Problem 12 - Thomas Calculus

Question

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. $$\sin x=-\frac{1}{2}, \quad x \in\left[\pi, \frac{3 \pi}{2}\right]$$

Step-by-Step Solution

Answer

\(\cos x = -\frac{\sqrt{3}}{2}, \tan x = \frac{\sqrt{3}}{3}\)

Step 1: Identify the Quadrant

Since \(x\) lies in the interval \(\left[\pi, \frac{3\pi}{2}\right]\), it is in the third quadrant. In this quadrant, the sine, cosine, and tangent functions have different signs. Specifically, \(\sin x\) and \(\cos x\) are negative, while \(\tan x\) is positive.

Step 2: Use the Pythagorean Identity for Cosine

We know \(\sin x = -\frac{1}{2}\). Using the Pythagorean identity \(\sin^2 x + \cos^2 x = 1\), substitute \(\sin x\) into the equation: \(-\left(-\frac{1}{2}\right)^2 + \cos^2 x = 1\). Simplifying gives \(\frac{1}{4} + \cos^2 x = 1\) and further solving for \(\cos^2 x\) results in \(\cos^2 x = \frac{3}{4}\).

Step 3: Determine the Sign of Cosine

Since we are in the third quadrant where the cosine function is negative, we take the negative root of \(\cos^2 x = \frac{3}{4}\). Thus, \(\cos x = -\sqrt{\frac{3}{4}} = -\frac{\sqrt{3}}{2}\).

Step 4: Calculate Tangent Using Sine and Cosine

The tangent of \(x\) is calculated using \(\tan x = \frac{\sin x}{\cos x}\). Substitute \(\sin x = -\frac{1}{2}\) and \(\cos x = -\frac{\sqrt{3}}{2}\) into the equation: \(\tan x = \frac{-\frac{1}{2}}{-\frac{\sqrt{3}}{2}} \). Simplify to obtain \(\tan x = \frac{1}{\sqrt{3}}\). Further simplification gives \(\tan x = \frac{\sqrt{3}}{3}\), since tangent is positive in the third quadrant.

Key Concepts

Calculating Cosine from Sine

When given a sine value, such as \( \sin x = -\frac{1}{2} \), you can find the corresponding cosine using the Pythagorean identity. This identity is a fundamental relationship among sine and cosine, which states that \[\sin^2 x + \cos^2 x = 1\]This equation allows you to determine one trigonometric function when you know the other. To find \( \cos x \), start by substituting the known sine value into the identity: \(-\left(-\frac{1}{2}\right)^2 + \cos^2 x = 1\). Simplify this equation to: \(\frac{1}{4} + \cos^2 x = 1\). Solve for \(\cos^2 x\) by subtracting \(\frac{1}{4}\) from both sides, resulting in:\[\cos^2 x = \frac{3}{4}\]To find \(\cos x\), take the square root of \(\cos^2 x\), leading to two potential solutions: \(\cos x = \pm \frac{\sqrt{3}}{2}\). However, selecting the correct sign requires understanding the specific quadrant of \(x\).

Third Quadrant Trigonometry

The sign of trigonometric functions depends on the quadrant in which the angle \(x\) is located. The interval \(x \in [\pi, \frac{3\pi}{2}]\) indicates that \(x\) is in the third quadrant.In this quadrant:

• \(\sin x\) is negative
• \(\cos x\) is also negative
• \(\tan x\), the ratio of \(\sin x\) to \(\cos x\), is positive because a negative divided by a negative is positive.

These sign rules guide the correct calculation of \(\cos x\), as described earlier. So, given \(\cos^2 x = \frac{3}{4}\), the negative value \(\cos x = -\frac{\sqrt{3}}{2}\) is chosen. Similarly, using \(\sin x = -\frac{1}{2}\) and \(\cos x = -\frac{\sqrt{3}}{2}\), the tangent can be found as \(\tan x = \frac{1}{\sqrt{3}}\), which simplifies to \(\frac{\sqrt{3}}{3}\). This confirms the positivity of tangent in the third quadrant.

Using Pythagorean Identity

The Pythagorean identity is a crucial tool for solving trigonometric problems, especially when determining unknown trigonometric values from a given one. It builds upon the Pythagorean theorem, linking square measures of sine and cosine to 1, as follows:\[\sin^2 x + \cos^2 x = 1\]When you know one function, like \(\sin x = -\frac{1}{2}\), you can find others by rearranging and substituting into this identity. Squaring \(\sin x\) gives \(\frac{1}{4}\), then setting \(\sin^2 x + \cos^2 x = 1\) implies \(\cos^2 x = 1 - \frac{1}{4} = \frac{3}{4}\). After determining \(\cos^2 x\), take the square root to find \(\cos x\), considering the appropriate sign based on the quadrant. This method exemplifies how fundamental trigonometric identities interconnect and allow for calculating unknown trig functions in various scenarios.