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(a^(2)-b^(2))sin Theta+2ab Cos Theta=a^(2)+b^(2)

5 min read Jun 16, 2024
(a^(2)-b^(2))sin Theta+2ab Cos Theta=a^(2)+b^(2)

Solving the Trigonometric Equation: (a^2 - b^2)sin θ + 2ab cos θ = a^2 + b^2

This article will guide you through the steps of solving the trigonometric equation: (a^2 - b^2)sin θ + 2ab cos θ = a^2 + b^2. We will explore the process of finding the solutions for θ within the specified domain, typically [0, 2π) or [0°, 360°).

Understanding the Equation

The equation involves trigonometric functions (sine and cosine) and constants (a and b). Our goal is to isolate θ to find its possible values.

Solution Approach

  1. Divide by (a^2 + b^2): Divide both sides of the equation by √(a^2 + b^2) to simplify the equation.

    (a^2 - b^2)/(√(a^2 + b^2)) sin θ + 2ab/(√(a^2 + b^2)) cos θ = √(a^2 + b^2)

  2. Introducing New Constants: Define two constants:

    • k = (a^2 - b^2)/(√(a^2 + b^2))
    • m = 2ab/(√(a^2 + b^2))

    The equation now becomes:

    k sin θ + m cos θ = √(a^2 + b^2)

  3. Trigonometric Identity: Use the trigonometric identity: sin(θ + α) = sin θ cos α + cos θ sin α.

    • We want to manipulate the left-hand side of our equation to match this identity. To achieve this, divide both sides of the equation by √(k^2 + m^2):

      [k/(√(k^2 + m^2))] sin θ + [m/(√(k^2 + m^2))] cos θ = √(a^2 + b^2)/(√(k^2 + m^2))

    • Notice that the coefficients of sin θ and cos θ now resemble the form of the identity. Define a new constant:

    α = tan⁻¹(m/k)

    • Now, substitute back into the equation:

    sin θ cos α + cos θ sin α = √(a^2 + b^2)/(√(k^2 + m^2))

    • Finally, apply the trigonometric identity:

    sin (θ + α) = √(a^2 + b^2)/(√(k^2 + m^2))

  4. Solving for θ:

    • Isolate sin(θ + α) by taking the inverse sine of both sides.

    • Find the general solution: Since sin function has a period of 2π, the general solution is:

      θ + α = sin⁻¹(√(a^2 + b^2)/(√(k^2 + m^2))) + 2nπ or θ + α = π - sin⁻¹(√(a^2 + b^2)/(√(k^2 + m^2))) + 2nπ, where n is an integer.

    • Solve for θ:

      θ = sin⁻¹(√(a^2 + b^2)/(√(k^2 + m^2))) + 2nπ - α or θ = π - sin⁻¹(√(a^2 + b^2)/(√(k^2 + m^2))) + 2nπ - α

  5. Finding Specific Solutions: Substitute the values of k, m, and α back into the general solution and determine the solutions for θ within the specified domain.

Example:

Let's solve the equation for a = 3, b = 4, and the domain [0, 2π):

  1. Calculate k and m:

    • k = (3^2 - 4^2)/(√(3^2 + 4^2)) = -7/5
    • m = 2 * 3 * 4 / (√(3^2 + 4^2)) = 24/5

  2. Calculate α:

    • α = tan⁻¹(m/k) = tan⁻¹(-24/7) ≈ -73.74°

  3. Substitute into the general solution:

    • θ ≈ sin⁻¹(5/(√(7^2 + 24^2))) + 2nπ + 73.74° ≈ 11.31° + 2nπ
    • θ ≈ π - sin⁻¹(5/(√(7^2 + 24^2))) + 2nπ + 73.74° ≈ 198.69° + 2nπ

  4. Find the solutions within the domain [0, 2π):

    • θ ≈ 11.31°, 371.31°
    • θ ≈ 198.69°, 558.69°

    Only the solutions 11.31° and 198.69° fall within the domain [0, 2π).

Conclusion:

Solving the trigonometric equation (a^2 - b^2)sin θ + 2ab cos θ = a^2 + b^2 involves strategic manipulation, utilizing trigonometric identities and finding the general solution for θ. By substituting specific values for a and b, and considering the specified domain, we can determine the particular solutions for θ within that range.

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