Problem 33
Question
Find the values of the six trigonometric functions of \(\theta\). Constraint \(\theta\) lies in Quadrant II. \(\theta\) lies in Quadrant III. \(\sin \theta < 0\) \(\cot \theta < 0\) \(0 \leq \theta \leq \pi\) \(\frac{\pi}{2} \leq \theta \leq \frac{3 \pi}{2}\) \(\frac{\pi}{2} \leq \theta \leq \frac{3 \pi}{2}\) \(\pi \leq \theta \leq 2 \pi\) Function Value $$\tan \theta=-\frac{15}{8}$$
Step-by-Step Solution
Verified Answer
The six trigonometric function values for \(\theta\) are \(\sin \theta = \frac{15}{17}\), \(\cos \theta = -\frac{8}{17}\), \(\csc \theta = \frac{17}{15}\), \(\sec \theta = -\frac{17}{8}\), \(\cot \theta = \frac{8}{15}\), and \(\tan \theta = -\frac{15}{8}\).
1Step 1: Determine \(\sin \theta\) and \(\cos \theta\) values
Since the exercise gives \(\tan \theta = -\frac{15}{8}\), and \(\tan \theta = \frac{\sin \theta}{\cos \theta}\), we first divise an association with a right triangle. The absolute values of the given \(\tan \theta\) can represent the opposite and adjacent sides of a right triangle, and so \(\sin \theta = \frac{15}{17}\) and \(\cos \theta = -\frac{8}{17}\) are derived through Pythagorean theorem. The negative cosine value arises because \(\theta\) is indicated to lie in the second quadrant where cosine is negative.
2Step 2: Determine \(\csc \theta\), \(\sec \theta\) and \(\cot \theta\) values
The co-functions can be found easily now as they are reciprocal of the basic trigonometric functions. Thus, \(\csc \theta = \frac{1}{\sin \theta} = \frac{17}{15}\), \(\sec \theta = \frac{1}{\cos \theta} = -\frac{17}{8}\), and \(\cot \theta = \frac{1}{\tan \theta} = \frac{8}{15}\).
3Step 3: Determine \(\tan \theta\) value
Since we started with it, the \(\tan \theta\) value is already known to be \(-\frac{15}{8}\).
Key Concepts
Quadrants in trigonometryPythagorean theorem applicationsTrigonometric identitiesReciprocal trigonometric functions
Quadrants in trigonometry
In trigonometry, the coordinate plane is divided into four sections called quadrants. These are labeled as I, II, III, and IV, starting from the positive x-axis, and moving counter-clockwise. Each quadrant has different rules for the signs of trigonometric functions.
For instance, in Quadrant I, all trigonometric functions are positive. In Quadrant II, where \(\theta\) lies for part of the exercise, only \(\sin\) and \(\csc\), its reciprocal function, are positive. Quadrant III, where \(\theta\) also lies for part of the exercise, is the environment where \(\tan\) and \(\cot\), another pair of reciprocals, are positive. \(\cos\) and \(\sin\) are both negative in Quadrant III. Understanding which quadrant an angle lies in is crucial, as this determines the sign of the trigonometric functions.
For instance, in Quadrant I, all trigonometric functions are positive. In Quadrant II, where \(\theta\) lies for part of the exercise, only \(\sin\) and \(\csc\), its reciprocal function, are positive. Quadrant III, where \(\theta\) also lies for part of the exercise, is the environment where \(\tan\) and \(\cot\), another pair of reciprocals, are positive. \(\cos\) and \(\sin\) are both negative in Quadrant III. Understanding which quadrant an angle lies in is crucial, as this determines the sign of the trigonometric functions.
Pythagorean theorem applications
The Pythagorean theorem is a principal theory in mathematics, especially in trigonometry; it states that in a right triangle, the square of the length of the hypotenuse (the side opposite the right angle) is equal to the sum of the squares of the lengths of the other two sides. This relationship is expressed as \(a^2 + b^2 = c^2\).
In trigonometric applications, this theorem helps to find the values of \(\sin\), \(\cos\), and \(\tan\) functions based on the sides of a right triangle. When we are given \(\tan \theta = -\frac{15}{8}\), we can think of a triangle with legs of 8 and 15, and use the Pythagorean theorem to find the hypotenuse. This is represented by \(8^2 + 15^2 = c^2\), or \(c = 17\), which is then used to find the other trigonometric functions.
In trigonometric applications, this theorem helps to find the values of \(\sin\), \(\cos\), and \(\tan\) functions based on the sides of a right triangle. When we are given \(\tan \theta = -\frac{15}{8}\), we can think of a triangle with legs of 8 and 15, and use the Pythagorean theorem to find the hypotenuse. This is represented by \(8^2 + 15^2 = c^2\), or \(c = 17\), which is then used to find the other trigonometric functions.
Trigonometric identities
Trigonometric identities are equations involving trigonometric functions that are true for all values of the occurring variables. They play a vital role in simplifying trigonometric expressions and solving equations. The most basic trigonometric identities are the reciprocal identities, quotient identities, and Pythagorean identities.
For instance, reciprocal identities express the relation between basic and reciprocal functions like \(\csc \theta = \frac{1}{\sin \theta}\) and \(\sec \theta = \frac{1}{\cos \theta}\). These identities reflect the inherent connection between the six trigonometric functions and are used to transition from one function to another without changing the value of the angle.
For instance, reciprocal identities express the relation between basic and reciprocal functions like \(\csc \theta = \frac{1}{\sin \theta}\) and \(\sec \theta = \frac{1}{\cos \theta}\). These identities reflect the inherent connection between the six trigonometric functions and are used to transition from one function to another without changing the value of the angle.
Reciprocal trigonometric functions
Reciprocal trigonometric functions are counterparts to the basic sine, cosine, and tangent functions. For every trigonometric function, there is another that is its reciprocal. Specifically, the reciprocal of \(\sin \theta\) is \(\csc \theta\), for \(\cos \theta\) it is \(\sec \theta\), and for \(\tan \theta\) it is \(\cot \theta\).
Understanding these relationships is fundamental in trigonometry because it expands the ways we can solve problems and analyze angles. For example, if \(\sin \theta\) is known, one can instantly find \(\csc \theta\) by taking its reciprocal, and the same is true for the other pairings. This concept is beautifully illustrated in the exercise, where after determining \(\sin \theta\) and \(\cos \theta\), the reciprocals are calculated to find \(\csc \theta\), \(\sec \theta\), and \(\cot \theta\) accordingly.
Understanding these relationships is fundamental in trigonometry because it expands the ways we can solve problems and analyze angles. For example, if \(\sin \theta\) is known, one can instantly find \(\csc \theta\) by taking its reciprocal, and the same is true for the other pairings. This concept is beautifully illustrated in the exercise, where after determining \(\sin \theta\) and \(\cos \theta\), the reciprocals are calculated to find \(\csc \theta\), \(\sec \theta\), and \(\cot \theta\) accordingly.
Other exercises in this chapter
Problem 33
When an airplane leaves the runway, its angle of climb is \(18^{\circ}\) and its speed is 275 feet per second. Find the plane's altitude after 1 minute.
View solution Problem 33
Use a graphing utility to graph the function (include two full periods). Graph the corresponding reciprocal function in the same viewing window. Describe and co
View solution Problem 33
Use a calculator to evaluate each function. Round your answers to four decimal places. (Be sure the calculator is in the correct angle mode.) (a) \(\sec 42^{\ci
View solution Problem 33
Sketch the graphs of \(f\) and \(g\) in the same coordinate plane. (Include two full periods.) $$\begin{aligned} &f(x)=\sin x\\\ &g(x)=-4 \sin x \end{aligned}$$
View solution