Decoding Phase Shifts Find The Correct Answer

Hey guys! Let's dive into the fascinating world of trigonometric functions and their transformations, specifically focusing on phase shifts. Ever wondered how a sine wave can slide left or right? That's the magic of phase shifts! We're going to break down the concept of phase shift, explore how it manifests in trigonometric equations, and, most importantly, nail down the correct answer to the question at hand: "Which function has a phase shift of $\frac{\pi}{2}$ to the right?"

Understanding Phase Shift

Let's get this straight: Phase shift is a horizontal transformation applied to a trigonometric function. Imagine the standard sine or cosine wave; the phase shift is what happens when you slide it along the x-axis. This shift is determined by a constant added or subtracted inside the trigonometric function's argument (the part inside the parentheses). It's super important to remember that the direction of the shift is opposite the sign you see in the equation. A negative sign inside the argument means a shift to the right, and a positive sign means a shift to the left. Let's visualize this to make it even clearer. Think of a sine wave starting at the origin. If we want to shift it to the right by $\frac{\pi}{2}$ units, we need to adjust the equation. The general form of a sine function with a phase shift is given by $y = A \sin(B(x - C))$, where C represents the phase shift. If C is positive, the shift is to the right; if C is negative, the shift is to the left. A represents the amplitude, which affects the vertical stretch, and B affects the period, which is the horizontal compression or stretch. In our case, we are only concerned with the phase shift, so we need to isolate the term that directly influences the horizontal shift. To identify the phase shift, we need to rewrite the given equations in the form $y = A \sin(B(x - C))$. This involves factoring out the coefficient of x inside the argument, which will reveal the true value and direction of the phase shift. It is also crucial to understand the relationship between the phase shift and the period of the function. The period of a standard sine function $y = \sin(x)$ is $2\pi$. When we have a function like $y = \sin(Bx)$, the period becomes $\frac{2\pi}{B}$. The phase shift tells us how much the function is shifted horizontally relative to its standard position. Grasping these fundamentals is key to correctly identifying phase shifts in various trigonometric functions. This knowledge will not only help us solve this specific question but also equip us with a solid understanding of trigonometric transformations in general. So, let's keep these concepts in mind as we analyze the given options and pinpoint the one that exhibits a phase shift of $\frac{\pi}{2}$ to the right.

Analyzing the Options

Alright, let's roll up our sleeves and carefully analyze each option to pinpoint the function with a phase shift of $\frac{\pi}{2}$ to the right. Remember, we're looking for an equation in the form $y = A \sin(B(x - C))$, where C should be equal to $\frac{\pi}{2}$.

Option A: $y=2 \sin \left(x+\frac{\pi}{2}\right)$

In option A, we have $y=2 \sin \left(x+\frac{\pi}{2}\right)$. Here, we can see that $B=1$ and the argument is in the form $(x + \frac{\pi}{2})$. This means that $C = -\frac{\pi}{2}$. Remember, a positive sign inside the parentheses indicates a shift to the left. So, this function has a phase shift of $\frac{\pi}{2}$ to the left, not the right. Therefore, option A is not the correct answer. It's a classic example of how the sign can trick you if you don't remember the rule about opposite directions. The graph of this function would be the standard sine wave shifted $\frac{\pi}{2}$ units to the left, which means it starts its cycle at $x = -\frac{\pi}{2}$ instead of $x = 0$. This shift can also be visualized by thinking about the unit circle and how the sine function relates to the y-coordinate of a point moving around the circle. Shifting to the left corresponds to starting the cycle earlier in terms of the angle, which is why we see the addition of $\frac{\pi}{2}$ inside the argument. But, as we established, we're looking for a shift to the right, so let's move on to the next option.

Option B: $y=2 \sin (2 x-\pi)$

Moving on to option B, we have $y=2 \sin (2 x-\pi)$. This looks a bit trickier, but don't worry, we can handle it! The key here is to factor out the coefficient of x from the argument. We need to rewrite $(2x - \pi)$ in the form $B(x - C)$. Factoring out the 2, we get $2(x - \frac{\pi}{2})$. Now the equation becomes $y = 2 \sin (2(x - \frac{\pi}{2}))$. Ah-ha! We see that C is indeed $\frac{\pi}{2}$. And since it's a negative sign inside the parentheses, this indicates a shift to the right. So, this function has a phase shift of $\frac{\pi}{2}$ to the right. Could this be our answer? It's looking promising! The factor of 2 in front of the x also affects the period of the function. Recall that the period of $y = \sin(Bx)$ is $\frac{2\pi}{B}$. In this case, the period is $\frac{2\pi}{2} = \pi$. This means the function completes one full cycle in an interval of length $\pi$, which is half the period of the standard sine function. So, option B not only has the correct phase shift but also a different period compared to the standard sine wave. This is an important observation because it highlights how multiple transformations can affect the graph of a trigonometric function. But, to be absolutely sure, let's examine the remaining options before we declare option B the winner.

Option C: $y=2 \sin (x-\pi)$

Let's dissect option C: $y=2 \sin (x-\pi)$. In this case, the argument is simply $(x - \pi)$. This directly tells us that the phase shift is$\pi$ to the right because $C = \pi$. While it's a shift to the right, it's not the $\frac{\pi}{2}$ shift we're looking for. So, option C is incorrect. It's important to note the difference in magnitude between $\pi$ and $\frac{\pi}{2}$. A phase shift of $\pi$ means the graph is shifted horizontally by a much larger amount compared to a shift of $\frac{\pi}{2}$. This can significantly alter the appearance of the sine wave, especially when considering its relationship to the period. The graph of this function would start its cycle at $x = \pi$, which is a full half-period away from the standard starting point at $x = 0$. This visual understanding of the magnitude of the phase shift is crucial in quickly assessing the behavior of trigonometric functions. But since we're after a specific shift of $\frac{\pi}{2}$, option C doesn't fit the bill. We're getting closer, though! Let's check the last option to be completely thorough.

Option D: $y=2 \sin \left(\frac{1}{2} x+\pi\right)$

Last but not least, we have option D: $y=2 \sin \left(\frac{1}{2} x+\pi\right)$. Just like in option B, we need to factor out the coefficient of x, which in this case is $\frac{1}{2}$. Factoring out $\frac{1}{2}$ from the argument, we get $\frac{1}{2}(x + 2\pi)$. So, the equation becomes $y = 2 \sin \left(\frac{1}{2}(x + 2\pi)\right)$. Here, we see that $C = -2\pi$. This indicates a phase shift of $2\pi$ to the left, not the right. Therefore, option D is also incorrect. The factor of $\frac{1}{2}$ in front of the x also affects the period of this function. The period is $\frac{2\pi}{\frac{1}{2}} = 4\pi$. This means the function completes one full cycle in an interval of length $4\pi$, which is twice the period of the standard sine function. The combination of the phase shift and the period change makes this function significantly different from the standard sine wave. But, as we've confirmed, the phase shift is to the left, not the right, so option D is not the answer we're looking for. We've now thoroughly analyzed all the options, and the correct answer is becoming clear.

The Decisive Answer

After carefully analyzing all the options, we've pinpointed the function that exhibits a phase shift of $\frac{\pi}{2}$ to the right. Remember, we were on the lookout for an equation in the form $y = A \sin(B(x - C))$, where C equals $\frac{\pi}{2}$.

Let's recap our findings:

• Option A: $y=2 \sin \left(x+\frac{\pi}{2}\right)$ – Phase shift of $\frac{\pi}{2}$ to the left. Incorrect.
• Option B: $y=2 \sin (2 x-\pi)$ – Factoring out the 2, we get $y = 2 \sin (2(x - \frac{\pi}{2}))$. Phase shift of $\frac{\pi}{2}$ to the right. A strong contender!
• Option C: $y=2 \sin (x-\pi)$ – Phase shift of $\pi$ to the right. Incorrect.
• Option D: $y=2 \sin \left(\frac{1}{2} x+\pi\right)$ – Factoring out the $\frac{1}{2}$, we get $y = 2 \sin \left(\frac{1}{2}(x + 2\pi)\right)$. Phase shift of $2\pi$ to the left. Incorrect.

It's crystal clear now: Option B, $y=2 \sin (2 x-\pi)$, is the correct answer! By factoring out the 2, we revealed the phase shift of $\frac{\pi}{2}$ to the right, precisely what we were searching for.

Final Answer:

The function with a phase shift of $\frac{\pi}{2}$ to the right is:

B. $y=2 \sin (2 x-\pi)$

Key Takeaways

So, what have we learned on this thrilling adventure through phase shifts? Let's jot down some key takeaways to solidify our understanding.

• Phase shift is a horizontal transformation of a trigonometric function, shifting the graph left or right.
• The general form $y = A \sin(B(x - C))$ is your best friend. C determines the phase shift. Positive C means a shift to the right, negative C means a shift to the left.
• Always factor out the coefficient of x inside the argument to correctly identify the phase shift.
• Don't be fooled by the sign! Remember, the shift direction is opposite the sign inside the parentheses.
• Understanding the period of the function, $\frac{2\pi}{B}$, helps in visualizing the overall transformation.

By keeping these points in mind, you'll be able to confidently tackle any phase shift problem that comes your way. Keep practicing, and you'll become a phase shift pro in no time!

Practice Problems

To really master this concept, let's tackle a few practice problems. Guys, working through these will help solidify your understanding and build your confidence.

1. Which function has a phase shift of $\frac{\pi}{4}$ to the left?
   • A. $y = \sin(x - \frac{\pi}{4})$
   • B. $y = \sin(x + \frac{\pi}{4})$
   • C. $y = \sin(2x - \frac{\pi}{2})$
   • D. $y = \sin(\frac{1}{2}x + \frac{\pi}{8})$
2. What is the phase shift of the function $y = 3\cos(2x + \pi)$?
3. Write an equation for a sine function with an amplitude of 2, a period of $\pi$, and a phase shift of $\frac{\pi}{3}$ to the right.

Work through these problems, and if you get stuck, revisit the concepts we've discussed. Remember, practice makes perfect!

Conclusion

We've successfully navigated the intriguing world of phase shifts, demystifying how trigonometric functions transform horizontally. By understanding the general form, the importance of factoring, and the sign convention, you're now well-equipped to identify and analyze phase shifts in various trigonometric equations. Keep practicing, keep exploring, and remember that math, like any adventure, is best enjoyed with a curious mind and a willingness to learn! You got this!