Math Problem Statement

Draw a horizontal, vertical, or diagonal line to represent the equation tan , theta, equals, 1, point, 8tan θ=1.8 and then use the line to help you solve that equation on 0, is less than or equal to, theta, is less than, 2, pi, .0≤θ<2π. Express your answer in radians to the nearest tenth.

Solution

To solve the equation tan(θ)=1.8\tan(\theta) = 1.8 on the interval 0θ<2π0 \leq \theta < 2\pi, we need to identify the values of θ\theta where the tangent of θ\theta equals 1.8.

Step 1: Set up the equation

We are given:

tan(θ)=1.8\tan(\theta) = 1.8

This is a trigonometric equation where we need to find the values of θ\theta that satisfy this condition within one full revolution (from 00 to 2π2\pi).

Step 2: Use the inverse tangent (arctan) function

To find the principal value of θ\theta, we can use the inverse tangent:

θ1=tan1(1.8)\theta_1 = \tan^{-1}(1.8)

Let’s compute this:

θ11.071 radians\theta_1 \approx 1.071 \text{ radians}

Step 3: General solution for the tangent function

The tangent function has a period of π\pi, meaning that for any solution θ1\theta_1, the general solutions are:

θ=θ1+nπwherenZ\theta = \theta_1 + n\pi \quad \text{where} \quad n \in \mathbb{Z}

Since we are asked to find solutions in the interval 0θ<2π0 \leq \theta < 2\pi, we consider n=0n = 0 and n=1n = 1.

  1. For n=0n = 0, we get the first solution:

θ1=1.071 radians\theta_1 = 1.071 \text{ radians}

  1. For n=1n = 1, we get the second solution:

θ2=1.071+π1.071+3.1416=4.2126 radians\theta_2 = 1.071 + \pi \approx 1.071 + 3.1416 = 4.2126 \text{ radians}

Step 4: Conclusion

Thus, the two solutions for tan(θ)=1.8\tan(\theta) = 1.8 in the interval 0θ<2π0 \leq \theta < 2\pi are:

θ11.1 radians\theta_1 \approx 1.1 \text{ radians} θ24.2 radians\theta_2 \approx 4.2 \text{ radians}

These are the solutions to the equation in radians, rounded to the nearest tenth.


Would you like more details on how the tangent function works, or do you have any other questions?

Here are some related questions you might find interesting:

  1. What happens to the solutions if tan(θ)=1.8\tan(\theta) = -1.8?
  2. How can you solve trigonometric equations using a unit circle?
  3. What is the period of the sine and cosine functions?
  4. How does the inverse tangent function relate to the unit circle?
  5. How do you graph the tangent function and its periodicity?

Tip: Remember that the tangent function has vertical asymptotes at θ=π2+nπ\theta = \frac{\pi}{2} + n\pi, where nn is an integer, so solutions that fall near these values must be carefully considered.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Tangent Function
Inverse Trigonometric Functions

Formulas

tan(θ) = 1.8
θ = tan^(-1)(1.8)
θ = θ₁ + nπ (General Solution for Tangent)

Theorems

Periodicity of Tangent Function

Suitable Grade Level

Grades 10-12