Problem 83 - Contemporary Precalculus

Question

Prove the identity. \(\sin ^{-1} x=\tan ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right) \quad(-1

Step-by-Step Solution

Answer

Question: Prove that \(\sin^{-1}x=\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\) for \(-1

Step 1: Defining the variable theta

Let's introduce a new variable, \(\theta\), and define it as: \(\theta = \sin^{-1}x\). In other words, we have \(x = \sin\theta\), where \(-1

Step 2: Express the sine and cosine functions in terms of theta

As we have defined \(\theta = \sin^{-1}x\), we can readily express the sine function as \(\sin\theta = x\). To express the cosine function in terms of \(\theta\), we can use the Pythagorean identity: \(\sin^2\theta + \cos^2\theta = 1\). So, \(\cos^2\theta = 1-\sin^2\theta = 1-x^2\) and consequently, \(\cos\theta = \sqrt{1-x^2}\).

Step 3: Define the tangent function in terms of theta

Now that we have expressions for \(\sin\theta\) and \(\cos\theta\), we can find the tangent function in terms of \(\theta\). Recall the identity: \(\tan\theta =\frac{\sin\theta}{\cos\theta}\). Plugging in the expressions for sine and cosine, we have \(\tan\theta = \frac{x}{\sqrt{1-x^2}}\).

Step 4: Find the inverse tangent

Since we have defined \(\tan\theta = \frac{x}{\sqrt{1-x^2}}\), we can find the inverse tangent of both sides: \(\tan^{-1}\left(\tan\theta\right) = \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\). By definition of inverse functions, we have \(\theta = \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\).

Step 5: Conclusion

As we have shown that \(\theta = \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\), and we initially defined \(\theta = \sin^{-1}x\), we can conclude that: $$\sin^{-1}x=\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right) \quad(-1

Key Concepts

Inverse Trigonometric Functions

When dealing with trigonometry, the notion of 'inverse functions' is crucial for solving various problems. The inverses of trigonometric functions allow us to work backwards from the ratio of sides in a right-angled triangle to the angles. For example, the inverse sine function, denoted as \(\sin^{-1}x\) or arcsin(x), returns the angle whose sine is x.

The domain for the inverse sine function is \( -1 \leq x \leq 1 \) and its range is \( -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \). Similarly, the inverse tangent function, \(\tan^{-1}x\) or arctan(x), gives us the angle whose tangent is x. Its domain is all real numbers, and its range is \( -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \), excluding \(\pm \frac{\pi}{2}\) where the tangent function is undefined.

These functions play a pivotal role in trigonometry, especially as we move from evaluating the trigonometric function of an angle to finding the angle itself based on known trigonometric values. They are extensively used in calculus, engineering, and physics.

Pythagorean Identity

One of the most fundamental concepts in trigonometry is the Pythagorean identity. It's an essential tool that relates the sine and cosine of an angle \(\theta\) to the number 1, adhering to the Pythagorean theorem’s essence. The identity is given by the equation \(\sin^2\theta + \cos^2\theta = 1\text{.}\)

This equation simplifies many trigonometric expressions and is the go-to strategy when needing to convert sine terms to cosine terms, or vice versa. In practice, if you know the value of the sine of an angle, you can use this identity to find the cosine of the angle, assuming \(\theta\) is acute. The identity is crucial in deriving other trigonometric identities and in solving equations involving trigonometric functions.

Application in Inverse Trigonometric Functions

For instance, when working with inverse trigonometric functions, the Pythagorean identity enables us to express one function in terms of another, which is exactly what was done in the steps to prove the exercise. When we know \(\sin\theta = x\text{,}\) we can find \(\cos\theta\) using this identity to assist in defining other trigonometric ratios, like the tangent, as seen in the solution.

Tangent Function

In trigonometry, the tangent function is one of the six fundamental trigonometric functions. It's defined as the ratio of the sine of an angle to the cosine of the same angle, written as \(\tan\theta = \frac{\sin\theta}{\cos\theta}\text{.}\)

It plays a pivotal role in both pure and applied mathematics, providing a connection between angles and their tangents, which can be thought of geometrically as the slope of the line through the circular point of the angle on the unit circle.

For angles less than \(\frac{\pi}{2}\text{,}\) the tangent is positive, and it increases as the angle approaches \(\frac{\pi}{2}\text{.}\) It's undefined exactly at \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\text{,}\) where the cosine of the angle is zero and thus the tangent would involve a division by zero. Tangent values repeat every \(\pi\) radians because of the periodic nature of the tangent function, often used in periodic motion such as waves.

In our exercise, we derived the tangent of \(\theta\) based on the sine value and used the concept of inverse functions to connect \(\sin^{-1}x\) and \(\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\text{,}\) compounding our understanding of trigonometric identities and how they are used to navigate between different functions.

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Problem 82

Next

Problem 84

Other exercises in this chapter

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

Problem 84

Prove the identity. \(\cos ^{-1} x=\frac{\pi}{2}-\tan ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right) \quad(-1

View solution

Problem 85

Use factoring, the quadratic formula, or identities to solve the equation. Find all solutions in the interval \([0,2 \pi)\). $$\cos ^{2} x-\sin ^{2} x+\sin x=0$

View solution

Key Concepts

Inverse Trigonometric Functions

When dealing with trigonometry, the notion of 'inverse functions' is crucial for solving various problems. The inverses of trigonometric functions allow us to work backwards from the ratio of sides in a right-angled triangle to the angles. For example, the inverse sine function, denoted as \(\sin^{-1}x\) or arcsin(x), returns the angle whose sine is x.

The domain for the inverse sine function is \( -1 \leq x \leq 1 \) and its range is \( -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \). Similarly, the inverse tangent function, \(\tan^{-1}x\) or arctan(x), gives us the angle whose tangent is x. Its domain is all real numbers, and its range is \( -\frac{\pi}{2} \leq \theta \leq \frac{\pi}{2} \), excluding \(\pm \frac{\pi}{2}\) where the tangent function is undefined.

These functions play a pivotal role in trigonometry, especially as we move from evaluating the trigonometric function of an angle to finding the angle itself based on known trigonometric values. They are extensively used in calculus, engineering, and physics.

Pythagorean Identity

One of the most fundamental concepts in trigonometry is the Pythagorean identity. It's an essential tool that relates the sine and cosine of an angle \(\theta\) to the number 1, adhering to the Pythagorean theorem’s essence. The identity is given by the equation \(\sin^2\theta + \cos^2\theta = 1\text{.}\)

This equation simplifies many trigonometric expressions and is the go-to strategy when needing to convert sine terms to cosine terms, or vice versa. In practice, if you know the value of the sine of an angle, you can use this identity to find the cosine of the angle, assuming \(\theta\) is acute. The identity is crucial in deriving other trigonometric identities and in solving equations involving trigonometric functions.

Application in Inverse Trigonometric Functions

For instance, when working with inverse trigonometric functions, the Pythagorean identity enables us to express one function in terms of another, which is exactly what was done in the steps to prove the exercise. When we know \(\sin\theta = x\text{,}\) we can find \(\cos\theta\) using this identity to assist in defining other trigonometric ratios, like the tangent, as seen in the solution.

Tangent Function

In trigonometry, the tangent function is one of the six fundamental trigonometric functions. It's defined as the ratio of the sine of an angle to the cosine of the same angle, written as \(\tan\theta = \frac{\sin\theta}{\cos\theta}\text{.}\)

It plays a pivotal role in both pure and applied mathematics, providing a connection between angles and their tangents, which can be thought of geometrically as the slope of the line through the circular point of the angle on the unit circle.

For angles less than \(\frac{\pi}{2}\text{,}\) the tangent is positive, and it increases as the angle approaches \(\frac{\pi}{2}\text{.}\) It's undefined exactly at \(\frac{\pi}{2}\) and \(\frac{3\pi}{2}\text{,}\) where the cosine of the angle is zero and thus the tangent would involve a division by zero. Tangent values repeat every \(\pi\) radians because of the periodic nature of the tangent function, often used in periodic motion such as waves.

In our exercise, we derived the tangent of \(\theta\) based on the sine value and used the concept of inverse functions to connect \(\sin^{-1}x\) and \(\tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right)\text{,}\) compounding our understanding of trigonometric identities and how they are used to navigate between different functions.