Function Equivalent To Y = -cot(x) Explained

Finding the equivalent function to $y = -\cot(x)$ involves understanding the relationships between trigonometric functions, particularly cotangent and tangent. This article delves into the properties of these functions, explores how they transform, and provides a step-by-step explanation to identify the correct equivalent function. Understanding these relationships is crucial for solving trigonometric equations and simplifying expressions.

Understanding Cotangent and Tangent

To identify which function is equivalent to $y = -\cot(x)$, let's first define the cotangent and tangent functions and explore their relationship. Cotangent, denoted as $\cot(x)$, is the reciprocal of the tangent function, $\tan(x)$. Mathematically, this can be expressed as:

$\cot(x) = \frac{1}{\tan(x)}$

The tangent function, on the other hand, is defined as the ratio of the sine function to the cosine function:

$\tan(x) = \frac{\sin(x)}{\cos(x)}$

Therefore, cotangent can also be written as the ratio of cosine to sine:

$\cot(x) = \frac{\cos(x)}{\sin(x)}$

These fundamental relationships are key to understanding how transformations affect these functions. The graphs of tangent and cotangent have vertical asymptotes, which occur where the denominator of their respective ratios is zero. For tangent, this happens when $\cos(x) = 0$, which occurs at $x = \frac{(2n+1)\pi}{2}$, where n is an integer. For cotangent, asymptotes occur where $\sin(x) = 0$, which happens at $x = n\pi$, where n is an integer. Furthermore, understanding the periodicity of tangent and cotangent is essential. Both functions have a period of $\pi$, meaning their values repeat every $\pi$ units.

Considering the negative sign in $y = -\cot(x)$, we need to analyze how this affects the function's behavior. Multiplying a function by -1 reflects it across the x-axis. Therefore, the graph of $y = -\cot(x)$ is a reflection of the graph of $y = \cot(x)$ across the x-axis. This transformation changes the intervals where the function is positive and negative but does not affect the asymptotes themselves.

The relationship between tangent and cotangent can be further explored through identities involving phase shifts. These identities are crucial for rewriting one trigonometric function in terms of another. Specifically, we will investigate identities that relate $\cot(x)$ to $\tan(x)$ with horizontal shifts, which will help us match the given expression to one of the provided options. Such transformations involve adding or subtracting constants within the argument of the trigonometric function, effectively shifting the graph horizontally. Understanding how these shifts affect the graph is vital for determining equivalent functions.

Analyzing the Options

Now, let's evaluate the given options to determine which one is equivalent to $y = -\cot(x)$. This involves comparing the properties and transformations of each function with the original function.

Option A: $y = -\tan(x)$

Option A presents the negative tangent function, $y = -\tan(x)$. While this function involves tangent, it's not immediately clear if it's equivalent to $y = -\cot(x)$. To verify, we need to understand how the negative sign affects the tangent function. The negative sign reflects the graph of $y = \tan(x)$ across the x-axis. However, the fundamental difference between tangent and cotangent, as reciprocals, means that simply negating the tangent function does not directly yield the negative cotangent function. The asymptotes and zeros of $- \tan(x)$ will be different from those of $- \cot(x)$. Therefore, this option is unlikely to be correct but serves as a useful point of comparison.

Option B: $y = -\tan\left(x+\frac{\pi}{2}\right)$

Option B, $y = -\tan\left(x+\frac{\pi}{2}\right)$, introduces a horizontal shift to the tangent function. The term $\left(x+\frac{\pi}{2}\right)$ inside the tangent function shifts the graph horizontally by $\frac{\pi}{2}$ units to the left. This transformation is critical because it directly relates tangent to cotangent. Recall the cofunction identity, which states that $\tan\left(x + \frac{\pi}{2}\right) = -\cot(x)$. Therefore, the given function becomes $y = -\tan\left(x+\frac{\pi}{2}\right) = -(-\cot(x)) = \cot(x)$. However, this is not equivalent to $y = -\cot(x)$. There is an extra negative sign that makes the equation contradictory.

Option C: $y = \tan(x)$

Option C, $y = \tan(x)$, is simply the tangent function. As discussed earlier, tangent and cotangent are reciprocals, so $\tan(x)$ is the reciprocal of $\cot(x)$, not the negative cotangent. Therefore, $y = \tan(x)$ is not equivalent to $y = -\cot(x)$. This option reinforces the basic difference between the two functions and why a direct equivalence without transformations is not possible.

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

Option D, $y = \tan\left(x+\frac{\pi}{2}\right)$, involves a similar horizontal shift as Option B but without the additional negative sign outside the tangent function. Using the cofunction identity $\tan\left(x + \frac{\pi}{2}\right) = -\cot(x)$, we can directly see that this option is equivalent to the negative cotangent function. This identity is derived from the properties of sine and cosine functions under phase shifts and provides a direct link between the tangent and cotangent functions. Therefore, $y = \tan\left(x+\frac{\pi}{2}\right)$ is the correct equivalent function.

Detailed Explanation of the Correct Option

The correct option, D. $y = \tan\left(x+\frac{\pi}{2}\right)$, is equivalent to $y = -\cot(x)$ due to the cofunction identity. This identity is a cornerstone of trigonometric transformations and is essential for simplifying and solving trigonometric equations. The identity can be derived using the sum-to-product formulas for sine and cosine, providing a rigorous mathematical basis for the equivalence. Understanding the derivation enhances comprehension and retention of this crucial concept.

To illustrate further, let's break down the transformation step by step:

1. Start with the given function: $y = \tan\left(x+\frac{\pi}{2}\right)$

2. Apply the tangent addition formula: $\tan(A + B) = \frac{\tan(A) + \tan(B)}{1 - \tan(A)\tan(B)}$

3. Substitute $A = x$ and $B = \frac{\pi}{2}$: $\tan\left(x + \frac{\pi}{2}\right) = \frac{\tan(x) + \tan(\frac{\pi}{2})}{1 - \tan(x)\tan(\frac{\pi}{2})}$

4. Recall that $\tan(\frac{\pi}{2})$ is undefined. Instead, we can use the sine and cosine definitions:

   $\tan\left(x + \frac{\pi}{2}\right) = \frac{\sin(x + \frac{\pi}{2})}{\cos(x + \frac{\pi}{2})}$

5. Apply the sine and cosine addition formulas:

   • $\sin\left(x + \frac{\pi}{2}\right) = \sin(x)\cos(\frac{\pi}{2}) + \cos(x)\sin(\frac{\pi}{2}) = \sin(x)(0) + \cos(x)(1) = \cos(x)$
   • $\cos\left(x + \frac{\pi}{2}\right) = \cos(x)\cos(\frac{\pi}{2}) - \sin(x)\sin(\frac{\pi}{2}) = \cos(x)(0) - \sin(x)(1) = -\sin(x)$

6. Substitute these back into the tangent expression:

   $\tan\left(x + \frac{\pi}{2}\right) = \frac{\cos(x)}{-\sin(x)} = -\frac{\cos(x)}{\sin(x)}$

7. Recognize that $\frac{\cos(x)}{\sin(x)} = \cot(x)$, so:

   $\tan\left(x + \frac{\pi}{2}\right) = -\cot(x)$

This step-by-step derivation clearly shows that $y = \tan\left(x+\frac{\pi}{2}\right)$ is indeed equivalent to $y = -\cot(x)$. The horizontal shift of $\frac{\pi}{2}$ combined with the inherent relationship between sine and cosine under this shift results in the cotangent function with a negative sign. This comprehensive explanation solidifies the understanding of why option D is the correct answer.

Conclusion

In conclusion, the function equivalent to $y = -\cot(x)$ is D. $y = \tan\left(x+\frac{\pi}{2}\right)$. This equivalence is derived from the cofunction identity, which links tangent and cotangent through a phase shift of $\frac{\pi}{2}$. Understanding this relationship and the underlying trigonometric identities is crucial for solving problems involving trigonometric functions and their transformations. The detailed explanation and step-by-step derivation provided in this article offer a comprehensive understanding of this concept, aiding in mastering trigonometric functions and their applications. This thorough exploration not only answers the specific question but also enhances overall proficiency in trigonometric manipulations and problem-solving.