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Chapter 7: Problem 68

Find each of the following. $$\sin x, \text { given } \cos 2 x=\frac{2}{3}, \text { with } \pi < x < \frac{3 \pi}{2}$$

Short Answer

Expert verified
\text{sin} x = - \frac{\text{sqrt}(1/6)}\.

Step by step solution

01

Use the double-angle identity for cosine

Recall that the double-angle identity for cosine is \(\text{cos } 2x = 2\text{cos}^2 x - 1 \). Use this identity to set up the equation: \(\frac{2}{3} = 2\text{cos}^2 x - 1\).
02

Solve for \(\text{cos}^2 x\)

Add 1 to both sides to isolate the term with \text{cos}^2 x\: \(\frac{2}{3} + 1 = 2\text{cos}^2 x\). Simplify to get \(\frac{5}{3} = 2\text{cos}^2 x\).
03

Isolate \text{cos} x\

Divide both sides by 2: \(\text{cos}^2 x = \frac{5}{6}\). Take the square root of both sides: \(\text{cos} x = \pm \frac{\text{sqrt}(5/6)} \).
04

Determine the sign of \text{cos} x\

Given the interval \( \pi < x < \frac{3 \pi}{2}\), which is in the third quadrant, and cosine is negative in the third quadrant, so \text{cos} x = - \frac{\text{sqrt}(5/6)}\.
05

Use the Pythagorean identity for sine

Recall the Pythagorean identity: \text{sin}^2 x + \text{cos}^2 x = 1\. Substitute \( \text{cos}^2 x = \frac{5}{6}\): \(\text{sin}^2 x + \frac{5}{6} = 1\).
06

Solve for \text{sin}^2 x\

Subtract \(\frac{5}{6}\) from 1: \( \text{sin}^2 x = 1 - \frac{5}{6} = \frac{1}{6}\).
07

Isolate \text{sin} x\

Take the square root of both sides: \( \text{sin} x = \pm \frac{\text{sqrt}(1/6)} \).
08

Determine the sign of \text{sin} x\

Given the interval \( \pi < x < \frac{3 \pi}{2}\), which is in the third quadrant, and sine is also negative in the third quadrant, so \text{sin} x = - \frac{\text{sqrt}(1/6)}\.

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Key Concepts

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

Double-Angle Identity
The double-angle identity is a crucial trigonometric formula. It helps us simplify expressions involving angles that are twice the size of a given angle. For cosine, the double-angle identity is given by:
\(\text{cos} \, 2x = 2 \, \text{cos}^2 x - 1\).
When you know \( \text{cos} \, 2x \,\) and need to find \(\text{cos} \, x\), follow these steps:
  • Start with the identity \( \text{cos} \, 2x = 2 \text{cos}^2 x - 1\).
  • Isolate \text{cos}^2 x by solving the equation.
  • Take the square root to find \( \text{cos} \, x \,\).
Remember, when solving for \( \text{cos} x\), two possible values exist: one positive and one negative. You need additional information about the angle's quadrant to determine the correct sign.
Pythagorean Identity
The Pythagorean identity is a fundamental equation in trigonometry, expressing the relationship between sine and cosine. It states: \(\text{sin}^2 x + \text{cos}^2 x = 1\).
This identity is helpful when we know the value of \( \text{cos}^2 x \,\) and need to find \( \text{sin}^2 x \,\) or vice versa. Here鈥檚 how you can use it:
  • Start with the Pythagorean identity: \( \text{sin}^2 x + \text{cos}^2 x = 1 \).
  • Substitute the given value (e.g., \(\text{cos}^2 x = \) something).
  • Isolate \( \text{sin}^2 x \) by subtracting to solve the equation.
Once you find \( \text{sin}^2 x\), take the square root to obtain \( \text{sin} x\). Just like with cosine, use the quadrant information to decide if the sine value is positive or negative.
Quadrant Angles
Understanding which quadrant an angle is in helps determine the signs of its trigonometric functions. The coordinate plane is divided into four quadrants:
  • Quadrant I: \(0 < x < \frac{\text{蟺}}{2}\) (sine and cosine are positive).
  • Quadrant II: \( \frac{\text{蟺}}{2} < x < \text{蟺}\) (sine is positive, cosine is negative).
  • Quadrant III: \( \text{蟺} < x < \frac{3\text{蟺}}{2}\) (sine and cosine are negative).
  • Quadrant IV: \( \frac{3\text{蟺}}{2} < x < 2\text{蟺}\) (sine is negative, cosine is positive).
Knowing the angle's quadrant allows you to assign the correct sign to the function values. For example, in the given exercise, since the angle is between \(\text{蟺}\) and \( \frac{3\text{蟺}}{2}\), it is in the third quadrant, where both sine and cosine are negative.

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