Problem 25
Question
find the exact value of each of the remaining trigonometric functions of \(\theta\) $$ \sin \theta=\frac{5}{13}, \quad \theta \text { in quadrant } 11 $$
Step-by-Step Solution
Verified Answer
The trigonometric function values are: \(cos \theta = -\frac{12}{13}\), \(tan \theta = -\frac{5}{12}\), \(cot \theta = -\frac{12}{5}\), \(sec \theta = -\frac{13}{12}\), and \(csc \theta = \frac{13}{5}\).
1Step 1: Calculate the Cosine
The Pythagorean identity states that \(sin^2 \theta + cos^2 \theta = 1\). We can rewrite this to find \(cos \theta\) as follows: \(cos^2 \theta =1 - sin^2 \theta\). Substituting \(sin \theta=\frac{5}{13}\) into this formula, we get: \(cos^2 \theta =1 - (\frac{5}{13})^2\). Evaluate that to get \(cos \theta = ±\sqrt{1 - (\frac{5}{13})^2} = ±\frac{12}{13}\). Since \(\theta\) is in quadrant II, \(cos \theta = -\frac{12}{13}\) because cosine is negative in quadrant II.
2Step 2: Calculate the Other Trigonometric Functions
Now we have both \(sin \theta\) and \(cos \theta\) values which we can use to find the other trigonometric function values. The remaining trigonometric functions are \(tan \theta\), \(cot \theta\), \(sec \theta\), and \(csc \theta\).Use the following formulas to find these values:\(tan \theta = \frac{sin \theta}{cos \theta}\),\(cot \theta = \frac{1}{tan \theta}\),\(sec \theta = \frac{1}{cos \theta}\),\(csc \theta = \frac{1}{sin \theta}\).Substitute \(sin \theta = \frac{5}{13}\) and \(cos \theta = -\frac{12}{13}\) into these formulas to calculate the respective function values:\(tan \theta = -\frac{5}{12}\),\(cot \theta = -\frac{12}{5}\),\(sec \theta = -\frac{13}{12}\),\(csc \theta = \frac{13}{5}\).
Key Concepts
Pythagorean IdentityQuadrant II in TrigonometryTrigonometric Function Values
Pythagorean Identity
The Pythagorean identity is a cornerstone in trigonometry, expressing a fundamental relationship between the sine and cosine of an angle. Specifically, it states that for any angle \(θ\), the square of the sine plus the square of the cosine will always equal one:
\[\sin^2(θ) + \cos^2(θ) = 1.\]
This identity comes directly from the Pythagorean theorem applied to a right triangle with a hypotenuse of length 1, known as a unit circle. In the context of solving trigonometric functions for an angle, it allows us to find one trigonometric function value when another is known.
For example, if we know that \(\sin(θ) = \frac{5}{13}\), we can substitute this value into the Pythagorean identity to find \(\cos(θ)\). We first square the sine value and subtract it from 1:
\[\cos^2(θ) = 1 - \left(\frac{5}{13}\right)^2.\]
After calculating, we get \(\cos^2(θ)\) and then find the cosine by taking the square root, considering the sign based on the quadrant in which the angle lies. This is a powerful tool enabling us to unlock other trigonometric function values.
\[\sin^2(θ) + \cos^2(θ) = 1.\]
This identity comes directly from the Pythagorean theorem applied to a right triangle with a hypotenuse of length 1, known as a unit circle. In the context of solving trigonometric functions for an angle, it allows us to find one trigonometric function value when another is known.
For example, if we know that \(\sin(θ) = \frac{5}{13}\), we can substitute this value into the Pythagorean identity to find \(\cos(θ)\). We first square the sine value and subtract it from 1:
\[\cos^2(θ) = 1 - \left(\frac{5}{13}\right)^2.\]
After calculating, we get \(\cos^2(θ)\) and then find the cosine by taking the square root, considering the sign based on the quadrant in which the angle lies. This is a powerful tool enabling us to unlock other trigonometric function values.
Quadrant II in Trigonometry
In trigonometry, the concept of quadrants is critical for determining the signs of trigonometric functions. The coordinate plane is divided into four parts called quadrants, each determined by the sign of the x (cosine) and y (sine) coordinates.
In Quadrant II, where angle \(θ\) resides, the sine function is positive and the cosine function is negative. This is because in this region, the x-coordinates (related to cosine) are negative, while the y-coordinates (related to sine) are positive. Knowing this, we can apply our understanding to the problem at hand. Given \(\sin(θ) = \frac{5}{13}\) and knowing the angle is in Quadrant II, we can deduce that \(\cos(θ)\) must be negative. Thus, after using the Pythagorean identity to find that \(\cos(θ) = ±\frac{12}{13}\), we select the negative value, making \(\cos(θ) = -\frac{12}{13}\). This understanding of quadrant-specific rules is essential for correctly solving trigonometric problems.
In Quadrant II, where angle \(θ\) resides, the sine function is positive and the cosine function is negative. This is because in this region, the x-coordinates (related to cosine) are negative, while the y-coordinates (related to sine) are positive. Knowing this, we can apply our understanding to the problem at hand. Given \(\sin(θ) = \frac{5}{13}\) and knowing the angle is in Quadrant II, we can deduce that \(\cos(θ)\) must be negative. Thus, after using the Pythagorean identity to find that \(\cos(θ) = ±\frac{12}{13}\), we select the negative value, making \(\cos(θ) = -\frac{12}{13}\). This understanding of quadrant-specific rules is essential for correctly solving trigonometric problems.
Trigonometric Function Values
Once the sine and cosine of an angle are known, we can determine the values of all the other trigonometric functions. This includes the tangent (tan), cotangent (cot), secant (sec), and cosecant (csc). These functions are all related to one another through sine and cosine.
Knowing that \(\sin(θ) = \frac{5}{13}\) and \(\cos(θ) = -\frac{12}{13}\) (considering angle \(θ\) is in Quadrant II), we can now calculate:
Knowing that \(\sin(θ) = \frac{5}{13}\) and \(\cos(θ) = -\frac{12}{13}\) (considering angle \(θ\) is in Quadrant II), we can now calculate:
- \(\tan(θ) = \frac{\sin(θ)}{\cos(θ)}\), which simplifies to \(-\frac{5}{12}\),
- \(\cot(θ) = \frac{1}{\tan(θ)}\), which simplifies to \(-\frac{12}{5}\),
- \(\sec(θ) = \frac{1}{\cos(θ)}\), which simplifies to \(-\frac{13}{12}\),
- \(\csc(θ) = \frac{1}{\sin(θ)}\), which simplifies to \(\frac{13}{5}\).
Other exercises in this chapter
Problem 25
Determine the amplitude, period, and phase shift of each function. Then graph one period of the function. $$y=-2 \sin \left(2 x+\frac{\pi}{2}\right)$$
View solution Problem 25
Use an identity to find the value of each expression. Do not use a calculator. $$ \sin 37^{\circ} \mathrm{csc} 37^{\circ} $$
View solution Problem 25
In Exercises \(21-28,\) convert each angle in radians to degrees. $$ \frac{7 \pi}{6} $$
View solution Problem 26
Use a calculator to find the value of each expression rounded to two decimal places. $$ \cos ^{-1} \frac{\sqrt{7}}{10} $$
View solution