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Chapter 7: Problem 43

Solve, finding all solutions in \([0,2 \pi)\). $$\cos (\pi-x)+\sin \left(x-\frac{\pi}{2}\right)=1$$

Short Answer

Expert verified
x = \frac{2\pi}{3}, \frac{4\pi}{3}

Step by step solution

01

- Use Trigonometric Identities

Rewrite \( \cos(\pi - x) \) and \( \sin\left(x - \frac{\pi}{2}\right) \) using trigonometric identities. We can do the following transforms: 1. \( \cos(\pi - x) = -\cos(x) \) 2. \( \sin(x - \frac{\pi}{2}) = -\cos(x) \)
02

- Substitute and Simplify

Substitute the transformed identities back into the equation: 1. \( -\cos(x) - \cos(x) = 1 \) Combine like terms: 2. \( -2\cos(x) = 1 \)
03

- Solve for \( \cos(x) \)

Isolate \( \cos(x) \) by dividing both sides of the equation by -2: 1. \( \cos(x) = -\frac{1}{2} \)
04

- Find Solutions in \( [0, 2\pi) \)

Determine the values of \( x \) for which \( \cos(x) = -\frac{1}{2} \)1. \( x = \frac{2\pi}{3} \) and \( x = \frac{4\pi}{3} \) are two solutions within the given range.

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Key Concepts

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

Cosine Function
The cosine function, denoted as \(\text{cos}(x)\), is one of the primary trigonometric functions. It measures the horizontal coordinate of a point on the unit circle as it rotates by an angle \(x\) from the positive x-axis.
The cosine function has several important properties:
  • Its range is \([-1, 1]\).
  • It is an even function, meaning that \( \text{cos}(-x) = \text{cos}(x) \).
  • It has a period of \(2 \pi \), so \(\text{cos}(x + 2 \pi) = \text{cos}(x)\).
For our exercise, recognizing transformations such as \(\text{cos}(\pi - x)\) helps in simplifying the problem. Specifically, \(\text{cos}(\pi - x) = -\text{cos}(x)\), thanks to the unit circle properties.
Sine Function
The sine function, denoted as \( \text{sin}(x) \), is another crucial trigonometric function. It measures the vertical coordinate of a point on the unit circle as it rotates by an angle \(x\) from the positive x-axis.
Key features of the sine function include:
  • Its range is \([-1, 1]\).
  • It is an odd function, meaning that \( \text{sin}(-x) = -\text{sin}(x) \).
  • It also has a period of \(2 \pi \), so \( \text{sin}(x + 2 \pi) = \text{sin}(x) \).
In our problem, noticing the transformation \(\text{sin}(x - \frac{\text{\text{蟺}}}{2})\), which equals \( -\text{cos}(x)\), is crucial. It allows us to substitute and simplify the equation smoothly.
Solving Trigonometric Equations
Solving trigonometric equations often involves using identities to simplify and isolate trigonometric functions like sine and cosine.
Here are the general steps:
  • Rewrite the equation using known trigonometric identities.
  • Simplify the expression to combine like terms.
  • Isolate the trigonometric function (e.g., \(\text{cos}(x)\) or \(\text{sin}(x)\)).
  • Determine the solutions for the function within the specified interval (e.g., \([0, 2 \pi)\)).
In the given exercise, we used identities to rewrite \( \text{cos}(\pi - x) \) and \( \text{sin}(x - \frac{\text{\text{蟺}}}{2})\). After simplifying, we isolated \( \text{cos}(x) \) and found its solutions within the interval.
Trigonometric Transformations
Trigonometric transformations help relate different trigonometric functions under various circumstances. These transformations leverage the properties of the unit circle.
Here are key transformations used in the problem:
  • Using \(\text{cos}(\pi - x) = -\text{cos}(x)\) simplifies terms involving a shifted cosine function.
  • Using \( \text{sin}(x - \frac{\text{\text{蟺}}}{2})=-\text{cos}(x)\) turns a shifted sine function into a cosine function.
These transformations streamline the solving process by reducing complex expressions into familiar trigonometric functions. They are essential for solving equations more efficiently and finding accurate solutions in the specified interval.

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Most popular questions from this chapter

Find the following. \(\cos \left(\frac{1}{2} \sin ^{-1} \frac{\sqrt{3}}{2}\right)\)

Consider the following functions ( \(a\) )- ( \(f\) ). Without graphing them, answer question. a) \(f(x)=2 \sin \left(\frac{1}{2} x-\frac{\pi}{2}\right)\) b) \(f(x)=\frac{1}{2} \cos \left(2 x-\frac{\pi}{4}\right)+2\) c) \(f(x)=-\sin \left[2\left(x-\frac{\pi}{2}\right)\right]+2\) d \(f(x)=\sin (x+\pi)-\frac{1}{2}\) e) \(f(x)=-2 \cos (4 x-\pi)\) f) \((x)=-\cos \left[2\left(x-\frac{\pi}{8}\right)\right]\) Which functions have a graph with a period of \(\pi ?\)

Line \(l_{1}\) contains the points \((-3,7)\) and \((-3,-2)\) Line \(l_{2}\) contains \((0,-4)\) and \((2,6) .\) Find the smallest positive angle from \(l_{1}\) to \(l_{2}\)

Use the given substitution to express the given radical expression as a trigonometric function without radicals. Assume that \(a>0\) and \(0<\theta<\pi / 2 .\) Then find expressions for the indicated trigonometric functions. Let \(x=3 \sec \theta\) in \(\sqrt{x^{2}-9}\). Then find \(\sin \theta\) and \(\cos \theta\)

Evaluate. \(\sin \left(\sin ^{-1} 0.6032+\cos ^{-1} 0.4621\right)\)
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