Sine And Cosine: Analyzing Sin(3x)cos(3x) Behavior

Hey everyone! Today, we're diving deep into the fascinating world of trigonometric functions, specifically looking at the behavior of sin(3x)cos(3x). If you've ever wondered how to figure out when a function is going up (increasing) or going down (decreasing), you're in the right place, guys. We're going to break down how to analyze sin(3x)cos(3x) and pinpoint those crucial intervals. This isn't just about memorizing formulas; it's about understanding the why behind the math, and trust me, it's way cooler than it sounds.

To really get a grip on when sin(3x)cos(3x) is increasing or decreasing, we need to leverage a super powerful tool in calculus: the first derivative. Think of the derivative as the instantaneous rate of change of a function. If the derivative is positive, the original function is increasing. If the derivative is negative, the original function is decreasing. Pretty straightforward, right? So, our main mission is to find the derivative of sin(3x)cos(3x), set it up, and analyze its sign.

Before we jump into the calculus, there's a neat little trigonometric identity that can seriously simplify our lives. Remember the double-angle formula for sine? It states that sin(2θ) = 2sin(θ)cos(θ). If we rearrange this, we get sin(θ)cos(θ) = (1/2)sin(2θ). Now, let's look at our function, f(x) = sin(3x)cos(3x). If we let θ = 3x, then our function becomes f(x) = (1/2)sin(2 * 3x) = (1/2)sin(6x). See how much cleaner that is? Working with (1/2)sin(6x) is going to be a breeze compared to the product rule we'd have to use on the original form. This simplification is a classic trick in trig calculus, and it's worth keeping in your mathematical toolkit.

So, our simplified function is f(x) = (1/2)sin(6x). Now, let's find its derivative, f'(x). Using the chain rule, the derivative of sin(u) is cos(u) * u', and the derivative of a constant times a function is just the constant times the derivative of the function. Here, u = 6x, so u' = 6. The derivative of (1/2)sin(6x) is therefore f'(x) = (1/2) * cos(6x) * 6. Simplifying this gives us f'(x) = 3cos(6x). This is the expression we'll use to determine the intervals where our original function, sin(3x)cos(3x), is increasing or decreasing.

Finding Critical Points: Where the Action Happens

Alright guys, now that we have our derivative, f'(x) = 3cos(6x), we need to find the critical points. Critical points are the x-values where the derivative is either zero or undefined. For 3cos(6x), it's never undefined, so we only need to worry about where it equals zero. Setting f'(x) = 0, we get 3cos(6x) = 0. Dividing by 3, we have cos(6x) = 0. Remember our unit circle? The cosine function is zero at angles of π/2, 3π/2, 5π/2, and so on. In general, cos(θ) = 0 when θ = (π/2) + nπ, where 'n' is any integer (..., -2, -1, 0, 1, 2, ...).

In our case, the angle is 6x. So, we set 6x = (π/2) + nπ. To find x, we divide everything by 6: x = (π/12) + (nπ/6). These are our critical points! They are the exact spots where the function might change from increasing to decreasing, or vice versa. These points divide the number line into intervals, and within each interval, the derivative will have a constant sign (either always positive or always negative). This is why finding these critical points is so important – they're the boundaries of our increasing and decreasing intervals.

Let's list a few of these critical points to get a feel for them. For n=0, x = π/12. For n=1, x = π/12 + π/6 = π/12 + 2π/12 = 3π/12 = π/4. For n=2, x = π/12 + 2π/6 = π/12 + 4π/12 = 5π/12. For n=3, x = π/12 + 3π/6 = π/12 + 6π/12 = 7π/12. For n=4, x = π/12 + 4π/6 = π/12 + 8π/12 = 9π/12 = 3π/4. And so on. We also need to consider negative integers for 'n' to cover the entire number line, like: For n=-1, x = π/12 - π/6 = π/12 - 2π/12 = -π/12. For n=-2, x = π/12 - 2π/6 = π/12 - 4π/12 = -3π/12 = -π/4.

These points are crucial because they are the potential turning points of our function sin(3x)cos(3x). Between these points, the slope of the tangent line to the graph of the function is constant. It's either positive (meaning the function is climbing uphill) or negative (meaning the function is sliding downhill).

Analyzing the Sign of the Derivative: Where Does it Go Up and Down?

Now for the fun part, guys! We need to pick test values within the intervals created by our critical points and plug them into our derivative, f'(x) = 3cos(6x), to see if the result is positive or negative. Let's consider the interval starting from -π/12 and going up to π/12. We can pick a test value like x = 0 (which lies between -π/12 and π/12). Plugging x=0 into f'(x), we get f'(0) = 3cos(60) = 3cos(0) = 31 = 3. Since 3 is positive, our function sin(3x)cos(3x) is increasing on the interval (-π/12, π/12).

Let's move to the next interval, from π/12 to 3π/12 (or π/4). A good test value here could be x = 2π/12 = π/6. Plugging this into the derivative: f'(π/6) = 3cos(6 * π/6) = 3cos(π) = 3(-1) = -3*. Since -3 is negative, our function sin(3x)cos(3x) is decreasing on the interval (π/12, 3π/12).

Let's keep going! The next interval is from 3π/12 (π/4) to 5π/12. Let's pick x = 4π/12 = π/3. Plugging this into the derivative: f'(π/3) = 3cos(6 * π/3) = 3cos(2π) = 3*1 = 3. Since 3 is positive, our function sin(3x)cos(3x) is increasing on the interval (3π/12, 5π/12).

We can see a pattern emerging here. The sign of cos(6x) alternates between positive and negative as we move through the intervals defined by our critical points. This is because the cosine function itself oscillates between 1 and -1. The general solutions for cos(6x) = 0 are 6x = π/2 + nπ, which means x = π/12 + nπ/6. These points represent the peaks and troughs of the cosine wave (and hence, the turning points of our original function).

To summarize the sign analysis for the derivative f'(x) = 3cos(6x):

• Interval (-π/12, π/12): Test x=0. f'(0) = 3cos(0) = 3 (Positive). f(x) is Increasing.
• Interval (π/12, 3π/12): Test x=π/6. f'(π/6) = 3cos(π) = -3 (Negative). f(x) is Decreasing.
• Interval (3π/12, 5π/12): Test x=π/3. f'(π/3) = 3cos(2π) = 3 (Positive). f(x) is Increasing.
• Interval (5π/12, 7π/12): Test x=π/2. f'(π/2) = 3cos(3π) = -3 (Negative). f(x) is Decreasing.

This pattern will continue indefinitely, repeating every π/6 interval because the period of cos(6x) is 2π/6 = π/3, and the critical points are spaced π/6 apart. The function sin(3x)cos(3x), which simplifies to (1/2)sin(6x), is essentially a sine wave with a period of 2π/6 = π/3. The peaks and valleys of this sine wave correspond to where the function changes from increasing to decreasing and vice versa.

Putting It All Together: The Final Intervals

So, to wrap it all up, guys, we've found that our function f(x) = sin(3x)cos(3x), which we cleverly simplified to f(x) = (1/2)sin(6x), has a derivative f'(x) = 3cos(6x). The critical points where f'(x) = 0 occur at x = π/12 + nπ/6, where 'n' is any integer.

Based on our sign analysis of the derivative, we can definitively state the intervals where sin(3x)cos(3x) is increasing and decreasing. Remember, the function is increasing when its derivative is positive, and decreasing when its derivative is negative.

The function f(x) = sin(3x)cos(3x) is INCREASING on the intervals:

• (π/12 + nπ/6, 3π/12 + nπ/6) for all integers 'n'.

This might look a little complex, so let's break it down. If we consider 'n=0', the interval is (π/12, 3π/12). Wait, I made a mistake in the previous section! Let's re-evaluate the intervals based on the critical points. The critical points are ..., -π/12, π/12, 3π/12, 5π/12, 7π/12, ....

Let's redo the sign analysis more carefully:

• Interval (-π/12, π/12): Test x=0. f'(0) = 3cos(0) = 3 (Positive). f(x) is Increasing.
• Interval (π/12, 3π/12): Test x=π/6. f'(π/6) = 3cos(π) = -3 (Negative). f(x) is Decreasing.
• Interval (3π/12, 5π/12): Test x=π/3. f'(π/3) = 3cos(2π) = 3 (Positive). f(x) is Increasing.
• Interval (5π/12, 7π/12): Test x=π/2. f'(π/2) = 3cos(3π) = -3 (Negative). f(x) is Decreasing.

My apologies, guys! It seems I got the intervals mixed up in the previous section. Let's correct this.

The function f(x) = sin(3x)cos(3x) is INCREASING when f'(x) = 3cos(6x) > 0. This occurs when cos(6x) is positive. The cosine function is positive in the first and fourth quadrants. So, 2kπ < 6x < π/2 + 2kπ and 3π/2 + 2kπ < 6x < 2π + 2kπ for any integer 'k'.

Dividing by 6, we get:

• kπ/3 < x < π/12 + kπ/3
• π/4 + kπ/3 < x < π/3 + kπ/3

Let's check if this aligns with our test points. For k=0:

• 0 < x < π/12 (Matches our first interval)
• π/4 < x < π/3 (This corresponds to the interval (3π/12, 4π/12) which is part of (3π/12, 5π/12). Let's check x=π/3 in the derivative: f'(π/3) = 3cos(2π) = 3 > 0. Yes, it is increasing here).

The function f(x) = sin(3x)cos(3x) is DECREASING when f'(x) = 3cos(6x) < 0. This occurs when cos(6x) is negative. The cosine function is negative in the second and third quadrants. So, π/2 + 2kπ < 6x < 3π/2 + 2kπ for any integer 'k'.

Dividing by 6, we get:

• π/12 + kπ/3 < x < π/4 + kπ/3

Let's check this for k=0:

• π/12 < x < π/4 (This corresponds to our interval (π/12, 3π/12), where we found it decreasing).

So, to present this clearly, let's use the general form of the critical points x = π/12 + nπ/6. These points mark the transitions.

The function f(x) = sin(3x)cos(3x) is INCREASING on the intervals:

• (-π/12 + nπ/3, π/12 + nπ/3) for all integers 'n'.

The function f(x) = sin(3x)cos(3x) is DECREASING on the intervals:

• (π/12 + nπ/3, 5π/12 + nπ/3) for all integers 'n'.

Let's verify with a few values of 'n':

For n=0:

• Increasing: (-π/12, π/12)
• Decreasing: (π/12, 5π/12)

For n=1:

• Increasing: (-π/12 + π/3, π/12 + π/3) = (3π/12, 5π/12). Oops, this contradicts my previous finding for (3π/12, 5π/12). Let's re-trace.

The critical points are x = π/12 + nπ/6. These are the points where the derivative is ZERO. The intervals are between these points.

Let's use a number line approach with the critical points: ..., -3π/12, -π/12, π/12, 3π/12, 5π/12, 7π/12, ...

Test points:

• Interval (-3π/12, -π/12): Let's use x = -2π/12 = -π/6. f'(-π/6) = 3cos(6*(-π/6)) = 3cos(-π) = 3*(-1) = -3. Decreasing.
• Interval (-π/12, π/12): Let's use x = 0. f'(0) = 3cos(0) = 3*1 = 3. Increasing.
• Interval (π/12, 3π/12): Let's use x = 2π/12 = π/6. f'(π/6) = 3cos(6*(π/6)) = 3cos(π) = 3*(-1) = -3. Decreasing.
• Interval (3π/12, 5π/12): Let's use x = 4π/12 = π/3. f'(π/3) = 3cos(6*(π/3)) = 3cos(2π) = 3*1 = 3. Increasing.

Okay, THIS pattern is consistent! My apologies for the confusion, guys. It's easy to get tangled up with trig intervals. The key is to be methodical.

So, the function f(x) = sin(3x)cos(3x) is INCREASING on the intervals (π/12 + nπ/6, 3π/12 + nπ/6) for all integers 'n'.

And the function f(x) = sin(3x)cos(3x) is DECREASING on the intervals (-π/12 + nπ/6, π/12 + nπ/6) for all integers 'n'.

Let's write out a few of these intervals for clarity:

Increasing Intervals:

• ..., (-π/12, π/12), (3π/12, 5π/12), (7π/12, 9π/12), ...
• Simplifying the fractions: ..., (-π/12, π/12), (π/4, 5π/12), (7π/12, 3π/4), ...

Decreasing Intervals:

• ..., (-3π/12, -π/12), (π/12, 3π/12), (5π/12, 7π/12), ...
• Simplifying the fractions: ..., (-π/4, -π/12), (π/12, π/4), (5π/12, 7π/12), ...

Understanding these intervals helps us visualize the graph of sin(3x)cos(3x). It's a wave that goes up, then down, then up again, and it does this repeatedly. The specific transformation sin(3x)cos(3x), which becomes (1/2)sin(6x), compresses the standard sine wave horizontally. The '6' inside the sine function means it completes 6 cycles in the usual 2π interval, making its period 2π/6 = π/3. The critical points we found are exactly where this oscillating function reaches its local maximums (where it changes from increasing to decreasing) and local minimums (where it changes from decreasing to increasing).

By mastering the use of the first derivative and trigonometric identities, you can confidently analyze the behavior of complex functions like sin(3x)cos(3x). Keep practicing, and these concepts will become second nature! Happy calculating, calculating, everyone!