Problem 9 - 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. $$ \cos x=\frac{1}{3}, \quad x \in\left[-\frac{\pi}{2}, 0\right] $$

Step-by-Step Solution

Answer

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

Step 1: Identify the Given Information

We are given that \( \cos x = \frac{1}{3} \) and that \( x \) is in the interval \([-\frac{\pi}{2}, 0]\). Our task is to find \( \sin x \) and \( \tan x \).

Step 2: Use the Pythagorean Identity

The identity \( \sin^2 x + \cos^2 x = 1 \) will help us find \( \sin x \). Substitute \( \cos x = \frac{1}{3} \) into the equation: \[ \sin^2 x + \left(\frac{1}{3}\right)^2 = 1 \] Simplify to get: \[ \sin^2 x + \frac{1}{9} = 1 \] Subtract \( \frac{1}{9} \) from both sides:\[ \sin^2 x = \frac{8}{9} \] Take the square root of both sides: \[ \sin x = \pm \sqrt{\frac{8}{9}} = \pm \frac{2\sqrt{2}}{3} \]

Step 3: Determine the Sign of sin x

Since \( x \in \left[-\frac{\pi}{2}, 0\right] \), \( x \) is in the fourth quadrant where the sine function is negative. Thus, \[ \sin x = -\frac{2\sqrt{2}}{3} \]

Step 4: Calculate tan x

The tangent function is the ratio of sine to cosine: \[ \tan x = \frac{\sin x}{\cos x} \]Using our found values:\[ \tan x = \frac{-\frac{2\sqrt{2}}{3}}{\frac{1}{3}} = -2\sqrt{2} \]

Key Concepts

Pythagorean Identity

The Pythagorean identity is one of the fundamental trigonometric identities and is very useful for solving many problems involving angles and trigonometric functions. It is given by the equation: \[ \sin^2 x + \cos^2 x = 1 \]This identity relates the squares of the sine and cosine functions of the same angle, which makes it an essential tool for finding one trigonometric function when another is known.
For example, if you know the value of \(\cos x\), you can find \(\sin x\) using this identity by rearranging the equation as:\[ \sin^2 x = 1 - \cos^2 x \]Then, you take the square root of both sides to solve for \(\sin x\), also considering whether the value should be positive or negative based on the angle's quadrant.
In our given exercise, knowing \(\cos x = \frac{1}{3}\) allowed us to find \(\sin x\) because both sine and cosine are linked through this identity.

Sine Function

The sine function, often denoted as \(\sin x\), is a crucial part of trigonometry that describes the ratio of the opposite side to the hypotenuse in a right triangle. It's a periodic function that varies between -1 and 1 as the angle moves from 0 to a full circle (or \(2\pi\) radians). In our specific task, we needed to determine the value of \(\sin x\) given \(\cos x = \frac{1}{3}\) and the angle located in the interval \([-\frac{\pi}{2}, 0]\). This interval indicates that our angle is in the fourth quadrant of the unit circle.
Since the sine function is negative in the fourth quadrant, when we calculated \(\sin x\), it resulted in \(-\frac{2\sqrt{2}}{3}\). This reasoning ensures the value aligns with both the magnitude and sign of sine in that specific region.

Cosine Function

The cosine function, or \(\cos x\), is similar to sine, as it describes the ratio of the adjacent side to the hypotenuse of a right triangle. Its range is also between -1 and 1, and like sine, it follows a periodic cycle. In this exercise, we were directly given \(\cos x = \frac{1}{3}\), allowing us to use this known value to determine \(\sin x\) via the Pythagorean identity.
Cosine values help in determining the quadrant in which the angle lies because of their periodicity and symmetry:

• In the first and fourth quadrants, \(\cos x\) is positive.
• In the second and third quadrants, \(\cos x\) is negative.

Knowing \(\cos x = \frac{1}{3}\) and the angle being in the interval \([-\frac{\pi}{2}, 0]\) confirms that we're working in the fourth quadrant.

Tangent Function

The tangent function, expressed as \(\tan x\), is the ratio of the sine function to the cosine function: \[ \tan x = \frac{\sin x}{\cos x} \]This identity is helpful when you need to find the tangent after determining both sine and cosine, as we did in the exercise. Once we found \(\sin x = -\frac{2\sqrt{2}}{3}\) and \(\cos x = \frac{1}{3}\), we easily calculated \(\tan x\): \[ \tan x = \frac{-\frac{2\sqrt{2}}{3}}{\frac{1}{3}} = -2\sqrt{2} \]The sign of \(\tan x\) depends on the signs of \(\sin x\) and \(\cos x\):

• If both are positive or negative, \(\tan x\) is positive.
• If one is positive and the other is negative, \(\tan x\) is negative.

In our calculations, since \(\sin x\) was negative and \(\cos x\) was positive in the fourth quadrant, \(\tan x\) became negative, as expected.