Problem 84 - Contemporary Precalculus

Question

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

Step-by-Step Solution

Answer

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

Step 1: Rewrite the right side of the identity in terms of sine and cosine

We are given \( \cos ^{-1} x = \frac{\pi}{2} - \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right) \) and we want to rewrite the right side of the equation in terms of sine and cosine. Since \(\tan^{-1}(a) = \frac{\pi}{2} - \cos^{-1}\left(\frac{1}{\sqrt{1+a^2}}\right)\), we have \( \frac{\pi}{2} - \tan^{-1}\left(\frac{x}{\sqrt{1-x^2}}\right) = \frac{\pi}{2} - \cos^{-1}\left(\frac{1}{\sqrt{1+\left(\frac{x}{\sqrt{1-x^2}}\right)^2 }}\right) \)

Step 2: Simplify the expression inside the inverse cosine

We can simplify the expression inside the inverse cosine as \begin{align*} \frac{1}{\sqrt{1+(\frac{x}{\sqrt{1-x^2}})^2}} &= \frac{1}{\sqrt{ \frac{1}{1-x^2}\left(1+(\frac{x}{\sqrt{1-x^2}})^2\right) }}\\ &= \frac{1}{\sqrt{ \frac{1}{1-x^2}\left(1+\frac{x^2}{1-x^2}\right) }} \\ &= \frac{1}{\sqrt{ \frac{1-x^2+x^2}{1-x^2} }} \\ &= \sqrt{1-x^2} \end{align*} Replacing this back into the original equation, we have \( \cos^{-1} x = \frac{\pi}{2} - \cos^{-1}\left(\sqrt{1-x^2}\right) \)

Step 3: Prove the identity

Now we'll show that both sides are equal. Consider the angles \(A\) and \(B\) such that \(A = \cos^{-1}(x)\) and \(B = \frac{\pi}{2} - \cos^{-1}\left(\sqrt{1-x^2}\right)\). We will find the sine and cosine of both \(A\) and \(B\) and show that \(\sin(A)=\sin(B)\) and \(\cos(A)=\cos(B)\). First, we know that \(\cos(A) = x\) and \(\sin(A) = \sqrt{1 - \cos^2(A)} = \sqrt{1 - x^2}\). Now, we find the sine and cosine of angle \(B\): \(\sin(B) = \sin\left(\frac{\pi}{2} - \cos^{-1}\left(\sqrt{1 - x^2}\right)\right) = \cos\left(\cos^{-1}\left(\sqrt{1 - x^2}\right)\right) = \sqrt{1-x^2}\) and \(\cos(B) = \cos\left(\frac{\pi}{2} - \cos^{-1}\left(\sqrt{1 - x^2}\right)\right) = \sin\left(\cos^{-1}\left(\sqrt{1-x^2}\right)\right) = x\). Since \(\sin(A) = \sin(B)\) and \(\cos(A) = \cos(B)\), we conclude that angle \(A\) and angle \(B\) are equal. So, we have proved the identity: $\cos ^{-1} x=\frac{\pi}{2}-\tan ^{-1}\left(\frac{x}{\sqrt{1-x^{2}}}\right) \quad(-1

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are essential for determining angles when given a trigonometric ratio. They are the inverse operations of the basic trigonometric functions. For example, the function \( \cos^{-1}(x) \) does the opposite of \( \cos(x) \), meaning it finds the angle whose cosine value is \( x \). These functions are particularly useful in solving equations and proving identities, just like the one in the original exercise where \( \cos^{-1}(x) \)is expressed through \( \tan^{-1} \).
Inverse trig functions have particular ranges to ensure they provide unique results: for instance, \( \cos^{-1}(x) \)ranges from 0 to \( \pi \). This is vital because it guarantees that every x-value within the domain corresponds to only one angle within the range. Understanding these restrictions can help when solving complex trigonometric equations by ensuring solutions fall within expected intervals.

Cosine Function

The cosine function is another cornerstone of trigonometry, representing the ratio of the adjacent side to the hypotenuse in a right triangle. When looking at the unit circle, the cosine of an angle gives the x-coordinate of the corresponding point on the circle. This property makes it incredibly useful in both geometry and calculus.In practice, this function can range between -1 and 1 for all angles, and it is periodic with a period of \( 2\pi \). This cyclical nature allows us to use cosine to model periodic phenomena like waves and oscillations.
In the original exercise, the value \( \cos^{-1}(x) \)translates back into an angle whose cosine is \( x \).
Using trigonometric identities effectively, such as rewriting \( \tan^{-1} \)in terms of cosine, reveals relationships between functions and angles that are key to solving more complex problems.

Sine Function

The sine function, another basic trigonometric function, relates to the opposite side and hypotenuse of a right triangle. On the unit circle, sine represents the y-coordinate of a point at a given angle. It shares many characteristics with cosine, being a periodic function with a range between -1 and 1. Its period is also \( 2\pi \).In trigonometric identities and transformations, sine plays a crucial role. This is evident in exercises that involve complementary angles, such as converting between \( \sin(\theta) \)and \( \cos(\pi/2 - \theta) \).
In the proof from the original exercise, \( \sin(A) \)and \( \sin(B) \)both simplify to\( \sqrt{1-x^2} \), linking directly with the Pythagorean identity.This relationship shows how understanding sine's properties can be used to confirm or discover new trigonometric principles.

Pythagorean Identity

One of the most fundamental identities in trigonometry is the Pythagorean Identity. Stemming from the Pythagorean theorem, this identity states\( \sin^2(\theta) + \cos^2(\theta) = 1 \).Its importance cannot be overstated as it provides a direct link between sine and cosine. This identity helps verify solutions when working with trigonometric expressions and identities. For instance, in our exercise, it is crucial for understanding why \( \sin(A) = \sqrt{1-x^2} \)when \( \cos(A) = x \).
Whenever you're dealing with angles and side lengths in trigonometry, the Pythagorean identity ensures that the relationships between the functions are preserved, no matter the context. It also allows for converting expressions between sine and cosine, making it a versatile tool for solving identities and equations in trigonometry.

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Problem 83

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Problem 85

Other exercises in this chapter

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 83

Prove the identity. \(\sin ^{-1} x=\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

Problem 85

Is it true that \(\tan ^{-1} x=\frac{\sin ^{-1} x}{\cos ^{-1} x}\) ? Justify your answer.

View solution

Key Concepts

Inverse Trigonometric Functions

Inverse trigonometric functions are essential for determining angles when given a trigonometric ratio. They are the inverse operations of the basic trigonometric functions. For example, the function \( \cos^{-1}(x) \) does the opposite of \( \cos(x) \), meaning it finds the angle whose cosine value is \( x \). These functions are particularly useful in solving equations and proving identities, just like the one in the original exercise where \( \cos^{-1}(x) \)is expressed through \( \tan^{-1} \).
Inverse trig functions have particular ranges to ensure they provide unique results: for instance, \( \cos^{-1}(x) \)ranges from 0 to \( \pi \). This is vital because it guarantees that every x-value within the domain corresponds to only one angle within the range. Understanding these restrictions can help when solving complex trigonometric equations by ensuring solutions fall within expected intervals.

Cosine Function

The cosine function is another cornerstone of trigonometry, representing the ratio of the adjacent side to the hypotenuse in a right triangle. When looking at the unit circle, the cosine of an angle gives the x-coordinate of the corresponding point on the circle. This property makes it incredibly useful in both geometry and calculus.In practice, this function can range between -1 and 1 for all angles, and it is periodic with a period of \( 2\pi \). This cyclical nature allows us to use cosine to model periodic phenomena like waves and oscillations.
In the original exercise, the value \( \cos^{-1}(x) \)translates back into an angle whose cosine is \( x \).
Using trigonometric identities effectively, such as rewriting \( \tan^{-1} \)in terms of cosine, reveals relationships between functions and angles that are key to solving more complex problems.

Sine Function

The sine function, another basic trigonometric function, relates to the opposite side and hypotenuse of a right triangle. On the unit circle, sine represents the y-coordinate of a point at a given angle. It shares many characteristics with cosine, being a periodic function with a range between -1 and 1. Its period is also \( 2\pi \).In trigonometric identities and transformations, sine plays a crucial role. This is evident in exercises that involve complementary angles, such as converting between \( \sin(\theta) \)and \( \cos(\pi/2 - \theta) \).
In the proof from the original exercise, \( \sin(A) \)and \( \sin(B) \)both simplify to\( \sqrt{1-x^2} \), linking directly with the Pythagorean identity.This relationship shows how understanding sine's properties can be used to confirm or discover new trigonometric principles.

Pythagorean Identity

One of the most fundamental identities in trigonometry is the Pythagorean Identity. Stemming from the Pythagorean theorem, this identity states\( \sin^2(\theta) + \cos^2(\theta) = 1 \).Its importance cannot be overstated as it provides a direct link between sine and cosine. This identity helps verify solutions when working with trigonometric expressions and identities. For instance, in our exercise, it is crucial for understanding why \( \sin(A) = \sqrt{1-x^2} \)when \( \cos(A) = x \).
Whenever you're dealing with angles and side lengths in trigonometry, the Pythagorean identity ensures that the relationships between the functions are preserved, no matter the context. It also allows for converting expressions between sine and cosine, making it a versatile tool for solving identities and equations in trigonometry.