Problem 81
Question
Prove the identity.
\(\tan ^{-1}(\cot x)=\pi / 2-x \quad(0
Step-by-Step Solution
Verified Answer
Based on the step-by-step solution above, the identity \(\tan ^{-1}(\cot x)=\pi / 2-x (0
1Step 1: Rewrite cotangent as cosine/sine
We first rewrite cotangent in terms of cosine and sine:
$$\tan ^{-1}(\cot x) = \tan^{-1}\left(\frac{\cos x}{\sin x}\right)$$
2Step 2: Replace tangent with sine and cosine
Next, we rewrite the arctangent function in terms of sine and cosine using the relationship \(\tan y = \sin y / \cos y\). So, we can write:
$$y = \tan^{-1}\left(\frac{\cos x}{\sin x}\right)$$
$$\tan y = \frac{\cos x}{\sin x}$$
3Step 3: Multiply by \(\sin y \cdot \sin x\) to eliminate the fractions
In order to eliminate the fractions, multiply both sides by \(\sin y \cdot \sin x\):
$$\sin y \cdot \tan y \cdot \sin x = \sin y \cdot \cos x$$
$$\sin y \cdot \frac{\sin y}{\cos y} \cdot \sin x = \sin y \cdot \cos x$$
4Step 4: Simplify
We can simplify the left side of the equation as:
$$\sin^2 y \cdot \sin x = \sin y \cdot \cos x$$
5Step 5: Replace \(\sin^2 y\) with \(1-\cos^2 y\)
We can replace \(\sin^2 y\) with \(1-\cos^2 y\) using the Pythagorean identity:
$$(1-\cos^2 y) \cdot \sin x = \sin y \cdot \cos x$$
6Step 6: Multiply by \(\cos^2 y\)
Now we make use of the equality between \(\cos x\) and \(\sin y\) and multiply both sides by \(\cos^2 y\):
$$\sin x \cdot \cos^2 y = \sin y \cdot \cos x$$
7Step 7: Express \(\cos^2 y\) in terms of \(\cos x\)
From step 3, we have the expression \(\tan y = \cos x / \sin x\) that can be restated as \(\cos x = \tan y \cdot \sin x\). Then we express \(\cos^2 y\):
$$\cos^2 y = \frac{\cos^2 x}{\sin^2 x}$$
8Step 8: Replace \(\cos^2 y\) in the equality
Replace the \(\cos^2 y\) with the expression that was found in step 7.
$$\sin x \cdot \frac{\cos^2 x}{\sin^2 x} = \sin y \cdot \cos x$$
9Step 9: Simplify
Now we can simplify the left side of the equation:
$$\frac{\sin x \cdot \cos^2 x}{\sin^2 x} = \sin y \cdot \cos x$$
10Step 10: Compare and conclude
If we look at the equality from step 8, we can see that:
$$\sin y \cdot \cos x = \sin x \cdot \cos^2 y$$
This is equivalent to:
$$\sin y \cdot \cos x = \sin(\pi / 2 - x)$$
Since \(0 < x < \pi\), we have \(0 < \pi / 2 - x < \pi / 2\), so we can safely apply the arcsine function to both sides, and we conclude:
$$y = \sin^{-1}(\sin(\pi / 2 - x))$$
$$y = \pi / 2 - x$$
Hence, \(\tan ^{-1}(\cot x) = \pi / 2 - x, \quad (0 < x < \pi)\).
Key Concepts
Trigonometric IdentitiesArctangent FunctionCotangent Function
Trigonometric Identities
Trigonometric identities are like a toolkit for solving various trigonometric equations. They help us transform and simplify expressions. One fundamental identity is the Pythagorean identity: \( \sin^2 y + \cos^2 y = 1 \). This identity allows us to express one function in terms of another, making substitutions easier. In the exercise, we used \( \sin^2 y = 1 - \cos^2 y \) to simplify our equation.
This substitution is key when trying to eliminate fractions or when expressing one function in terms of another. Without these identities, solving complex trigonometric problems would be much more difficult.
This substitution is key when trying to eliminate fractions or when expressing one function in terms of another. Without these identities, solving complex trigonometric problems would be much more difficult.
- They provide a way to handle complex expressions.
- They assist in proving other identities.
- They simplify calculations involving trigonometric functions.
Arctangent Function
The arctangent function, denoted as \( \tan^{-1} x \), is the inverse of the tangent function. It helps us find the angle whose tangent value is a given number. This function becomes essential when solving for angles in various equations.
In the exercise, the arctangent function is used to express the relationship: \( y = \tan^{-1}(\cot x) \). We're finding an angle whose tangent would give us \( \cot x \). This understanding is crucial in transforming and simplifying expressions in terms of known angles.
In the exercise, the arctangent function is used to express the relationship: \( y = \tan^{-1}(\cot x) \). We're finding an angle whose tangent would give us \( \cot x \). This understanding is crucial in transforming and simplifying expressions in terms of known angles.
- The range of \( \tan^{-1} \) is typically \((-\pi/2, \pi/2)\).
- It's used to revert values from tangent back into angles.
- Helps to understand the geometry and linkage between angles and ratios.
Cotangent Function
The cotangent function, \( \cot x \), is the reciprocal of the tangent function, meaning that \( \cot x = \frac{1}{\tan x} = \frac{\cos x}{\sin x} \). This is a vital concept in trigonometry since it provides a different perspective on the sine and cosine functions.
Understanding how \( \cot x \) connects to \( \tan x \) allows us to manipulate and transform equations and identities effectively. By rewriting the cotangent in terms of sine and cosine, as shown in the exercise, we set the stage for using arctangent. This linkage is essential in proving relationships such as \( \tan^{-1}(\cot x) = \pi / 2 - x \).
Understanding how \( \cot x \) connects to \( \tan x \) allows us to manipulate and transform equations and identities effectively. By rewriting the cotangent in terms of sine and cosine, as shown in the exercise, we set the stage for using arctangent. This linkage is essential in proving relationships such as \( \tan^{-1}(\cot x) = \pi / 2 - x \).
- Cotangent is often used in reciprocal relationships.
- It simplifies expressions involving division by tangent.
- Capable of transforming angles for better analysis.
Other exercises in this chapter
Problem 80
Prove the identity. $$\frac{\tan x-\tan y}{\cot x-\cot y}=-\tan x \tan y$$
View solution Problem 81
Use factoring, the quadratic formula, or identities to solve the equation. Find all solutions in the interval \([0,2 \pi)\). $$\sin 2 x+\cos x=0$$
View solution Problem 81
Prove the identity. $$\frac{\cos x-\sin y}{\cos y-\sin x}=\frac{\cos y+\sin x}{\cos x+\sin y}$$
View solution Problem 82
Prove the identity. $$\frac{\tan x+\tan y}{\cot x+\cot y}=\frac{\tan x \tan y-1}{1-\cot x \cot y}$$
View solution