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Chapter 1: Problem 8

In Exercises \(7-12,\) one of \(\sin x, \cos x,\) and \(\tan x\) is given. Find the other two if \(x\) lies in the specified interval. $$\tan x=2, \quad x \in\left[0, \frac{\pi}{2}\right]$$

Short Answer

Expert verified
\( \sin x = \frac{2}{\sqrt{5}}, \cos x = \frac{1}{\sqrt{5}} \)

Step by step solution

01

Identify Given Information

We are given that \( \tan x = 2 \) and must find \( \sin x \) and \( \cos x \) within the interval \( x \in \left[ 0, \frac{\pi}{2} \right] \). This tells us \( x \) is in the first quadrant.
02

Use the Definition of Tangent

Recall that \( \tan x = \frac{\sin x}{\cos x} \). We have \( \tan x = 2 = \frac{\sin x}{\cos x} \), which implies \( \sin x = 2 \cos x \).
03

Solve for \( \sin x \) and \( \cos x \) using the Pythagorean Identity

The Pythagorean identity tells us \( \sin^2 x + \cos^2 x = 1 \). Substitute \( \sin x = 2 \cos x \) into this identity: \( (2 \cos x)^2 + \cos^2 x = 1 \).
04

Simplify and Solve the Equation

Simplifying gives \( 4 \cos^2 x + \cos^2 x = 1 \), or \( 5 \cos^2 x = 1 \). Solve for \( \cos^2 x \) to get \( \cos^2 x = \frac{1}{5} \), so \( \cos x = \frac{1}{\sqrt{5}} \).
05

Find \( \sin x \) Using \( \sin x = 2 \cos x \)

Substitute \( \cos x = \frac{1}{\sqrt{5}} \) into \( \sin x = 2 \cos x \) to find \( \sin x = 2 \times \frac{1}{\sqrt{5}} = \frac{2}{\sqrt{5}} \).

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Key Concepts

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

Tangent Function
The tangent function, denoted as \( \tan x \), is one of the primary trigonometric functions and represents the ratio of the sine function to the cosine function. This can be expressed as:
  • \( \tan x = \frac{\sin x}{\cos x} \)
This definition allows us to understand that whenever we know the value of \( \tan x \), we can relate it to \( \sin x \) and \( \cos x \).
In the exercise above, given \( \tan x = 2 \), this means that \( \sin x = 2 \cdot \cos x \).
All tangent values are positive in the first quadrant, making the computation straightforward within the interval \([0, \frac{\pi}{2}]\).
Understanding the tangent function helps us derive the missing trigonometric values when some are known.
Sine Function
The sine function, expressed as \( \sin x \), represents the ratio of the opposite side to the hypotenuse in a right triangle. It's a primary function and fundamental for trigonometric calculations. In our exercise, \( \sin x \) is involved as part of the given equation for \( \tan x \):
  • If \( \tan x = \frac{\sin x}{\cos x} = 2 \), then \( \sin x = 2 \cdot \cos x \).
Knowing that \( x \) is in the first quadrant, we ensure that \( \sin x \) is positive within the range. After finding \( \cos x = \frac{1}{\sqrt{5}} \), we substitute to get \( \sin x = 2 \cdot \frac{1}{\sqrt{5}} = \frac{2}{\sqrt{5}} \).
This calculation emphasizes the importance of the sine function's relationship with both the cosine and tangent functions.
Cosine Function
The cosine function, denoted \( \cos x \), indicates the ratio of the adjacent side to the hypotenuse in a right triangle. Cosine is another primary trigonometric function that is essential alongside the sine and tangent functions. In the context of this exercise, knowing \( \tan x = 2 \) leads us through the equation \( \sin x = 2 \cdot \cos x \) to explore:
  • Using the Pythagorean identity, \( \sin^2 x + \cos^2 x = 1 \), allows us to find \( \cos x \).
By substituting \( \sin x = 2 \cos x \), the equation becomes \( 4\cos^2 x + \cos^2 x = 1 \) or \( 5\cos^2 x = 1 \). Thus, solving gives \( \cos x = \frac{1}{\sqrt{5}} \).
This illustrates how we can derive one trigonometric function from another by employing known identities and relationships.
Pythagorean Identity
The Pythagorean identity is a key relationship in trigonometry, stating that:
  • \( \sin^2 x + \cos^2 x = 1 \)
This identity is crucial as it connects \( \sin x \) and \( \cos x \) in a relationship that always holds true for any angle \( x \).
In the given exercise, we leveraged this identity to solve for \( \cos^2 x \) using \( \tan x \):\( \sin x = 2 \cos x \), allowing us to use \( (2\cos x)^2 + \cos^2 x = 1 \).
Simplifying, we arrived at \( 5\cos^2 x = 1 \), and so \( \cos x = \frac{1}{\sqrt{5}} \).
The ease of use of the Pythagorean identity lies in its universal applicability for computing sine and cosine values when one or both are unknown.

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Most popular questions from this chapter

Exercises \(27-36\) tell how many units and in what directions the graphs of the given equations are to be shifted. Give an equation for the shifted graph. Then sketch the original and shifted graphs together, labeling each graph with its equation. $$ x^{2}+y^{2}=25 \quad \text { Up } 3, \text { left } 4 $$

Let \(f(x)=x-3, \quad g(x)=\sqrt{x}, \quad h(x)=x^{3},\) and \(j(x)=2 x .\) Express each of the functions in Exercises 11 and 12 as a composition involving one or more of \(f, g, h,\) and \(j\) $$\begin{array}{ll}{\text { a. } y=\sqrt{x}-3} & {\text { b. } y=2 \sqrt{x}} \\\ {\text { c. } y=x^{1 / 4}} & {\text { d. } y=4 x} \\ {\text { e. } y=\sqrt{(x-3)^{3}}} & {\text { f. } y=(2 x-6)^{3}}\end{array}$$

Graph the functions in Exercises \(37-56\) $$ y=|x-2| $$

In Exercises 17 and \(18,(\) a) write formulas for \(f \circ g\) and \(g \circ f\) and find the (b) domain and (c) range of each. $$ f(x)=x^{2}, g(x)=1-\sqrt{x} $$

If \(f(x)=x-1\) and \(g(x)=1 /(x+1),\) find the following. $$ \begin{array}{ll}{\text { a. }} & {f(g(1 / 2))} & {\text { b. } g(f(1 / 2))} \\\ {\text { c. }} & {f(g(x))} & {\text { d. } g(f(x))} \\ {\text { e. }} & {f(f(2))} & {\text { f. } g(g(2))} \\ {\text { g. }} & {f(f(x))} & {\text { h. } g(g(x))}\end{array} $$
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