period of $\frac{2\pi}{6} = \frac{\pi}{3}$.

When $f'(x) = 3\cos(6x)$ is positive, the graph of $f(x)$ is going upwards. This happens when $\cos(6x)$ is positive, which is on intervals like $\left(-\frac{\pi}{12}, \frac{\pi}{12}\right)$, $\left(\frac{3\pi}{12}, \frac{5\pi}{12}\right)$, etc. (using $n=0, 1$ in our general formulas).

When $f'(x) = 3\cos(6x)$ is negative, the graph of $f(x)$ is going downwards. This happens when $\cos(6x)$ is negative, which is on intervals like \left(rac{\pi}{12}, \frac{\pi}{4}\right), $\left(\frac{5\pi}{12}, \frac{7\pi}{12}\right)$, etc.

Notice how the critical points $x = \frac{\pi}{12} + \frac{n\pi}{6}$ correspond to the peaks and valleys (local maxima and minima) of the sine wave. At these points, the derivative is zero, meaning the tangent line to the curve is horizontal. These are the turning points where the function transitions from increasing to decreasing, or vice versa.

If you were to plot $y = \frac{1}{2} \sin(6x)$, you'd see a smooth, wave-like pattern. Between x = -rac{\pi}{12} and x = rac{\pi}{12}, the graph rises. Between x = rac{\pi}{12} and x = rac{\pi}{4}, it falls. This pattern repeats every $\frac{\pi}{3}$ units along the x-axis, which is the period of our function. This visual confirmation is super satisfying and helps solidify the calculus concepts. It’s like seeing the map and then actually walking the path – the two confirm each other beautifully.

Conclusion

So there you have it, folks! We've successfully navigated the process of determining the increasing and decreasing intervals for the function $\sin(3x)\cos(3x)$. By leveraging the power of trigonometric identities to simplify the function to $f(x) = \frac{1}{2} \sin(6x)$, we made finding its derivative, $f'(x) = 3\cos(6x)$, a breeze. Analyzing the sign of this derivative revealed that:

• The function is increasing on the intervals $\left(-\frac{\pi}{12} + \frac{n\pi}{3}, \frac{\pi}{12} + \frac{n\pi}{3}\right)$ for any integer $n$.
• The function is decreasing on the intervals $\left(\frac{\pi}{12} + \frac{n\pi}{3}, \frac{\pi}{4} + \frac{n\pi}{3}\right)$ for any integer $n$.

The critical points, where $f'(x)=0$, are $x = \frac{\pi}{12} + \frac{n\pi}{6}$, marking the turning points of the function's graph.

This whole process underscores how crucial it is to recognize and apply trigonometric identities. They can transform complex problems into manageable ones. Remember, understanding when a function increases or decreases is fundamental in calculus, helping us sketch graphs, find maximum and minimum values, and analyze behavior. Keep practicing these techniques, and you'll become a calculus whiz in no time! Stay curious, keep exploring, and I'll catch you in the next one!