Solving for 2cosθ - 1: When tan2θtan3θ 1 and θ is a Positive Acute Angle

In this article, we will delve into solving the equation 2cosθ - 1 given the condition tan2θ tan3θ 1. This problem involves a positive acute angle θ, which implies that both 2θ and 3θ are angles within a specific range. We will use trigonometric identities and algebraic manipulation to derive the value of 2cosθ - 1.

Solving the Equation: tan2θtan3θ 1

We begin by setting up the given condition: tan2θ tan3θ 1. To proceed, we will assume that tanθ k. This approach simplifies our calculations and allows us to express tan2θ and tan3θ in terms of k.

Recall the double-angle and triple-angle formulas for tangent:

Substitute these expressions into the given equation:

(2k / (1 - k2)) * ((3k - k3) / (1 - 3k2)) 1

Multiply the numerators and denominators:

(2k * (3k - k3)) / ((1 - k2)(1 - 3k2)) 1

Expand and simplify the equation:

(6k2 - 2k?) / (1 - 4k2 3k?) 1

Multiply both sides by the denominator to clear the fraction:

6k2 - 2k? 1 - 4k2 3k?

Combine like terms and rearrange:

5k? - 10k2 1 0

This is a quadratic equation in terms of k2. Let's substitute x k2:

5x2 - 1 1 0

Solve this quadratic equation using the quadratic formula x [-b ± √(b2 - 4ac)] / 2a:

x [10 ± √(100 - 20)] / 10

x [10 ± √80] / 10

x [10 ± 4√5] / 10

x 1 ± 2√5 / 5

Since x k2, we have:

Deriving 2cosθ - 1

Now that we have the values for k2, we can find the value of 2cosθ - 1. Recall that cosθ 1 / √(1 tan2θ) 1 / √(1 k2):

For k2 1 2√5 / 5:

For k2 1 - 2√5 / 5:

Thus, we have two possible values for 2cosθ - 1: