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Chapter 5: Problem 127

Find the five remaining trigonometric finction values for each angle. \(\tan \theta=-\frac{15}{8},\) and \(\theta\) is in quadrant II.

Short Answer

Expert verified
\( \sin \theta = \frac{15}{17}, \cos \theta = \frac{-8}{17}, \csc \theta = \frac{17}{15}, \sec \theta = -\frac{17}{8}, \cot \theta = -\frac{8}{15} \)

Step by step solution

01

- Identify the Given Information

It is given that \(\tan \theta = -\frac{15}{8} \) and \( \theta \) is in quadrant II. Note that in quadrant II, sine is positive, cosine is negative, and tangent is negative.
02

- Compute the Hypotenuse

Use the Pythagorean identity to find the hypotenuse \( r \). Let the opposite side be \( y = 15 \) and the adjacent side be \( x = -8 \) (keeping track of signs in quadrant II).\[ r = \sqrt{x^2 + y^2} = \sqrt{(-8)^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17 \]
03

- Compute Sine

Use the definition of sine: \(\frac{\text{opposite}}{\text{hypotenuse}} = \frac{y}{r} \).\[ \sin \theta = \frac{15}{17} \]
04

- Compute Cosine

Use the definition of cosine: \(\frac{\text{adjacent}}{\text{hypotenuse}} = \frac{x}{r} \).\[ \cos \theta = \frac{-8}{17} \]
05

- Compute Cosecant

The cosecant is the reciprocal of sine: \(\frac{1}{\text{sin}} \).\[ \csc \theta = \frac{1}{\frac{15}{17}}= \frac{17}{15} \]
06

- Compute Secant

The secant is the reciprocal of cosine: \(\frac{1}{\text{cos}} \).\[ \sec \theta = \frac{1}{\frac{-8}{17}} = -\frac{17}{8} \]
07

- Compute Cotangent

The cotangent is the reciprocal of tangent: \(\frac{1}{\text{tan}} \).\[ \cot \theta = \frac{1}{-\frac{15}{8}} = -\frac{8}{15} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

tangent
The tangent of an angle in a right triangle is the ratio of the length of the opposite side to the length of the adjacent side. It is represented as: \(\tan \theta = \frac{\text{opposite}}{\text{adjacent}} \). In the given problem, \(\tan \theta = -\frac{15}{8}\). This ratio helps us find other trigonometric functions based on the given angle.
quadrant II
In the coordinate system, each quadrant has specific signs for trigonometric functions. In quadrant II:
  • Sine is positive,
  • Cosine is negative,
  • Tangent is negative.
Understanding the signs is essential for solving trigonometric problems and ensuring correct values for functions.
Pythagorean identity
The Pythagorean identity is a fundamental relation in trigonometry. It states: \[ \text{sin}^2 \theta + \text{cos}^2 \theta = 1 \]. Using this identity helps us find the hypotenuse when we know the opposite and adjacent sides. In the problem, we find the hypotenuse as: \[ r = \sqrt{x^2 + y^2} = 17 \].
sine
Sine of an angle is defined as the ratio of the length of the opposite side to the hypotenuse: \(\text{sin} \theta = \frac{\text{opposite}}{\text{hypotenuse}}\). For our problem, with the opposite side at 15 and hypotenuse at 17:
\[ \text{sin} \theta = \frac{15}{17} \].
cosine
Cosine of an angle is the ratio of the length of the adjacent side to the hypotenuse: \(\text{cos} \theta = \frac{\text{adjacent}}{\text{hypotenuse}}\). Here, with the adjacent side of -8 and hypotenuse of 17:
\[ \text{cos} \theta = \frac{-8}{17} \].
cosecant
Cosecant is the reciprocal of sine: \(\text{csc} \theta = \frac{1}{\text{sin} \theta}\). For the given problem:
\[ \text{csc} \theta = \frac{1}{\frac{15}{17}} = \frac{17}{15} \]. It helps us understand sine in terms of its reciprocal value.
secant
Secant is the reciprocal of cosine: \(\text{sec} \theta = \frac{1}{\text{cos} \theta}\). For \(\theta\) in our problem:
\[ \text{sec} \theta = \frac{1}{\frac{-8}{17}} = -\frac{17}{8} \]. This reciprocal relationship helps in comprehending the negative value of cosine.
cotangent
Cotangent is defined as the reciprocal of tangent: \(\text{cot} \theta = \frac{1}{\text{tan} \theta}\). Considering the provided \(\tan \theta = -\frac{15}{8}\):
\[ \text{cot} \theta = \frac{1}{-\frac{15}{8}} = -\frac{8}{15} \]. Using cotangent helps in further confirming the negative tangent value.

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Most popular questions from this chapter

Work each problem. In these exercises, assume the course of a plane or ship is on the indicated bearing. The bearing from Winston-Salem, North Carolina, to Danville, Virginia, is \(N 42^{\circ}\) E. The bearing from Danville to Goldsboro, North Carolina, is \(S 48^{\circ}\) E. A car driven by Ellen Winchell, traveling at 65 mph, takes \(1.1 \mathrm{hr}\) to go from Winston-Salem to Danville and \(1.8 \mathrm{hr}\) to go from Danville to Goldsboro. Find the distance from Winston-Salem to Goldsboro.

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