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Chapter 10: Problem 38

Write expression as a single trigonometric function or a power of a trigonometric function. (You may wish to use a graph to support your result.) $$\frac{1}{\tan ^{2} \alpha}+\cot \alpha \tan \alpha$$

Short Answer

Expert verified
The expression simplifies to \( \csc^2 \alpha \).

Step by step solution

01

Simplify the First Term

The given expression is \( \frac{1}{\tan^2 \alpha} + \cot \alpha \tan \alpha \). First, simplify \( \frac{1}{\tan^2 \alpha} \). Remember that \( \tan \alpha = \frac{\sin \alpha}{\cos \alpha} \), so \( \tan^2 \alpha = \frac{\sin^2 \alpha}{\cos^2 \alpha} \). Therefore, \( \frac{1}{\tan^2 \alpha} = \frac{\cos^2 \alpha}{\sin^2 \alpha} = \cot^2 \alpha \).
02

Simplify the Second Term

Now, simplify \( \cot \alpha \tan \alpha \). We know that \( \cot \alpha = \frac{1}{\tan \alpha} \), thus \( \cot \alpha \tan \alpha = 1 \).
03

Combine the Terms

Combine the simplified terms from Steps 1 and 2. We have \( \cot^2 \alpha + 1 \) from the expression, since we simplified \( \frac{1}{\tan^2 \alpha} \) to \( \cot^2 \alpha \) and \( \cot \alpha \tan \alpha \) to 1.
04

Recognize the Trigonometric Identity

Use the Pythagorean identity \( \cot^2 \alpha + 1 = \csc^2 \alpha \). Hence, the expression \( \frac{1}{\tan^2 \alpha} + \cot \alpha \tan \alpha \) simplifies to \( \csc^2 \alpha \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Trigonometric Functions
Trigonometric functions are fundamental in understanding relationships in right-angled triangles. They include sine (\( \sin \)), cosine (\( \cos \)), and tangent (\( \tan \)), among others. These functions help describe the angle measures and side lengths in a triangle. In terms of a unit circle:
  • The sine function represents the y-coordinate of the point on the circle.
  • The cosine function represents the x-coordinate.
  • The tangent function is the ratio of the sine to the cosine, \( \tan \theta = \frac{\sin \theta}{\cos \theta} \).
Using these basic functions, we can derive others like secant (\( \sec \)), cosecant (\( \csc \)), and cotangent (\( \cot \)). Understanding these relationships is key to simplifying trigonometric expressions, like in our task.
Pythagorean Identity
The Pythagorean identity is a fundamental concept in trigonometry. It relates the square of sine and cosine to one. The most common identity is:
  • \( \sin^2 \theta + \cos^2 \theta = 1 \)
Other variations are derived by dividing this basic identity by either sine squared or cosine squared:
  • \( 1 + \cot^2 \theta = \csc^2 \theta \)
  • \( \tan^2 \theta + 1 = \sec^2 \theta \)
In the exercise, recognizing the identity \( \cot^2 \alpha + 1 = \csc^2 \alpha \) allows us to simplify the expression effectively. These identities are useful for transforming and simplifying complex trigonometric expressions.
Cotangent
The cotangent function, denoted as \( \cot \theta \), is another key trigonometric function. Cotangent is the reciprocal of tangent:
  • \( \cot \theta = \frac{1}{\tan \theta} = \frac{\cos \theta}{\sin \theta} \)
This relation shows that the cotangent function is the ratio of the cosine function to the sine function. In our step-by-step solution, we used this property to simplify terms like \( \frac{1}{\tan^2 \alpha} \) into \( \cot^2 \alpha \). Recognizing these reciprocal relationships is crucial for manipulating and solving trigonometric equations.
Cosecant
Cosecant, represented by \( \csc \theta \), is one of the reciprocal trigonometric functions. It is the inverse of the sine function:
  • \( \csc \theta = \frac{1}{\sin \theta} \)
Cosecant is particularly useful when dealing with vertical lines on the unit circle or when the sine value is small, as it magnifies this ratio. In our example, using the Pythagorean identity \( \cot^2 \alpha + 1 = \csc^2 \alpha \) allowed us to transform the original expression into a single cosecant function. Understanding how and when to utilize these reciprocal functions can simplify solving trigonometric problems, making them less daunting.

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Most popular questions from this chapter

Solve each equation ( \(x\) in radians and \(\theta\) in degrees) for all exact solutions where appropriate. Round approximale values in radians to four decimal places and approximate values in degrees to the nearest tenth. Write answers using the least possible nonnegative angle measures. $$2 \cos ^{2} 2 \theta=1-\cos 2 \theta$$

Write each expression as an algebraic expression in \(u, u>0\). $$\cot \left(\tan ^{-1} u\right)$$

Sound Waves Sound is a result of waves applying pressure to a person's eardrum. For a pure sound wave radiating outward in a spherical shape, the trigonometric function $$P=\frac{a}{r} \cos \left(\frac{2 \pi r}{\lambda}-c t\right)$$ can be used to model the sound pressure \(P\) at a radius of \(r\) feet from the source, where \(t\) is time in seconds, \(\lambda\) is length of the sound wave in feet, \(c\) is speed of sound in feet per second, and \(a\) is maximum sound pressure at the source measured in pounds per square foot. (Source: Beranek, L.., Noise and Vibration Control, Institute of Noise Control Engineering. Washington, DC.) Let \(\lambda=4.9\) feet and \(c=1026\) feet per second. (IMAGE CANNOT COPY) (a) Let \(a=0.4\) pound per square foot. Graph the sound pressure at a distance \(r=10\) feet from its source over the interval \(0 \leq t \leq 0.05 .\) Describe \(P\) at this distance. (b) Now let \(a=3\) and \(t=10 .\) Graph the sound pressure for \(0 \leq r \leq 20 .\) What happens to the pressure \(P\) as the radius \(r\) increases? (c) Suppose a person stands at a radius \(r\) so that $$r=n \lambda$$ where \(n\) is a positive integer. Use the difference identity for cosine to simplify \(P\) in this situation.

Write each expression as an algebraic expression in \(u, u>0\). $$\sec \left(\cos ^{-1} u\right)$$

Give the exact real number value of each expression. Do not use a calculator. $$\cos \left(\cos ^{-1} \frac{3}{5}+\sin ^{-1} \frac{5}{13}\right)$$
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