Math Problem Statement

Find the exact value of each of the remaining trigonometric functions of theta. Rationalize denominators when applicable. tan theta =8/15, given that theta is in quadrant III

Solution

We are given that tan(θ)=815\tan(\theta) = \frac{8}{15} and that θ\theta is in Quadrant III. We need to find the exact values of the other trigonometric functions: sin(θ)\sin(\theta), cos(θ)\cos(\theta), sec(θ)\sec(\theta), csc(θ)\csc(\theta), and cot(θ)\cot(\theta).

Step 1: Use the identity tan(θ)=sin(θ)cos(θ)\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}

We know: tan(θ)=sin(θ)cos(θ)=815\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} = \frac{8}{15} So we have: sin(θ)cos(θ)=815\frac{\sin(\theta)}{\cos(\theta)} = \frac{8}{15}

Step 2: Use the Pythagorean identity

The Pythagorean identity states: sin2(θ)+cos2(θ)=1\sin^2(\theta) + \cos^2(\theta) = 1

To use this identity, we need to express sin(θ)\sin(\theta) and cos(θ)\cos(\theta) in terms of a hypotenuse.

Step 3: Find sin(θ)\sin(\theta) and cos(θ)\cos(\theta) using a right triangle

Since tan(θ)=oppositeadjacent\tan(\theta) = \frac{\text{opposite}}{\text{adjacent}}, we can construct a right triangle where the opposite side is 8 and the adjacent side is 15. Using the Pythagorean theorem to find the hypotenuse rr: r=82+152=64+225=289=17r = \sqrt{8^2 + 15^2} = \sqrt{64 + 225} = \sqrt{289} = 17

Step 4: Determine the signs of sin(θ)\sin(\theta) and cos(θ)\cos(\theta)

Since θ\theta is in Quadrant III, both sin(θ)\sin(\theta) and cos(θ)\cos(\theta) are negative. Therefore: sin(θ)=817,cos(θ)=1517\sin(\theta) = \frac{-8}{17}, \quad \cos(\theta) = \frac{-15}{17}

Step 5: Calculate the remaining trigonometric functions

Now, we can find the remaining trigonometric functions:

  1. sec(θ)\sec(\theta): Reciprocal of cos(θ)\cos(\theta) sec(θ)=1cos(θ)=11517=1715\sec(\theta) = \frac{1}{\cos(\theta)} = \frac{1}{-\frac{15}{17}} = -\frac{17}{15}

  2. csc(θ)\csc(\theta): Reciprocal of sin(θ)\sin(\theta) csc(θ)=1sin(θ)=1817=178\csc(\theta) = \frac{1}{\sin(\theta)} = \frac{1}{-\frac{8}{17}} = -\frac{17}{8}

  3. cot(θ)\cot(\theta): Reciprocal of tan(θ)\tan(\theta) cot(θ)=1tan(θ)=1815=158\cot(\theta) = \frac{1}{\tan(\theta)} = \frac{1}{\frac{8}{15}} = \frac{15}{8}

Final Results

  • sin(θ)=817\sin(\theta) = \frac{-8}{17}
  • cos(θ)=1517\cos(\theta) = \frac{-15}{17}
  • sec(θ)=1715\sec(\theta) = \frac{-17}{15}
  • csc(θ)=178\csc(\theta) = \frac{-17}{8}
  • cot(θ)=158\cot(\theta) = \frac{15}{8}

Would you like any further details on these calculations?


Related Questions:

  1. What is the geometric interpretation of trigonometric functions in each quadrant?
  2. How does the Pythagorean identity relate to the unit circle?
  3. Why are sine and cosine both negative in Quadrant III?
  4. What are the general formulas for trigonometric functions in different quadrants?
  5. How does the reciprocal identity help in finding trigonometric functions?

Tip: When solving trigonometric problems in different quadrants, always remember the sign rules for each function in the four quadrants.

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Math Problem Analysis

Mathematical Concepts

Trigonometry

Formulas

Pythagorean identity
Reciprocal identities

Theorems

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Suitable Grade Level

High school