Problem 74

Question

Verify the Identity. $$\arccos (-x)=\pi-\arccos x$$

Step-by-Step Solution

Verified

Answer

The identity $ \arccos (-x) = \pi - \arccos x $ is verified using properties of the cosine function.

1Step 1: Understand the Problem

We are asked to verify the trigonometric identity: $ \arccos(-x) = \pi - \arccos(x) $. This means that we need to show that both sides of the equation give the same result for any value of $ x $ within the valid range.

2Step 2: Consider the Domain and Range of $ \arccos(x) $

The function $ \arccos(x) $ is defined for $ -1 \leq x \leq 1 $ and gives values between $ 0 $ and $ \pi $. Similarly, $ \arccos(-x) $ is defined by flipping the sign of $ x $ and outputs values in the same range.

3Step 3: Apply the Property of Cosine

Recall that the cosine function is even, meaning $ \cos(\theta) = \cos(-\theta) $. For an angle $ \theta $, if $ \cos(\theta) = x $, then $ \cos(\pi - \theta) = -x $. This implies that $ \arccos(-x) $ gives the angle equivalent to $ \pi - \arccos(x) $.

4Step 4: Rewrite $ \arccos(-x) $

Since $ \cos(\theta) = \cos(\pi - \theta) $, we can say $ \arccos(-x) = \theta' $, where $ \theta' = \pi - \theta $.

5Step 5: Conclusion

Because $ \arccos(x) = \theta $, then $ \arccos(-x) = \pi - \arccos(x) $. This verifies that the identity is true.

Key Concepts

arccos functioncosine propertydomain and range

arccos function

The inverse cosine function, commonly denoted as $ \arccos(x) $, plays a pivotal role in trigonometry. It reverses the action of the cosine function by returning the original angle given a cosine value. This function is particularly useful when solving problems where you have the cosine of an angle and need to determine the angle itself.
The concept behind $ \arccos(x) $ is straightforward: If $ \cos(\theta) = x $, then $ \arccos(x) = \theta $. It is important to remember that this function only works for values of $ x $ within the range $[-1, 1]$.

For $ x = 1 $, $ \arccos(x) = 0 $, since $ \cos(0) = 1 $.
For $ x = 0 $, $ \arccos(x) = \frac{\pi}{2} $, since $ \cos\left(\frac{\pi}{2}\right) = 0 $.
For $ x = -1 $, $ \arccos(x) = \pi $, since $ \cos(\pi) = -1 $.

Exploring the $ \arccos $ function helps us verify identities by finding specific angles related to given cosine values.

cosine property

Understanding the properties of the cosine function is key to exploring trigonometric identities. The cosine function is even, which means that the value of $ \cos(\theta) $ is the same as $ \cos(-\theta) $. This property simplifies many calculations and transformations.
To delve deeper, consider the identity $ \arccos(-x) = \pi - \arccos(x) $. Here’s why it holds true:

The cosine of an angle $ \theta $ remains unchanged whether $ \theta $ is negative or not: $ \cos(\theta) = \cos(-\theta) $.
If we know $ \cos(\theta) = x $, the angle $ \theta $ is equivalent to $ \arccos(x) $.
Hence, for $ x $, the angle $ \pi - \arccos(x) $ gives us $ -x $ in terms of the cosine function.

This demonstrates how the cosine property underpins transformation and verification of trigonometric identities, allowing us to manipulate angles in equations confidently.

domain and range

The concept of domain and range is crucial in understanding functions like $ \arccos $. The domain specifies the acceptable input values, while the range describes possible output values:

For the function $ \arccos(x) $, the domain is $[-1, 1]$, meaning it accepts any $ x $ within this interval, as no other input will yield a real number output.
The range for $ \arccos(x) $ spans from 0 to $ \pi $, providing angles in radians that align with typical trigonometric outputs when you think of a full rotation as $ 2\pi $ radians.

Understanding the constraints of domain and range helps to avoid errors in calculation or assumption. By knowing the valid input and output values, one can analyze trigonometric functions, like $ \arccos(x) $, more effectively, ensuring solutions are within the realm of mathematical reality.

Problem 73

Problem 74

Other exercises in this chapter

Problem 73

Make the trigonometric substitution $$x=a \sec \theta \quad \text { for } 00.$$ Simplify the resulting expression. $$\sqrt{x^{2}-a^{2}}$$

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Problem 73

On a clear day with $D$ hours of daylight, the intensity of sunlight $I$ (in calories/cm^) may be approximated by $$ I=I_{M} \sin ^{3} \frac{\pi t}{D} \quad

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Problem 74

Make the trigonometric substitution $$x=a \sec \theta \quad \text { for } 00.$$ Simplify the resulting expression. $$\frac{1}{x^{2} \sqrt{x^{2}-a^{2}}}$$

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Problem 75

Verify the Identity. $$\arctan x+\arctan \frac{1}{x}=\frac{\pi}{2}, x>0$$

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