Problem 18 - Precalculus with Limits

Question

In Exercises 11 - 24, use the given values to evaluate (if possible)all six trigonometric functions. \( \cos\left(\dfrac{\pi}{2} - x\right) = \dfrac{3}{5} \), \( \cos x = \dfrac{4}{5} \)

Step-by-Step Solution

Answer

All six trigonometric functions for \( x \) and \( \dfrac{\pi}{2} - x \) are as follows: \n 1. For \( x \), we have \( \sin x = \dfrac{3}{5}, \cos x = \dfrac{4}{5}, \tan x = \dfrac{3}{4}, \cot x = \dfrac{4}{3}, \csc x = \dfrac{5}{3}, \sec x = \dfrac{5}{4} \) \n 2. For \( \dfrac{\pi}{2} - x \), we have \( \sin\left(\dfrac{\pi}{2} - x\right) = \dfrac{4}{5}, \cos\left(\dfrac{\pi}{2} - x\right) = \dfrac{3}{5}, \tan\left(\dfrac{\pi}{2} - x\right) = \dfrac{4}{3}, \cot\left(\dfrac{\pi}{2} - x\right) = \dfrac{3}{4}, \csc\left(\dfrac{\pi}{2} - x\right) = \dfrac{5}{4}, \sec\left(\dfrac{\pi}{2} - x\right) = \dfrac{5}{3} \)

Step 1: Calculate \( \sin \left(\dfrac{\pi}{2} - x\right) \) and \( \sin x \)

Using the Pythagorean identity \( \sin^2\theta + \cos^2\theta = 1 \), we can solve for \( \sin\theta \) by subtracting \( \cos^2\theta \) from both sides and then taking the square root. We have \( \sin\left(\dfrac{\pi}{2} - x\right) = \sqrt{1 - \cos^2\left(\dfrac{\pi}{2} - x\right)} = \sqrt{1 - \left(\dfrac{3}{5}\right)^2} = \dfrac{4}{5} \), \n and \( \sin x = \sqrt{1 - \cos^2x} = \sqrt{1 - \left(\dfrac{4}{5}\right)^2} = \dfrac{3}{5} \)

Step 2: Compute the remaining trigonometric functions of \( x \) and \( \dfrac{\pi}{2} - x \)

The remaining trigonometric functions can be calculated as follows: \n 1. \( \tan\left(\dfrac{\pi}{2} - x\right) = \dfrac{\sin\left(\dfrac{\pi}{2} - x\right)}{\cos\left(\dfrac{\pi}{2} - x\right)} = \dfrac{4}{3} \) \n 2. \( \cot\left(\dfrac{\pi}{2} - x\right) = \dfrac{1}{\tan\left(\dfrac{\pi}{2} - x\right)} = \dfrac{3}{4} \) \n 3. \( \csc\left(\dfrac{\pi}{2} - x\right) = \dfrac{1}{\sin\left(\dfrac{\pi}{2} - x\right)} = \dfrac{5}{4} \) \n 4. \( \sec\left(\dfrac{\pi}{2} - x\right) = \dfrac{1}{\cos\left(\dfrac{\pi}{2} - x\right)} = \dfrac{5}{3} \) \n 5. \( \tan x = \dfrac{\sin x}{\cos x} = \dfrac{3}{4} \) \n 6. \( \cot x = \dfrac{1}{\tan x} = \dfrac{4}{3} \) \n 7. \( \csc x = \dfrac{1}{\sin x} = \dfrac{5}{3} \) \n 8. \( \sec x = \dfrac{1}{\cos x} = \dfrac{5}{4} \)

Key Concepts

Pythagorean Identity

The Pythagorean identity is a fundamental relation in trigonometry that describes a link between the sine and cosine functions for any angle \theta. The identity is expressed as:\
\
\[\begin{equation} \sin^2\theta + \cos^2\theta = 1 \end{equation}\]\
\This means that for any angle, the square of the sine plus the square of the cosine of that angle will always equal 1. This relationship is derived from the Pythagorean theorem, which applies to right triangles.\
\In the step by step solution provided, the Pythagorean identity is used to calculate the sine values from the given cosine values. Since we know that \(\cos\left(\frac{\pi}{2} - x\right)\) is \(\frac{3}{5}\) and \(\cos x\) is \(\frac{4}{5}\), we can use the identity to solve for their respective sine values by subtracting the square of the cosine from 1 and taking the square root:\
\begin{itemize}\

\(\sin\left(\frac{\pi}{2} - x\right) = \sqrt{1 - \left(\frac{3}{5}\right)^2}\)

\

\(\sin x = \sqrt{1 - \left(\frac{4}{5}\right)^2}\)

\begin{itemize}\
\These calculated sine values are essential in finding other trigonometric functions, such as tangent, cotangent, cosecant, and secant for the angles \(x\) and \(\frac{\pi}{2} - x\).

Evaluating Trigonometric Functions

Evaluating trigonometric functions involves finding the values of sine, cosine, tangent, and other trigonometric ratios for a given angle. The process may require using basic definitions, identities like the Pythagorean identity, and sometimes properties of inverse trigonometric functions.\
\In our exercise, we begin with known values of cosine for angles \(x\) and \(\frac{\pi}{2} - x\), from which we can determine the values of the remaining trigonometric functions. Once the sine is found using the Pythagorean identity, as shown in the previous section, the tangent is calculated by dividing sine by cosine, cotangent as the reciprocal of tangent, secant as the reciprocal of cosine, and cosecant as the reciprocal of sine:\
\begin{itemize}\

\(\tan\theta = \frac{\sin\theta}{\cos\theta}\)

\

\(\cot\theta = \frac{1}{\tan\theta}\)

\

\(\csc\theta = \frac{1}{\sin\theta}\)

\

\(\sec\theta = \frac{1}{\cos\theta}\)

\begin{itemize}\
\For students aiming to understand these relationships better, it is advised to practice sketching the unit circle and familiarize themselves with the sine and cosine values for commonly known angles, which can then be utilized for evaluating the functions for related angles.

Inverse Trigonometric Functions

Inverse trigonometric functions are key when we wish to find the angle that corresponds to a specific trigonometric ratio. The main inverse functions include \(\arcsin\), \(\arccos\), and \(\arctan\), which are the inverses of sine, cosine, and tangent, respectively.\
\For example, if you know \(\sin x = \frac{1}{2}\), then \(x = \arcsin\left(\frac{1}{2}\right)\) will give you the value of angle \(x\) whose sine is \(\frac{1}{2}\). It's important to remember that because there are multiple angles with the same sine, cosine, or tangent values (due to periodicity), the outputs of inverse trigonometric functions are generally restricted to specific ranges to provide a single value.\
\

Real-World Application

\Inverse trigonometric functions are often used in various real-world scenarios, such as in physics to determine angles of vectors, in geometry to solve for angles in a triangle, and in engineering for control systems and signal analysis. Being proficient in these functions equips students with the mathematical tools necessary to solve practical problems involving angles and ratios.\
\The inverse trigonometric functions were not directly used in the step by step solution for our exercise, but understanding them is crucial when working with trigonometry as they offer a method to work backwards from knowing a function's value to finding the angle.