Understanding Trigonometric Equations and Identities: cos2α

In this article, we explore the solution to a particular trigonometric equation and the application of trigonometric identities, specifically focusing on the value of cos2α. We will solve the equation (frac{4sin alpha - cos alpha}{2sin alpha cos alpha} 3) and derive the value of cos2α. Let's delve into the step-by-step process and understand the underlying concepts.

Solving the Given Equation

We start with the equation:

(frac{4sin alpha - cos alpha}{2sin alpha cos alpha} 3)

Multiplying both sides by (2sin alpha cos alpha), we get:

4sin alpha - cos alpha 6sin alpha cos alpha)

Shifting all terms to one side:

4sin alpha - 6sin alpha cos alpha - cos alpha 0)

Factoring out common terms, we obtain:

(4sin alpha - cos alpha 6sin alpha cos alpha)

Deriving tanα

A simpler approach to solving the equation is by dividing both the numerator and the denominator by cosα:

(frac{4sin alpha - cos alpha}{2sin alpha cos alpha} frac{4tan alpha - 1}{2tan alpha} 3)

Now, solving for (tan alpha):

(frac{4tan alpha - 1}{2tan alpha} 3)

Multiplying both sides by 2tan alpha:

4tan alpha - 1 6tan alpha)

Shifting terms:

-2tan alpha 1)

Thus, (tan alpha -frac{1}{2})

Determining cos2α

With (tan alpha -frac{1}{2}), we can use the double-angle identity for cosine, which states:

(cos 2alpha frac{1 - tan^2 alpha}{1 tan^2 alpha})

Substituting (tan alpha -frac{1}{2}):

(cos 2alpha frac{1 - left(-frac{1}{2}right)^2}{1 left(-frac{1}{2}right)^2} frac{1 - frac{1}{4}}{1 frac{1}{4}} frac{frac{3}{4}}{frac{5}{4}} frac{3}{5})

Alternative Methods and Verification

Let's also verify the solution using the alternative method from the given equations:

(4sin x - cos x 6sin x cos x)

Dividing both sides by cosx:

(frac{4sin x - cos x}{cos x} 6sin x)

(4tan x - 1 6sin x)

Since (tan x -2), we get:

(4(-2) - 1 6sin x)

(-8 - 1 6sin x)

(-9 6sin x)

(sin x -frac{3}{2})

However, this leads to a contradiction since (sin x) must be within the range [-1, 1]. The correct approach is to find (tan x -2). From this, we can derive:

(tan^2 x 4)

(1 tan^2 x 5)

(sec^2 x 5)

(cos^2 x frac{1}{5})

(cos 2x 2cos^2 x - 1 2left(frac{1}{5}right) - 1 frac{2}{5} - 1 -frac{3}{5})

Thus, (cos 2alpha -frac{3}{5})

Conclusion

In this article, we have explored the solution to the trigonometric equation (frac{4sin alpha - cos alpha}{2sin alpha cos alpha} 3) and derived the value of cos2α. We have used both the direct method of factoring and the alternative method of using the double-angle identity for cosine. Understanding these identities and methods is crucial for solving complex trigonometric equations and deriving specific trigonometric values like cos2α.

Keywords: trigonometric identities, trigonometric equations, cosine double angle