Math Problem Statement

Find the value of each of the six trigonometric functions of the angle θ in the figure.

Solution

The given image shows a right triangle, where the hypotenuse is 13\sqrt{13}, the adjacent side to angle θ\theta is 2, and the opposite side can be determined using the Pythagorean theorem. Let’s calculate the six trigonometric functions for θ\theta.

Step 1: Use the Pythagorean Theorem

The formula is: Hypotenuse2=(Adjacent side)2+(Opposite side)2\text{Hypotenuse}^2 = (\text{Adjacent side})^2 + (\text{Opposite side})^2 Substituting the values: (13)2=22+(Opposite side)2(\sqrt{13})^2 = 2^2 + (\text{Opposite side})^2 13=4+(Opposite side)213 = 4 + (\text{Opposite side})^2 Opposite side2=9Opposite side=3\text{Opposite side}^2 = 9 \quad \Rightarrow \quad \text{Opposite side} = 3

Now the triangle's sides are:

  • Adjacent: 22
  • Opposite: 33
  • Hypotenuse: 13\sqrt{13}

Step 2: Trigonometric Functions

  1. Sine (sinθ\sin \theta): sinθ=OppositeHypotenuse=313=31313(rationalized)\sin \theta = \frac{\text{Opposite}}{\text{Hypotenuse}} = \frac{3}{\sqrt{13}} = \frac{3\sqrt{13}}{13} \quad (\text{rationalized})

  2. Cosine (cosθ\cos \theta): cosθ=AdjacentHypotenuse=213=21313(rationalized)\cos \theta = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{2}{\sqrt{13}} = \frac{2\sqrt{13}}{13} \quad (\text{rationalized})

  3. Tangent (tanθ\tan \theta): tanθ=OppositeAdjacent=32\tan \theta = \frac{\text{Opposite}}{\text{Adjacent}} = \frac{3}{2}

  4. Cosecant (cscθ\csc \theta): cscθ=1sinθ=133\csc \theta = \frac{1}{\sin \theta} = \frac{\sqrt{13}}{3}

  5. Secant (secθ\sec \theta): secθ=1cosθ=132\sec \theta = \frac{1}{\cos \theta} = \frac{\sqrt{13}}{2}

  6. Cotangent (cotθ\cot \theta): cotθ=1tanθ=23\cot \theta = \frac{1}{\tan \theta} = \frac{2}{3}

Final Answer:

sinθ=31313,cosθ=21313,tanθ=32\sin \theta = \frac{3\sqrt{13}}{13}, \quad \cos \theta = \frac{2\sqrt{13}}{13}, \quad \tan \theta = \frac{3}{2} cscθ=133,secθ=132,cotθ=23\csc \theta = \frac{\sqrt{13}}{3}, \quad \sec \theta = \frac{\sqrt{13}}{2}, \quad \cot \theta = \frac{2}{3}

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Related Questions:

  1. How can you find trigonometric functions for a given angle in any triangle?
  2. What is the significance of rationalizing fractions in trigonometry?
  3. How are the reciprocal functions (e.g., secant, cosecant) related to sine and cosine?
  4. How can you use trigonometric ratios to solve for missing angles in a triangle?
  5. What are some real-world applications of trigonometry?

Tip:

Always verify the triangle’s side lengths using the Pythagorean theorem to avoid errors in trigonometric calculations.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Right Triangle
Trigonometric Ratios

Formulas

Pythagorean Theorem: a² + b² = c²
sin(θ) = Opposite / Hypotenuse
cos(θ) = Adjacent / Hypotenuse
tan(θ) = Opposite / Adjacent
csc(θ) = 1 / sin(θ)
sec(θ) = 1 / cos(θ)
cot(θ) = 1 / tan(θ)

Theorems

Pythagorean Theorem

Suitable Grade Level

Grades 9-12