Cosine Function Equation: Amplitude, Period & Shift

Dec 11, 2025 by Admin 52 views

Hey everyone, let's dive into the nitty-gritty of cosine functions today, specifically how to nail down their general equation when you've got the amplitude, period, and horizontal shift all figured out. You know, understanding these components is super crucial for graphing and analyzing these wave-like functions. We're going to break down a specific problem and figure out the correct equation, so stick around!

Decoding the Cosine Function's DNA

Alright, guys, let's get down to basics. The general equation for a cosine function looks something like this: $y = A abla extbf{cos} (B(x - C)) + D$ . Now, each of these letters represents a key characteristic of the cosine wave. Let's break 'em down:

A (Amplitude): This bad boy tells you how high and low the wave goes from its center line. Think of it as the 'height' of the wave. If A is 3, the wave goes up 3 units and down 3 units from the midline. It's all about the vertical stretch or compression.
B (Affects Period): This one's a bit trickier. 'B' directly influences the period of the function, which is the horizontal length of one complete cycle of the wave. The relationship is: Period = $2 abla oldsymbol{\pi} / B$ . So, if you know the period, you can easily find B by rearranging the formula: $B = 2 abla oldsymbol{\pi} / ext{Period}$ . A smaller B means a longer, stretched-out wave, and a larger B means a shorter, more squished wave.
C (Horizontal Shift/Phase Shift): This is your 'side-to-side' move. 'C' tells you how much the entire graph is shifted to the left or right. If it's $(x - C)$ , the shift is to the right by 'C' units. If it's $(x + C)$ (which is the same as $(x - (-C))$ ), the shift is to the left by 'C' units. It's all about where the starting point of your standard cosine wave (which usually begins at its peak) gets moved.
D (Vertical Shift): This one's pretty straightforward. 'D' shifts the entire graph up or down. If D is positive, the graph moves up; if D is negative, it moves down. This effectively changes the midline of the wave from the x-axis ( $y=0$ ) to $y=D$ .

Now, for the problem at hand, we're given specific values for amplitude, period, and horizontal shift. We need to plug these into the general equation and see which of the options fits. Remember, we're only dealing with A, B, and C in this particular scenario, as there's no mention of a vertical shift (so D will be 0).

Putting the Pieces Together: Amplitude = 3

The amplitude is the easiest to spot, guys. It's the coefficient right out front of the cosine function. In our problem, the amplitude is 3. This means that our 'A' value is 3. So, our equation will start with $y = 3 abla extbf{cos} (...)$ . This immediately tells us that any options not starting with a coefficient of 3 are likely incorrect. Let's keep that in mind as we move forward. The amplitude dictates the wave's height from its midline. A larger amplitude means a more pronounced wave, while a smaller one signifies a more subdued fluctuation. For instance, a cosine function with an amplitude of 10 would oscillate much more dramatically than one with an amplitude of 0.5. It's the factor that scales the vertical extent of the function, ensuring that the peak value is A and the trough value is -A, relative to the vertical shift.

Calculating the Period and Finding 'B'

Next up, we've got the period of $4 abla oldsymbol{\pi}$ . Remember our formula relating B and the period? Period = $2 abla oldsymbol{\pi} / B$ . We need to solve for B. So, we rearrange it to $B = 2 abla oldsymbol{\pi} / ext{Period}$ .

Plugging in our given period:

$B = 2 abla oldsymbol{\pi} / (4 abla oldsymbol{\pi})$

$B = 2/4$

$B = 0.5$

So, our 'B' value is 0.5. This means the part inside our cosine function will involve $0.5(x - C)$ . A B-value of 0.5 tells us that the cycle is stretched out compared to a standard cosine function (where $B=1$ and the period is $2 abla oldsymbol{\pi}$ ). In fact, our period is twice as long as the standard period because B is half of 1. This means it takes twice as long for the function to complete one full oscillation. This stretching is fundamental to how trigonometric functions are manipulated to model real-world phenomena that have longer cycles, such as seasonal changes or longer-term economic trends. The value of B directly dictates this horizontal scaling. If B were larger, say 2, the period would be $2 abla oldsymbol{\pi} / 2 = abla oldsymbol{\pi}$ , meaning the cycle would be compressed, happening twice as fast as the standard function.

Handling the Horizontal Shift ('C')

Finally, we have a horizontal shift of $- abla oldsymbol{\pi}$ . Remember, the general form is $(x - C)$ . If our shift is $- abla oldsymbol{\pi}$ , it means that $C = - abla oldsymbol{\pi}$ to achieve a shift of $- abla oldsymbol{\pi}$ (because it's $x - (- abla oldsymbol{\pi})$ which equals $x + abla oldsymbol{\pi}$ ).

So, the term inside the cosine function will be $(x - (- abla oldsymbol{\pi}))$ , which simplifies to $(x + abla oldsymbol{\pi})$ . This means the graph of our cosine function is shifted $abla oldsymbol{\pi}$ units to the left. The standard cosine graph starts its cycle at a peak when $x=0$ . With a horizontal shift of $- abla oldsymbol{\pi}$ , the peak will now occur at $x = - abla oldsymbol{\pi}$ . This shift is critical for aligning the function's behavior with observed data that doesn't start its cycle at the origin. For example, if you're modeling tidal patterns, the high tide might not occur at the start of your observation period, necessitating a horizontal shift to match the real-world timing. A positive horizontal shift (e.g., $x-C$ where C is positive) moves the graph to the right, meaning the standard starting point of the cycle shifts to a positive x-value. Conversely, a negative shift (like the one in our problem) moves it to the left, aligning the function's cyclical behavior with earlier time points or initial conditions.

Assembling the Final Equation

Now, let's put all our pieces together. We have:

$A = 3$
$B = 0.5$
$C = - abla oldsymbol{\pi}$ (which gives us $(x + abla oldsymbol{\pi})$ inside the parenthesis)

Plugging these into our general equation $y = A abla extbf{cos} (B(x - C)) + D$ (and knowing D=0):

$y = 3 abla extbf{cos} (0.5(x - (- abla oldsymbol{\pi})))$

$y = 3 abla extbf{cos} (0.5(x + abla oldsymbol{\pi}))$

And there you have it! This is the equation that perfectly matches the given amplitude, period, and horizontal shift.

Checking the Options

Let's look at the options provided and see which one matches our derived equation:

A. $y=4 abla oldsymbol{\pi} abla extbf{cos} (3(x- abla oldsymbol{\pi}))$

B. $y=3 abla extbf{cos} (4 abla oldsymbol{\pi}(x+ abla oldsymbol{\pi}))$

C. $y=3 abla extbf{cos} (0.5(x+ abla oldsymbol{\pi}))$

D. $y=4 abla extbf{cos} (n .2(x+ abla oldsymbol{\pi}))$

Comparing our equation, $y = 3 abla extbf{cos} (0.5(x + abla oldsymbol{\pi}))$ , with the options, it's clear that Option C is the correct one. It has the amplitude of 3, the B value of 0.5 (which gives the correct period), and the correct horizontal shift leading to $(x+ abla oldsymbol{\pi})$ .

Let's quickly see why the others are wrong:

Option A: The amplitude is $4 abla oldsymbol{\pi}$ and B is 3. This doesn't match our given values.
Option B: The amplitude is 3, which is good, but B is $4 abla oldsymbol{\pi}$ . If B is $4 abla oldsymbol{\pi}$ , the period would be $2 abla oldsymbol{\pi} / (4 abla oldsymbol{\pi}) = 0.5$ . This is not the period of $4 abla oldsymbol{\pi}$ we were given.
Option D: This option seems to have a typo with 'n.2' and the amplitude is 4. It's definitely not correct based on our calculations.

So, it's solid proof that Option C is the winner! Mastering these components – amplitude, period, and horizontal shift – allows you to accurately represent and visualize cyclical phenomena using cosine functions. Keep practicing, guys, and you'll be an equation-building pro in no time!