Understanding the Minimum Values of Trigonometric Functions: 2cos x 8sec^2 x

In this article, we will delve into the process of finding the minimum value of the function 2cos x 8sec^2 x. We'll explore the key concepts in trigonometry and calculus, using step-by-step methods to solve this problem effectively.

Introduction to Trigonometric Functions

Trigonometric functions are a fundamental part of mathematics, covering various applications in science, engineering, and everyday life. The function we are analyzing, 2cos x 8sec^2 x, involves two trigonometric functions: the cosine and the secant.

Maximum and Minimum Values

Understanding the maximum and minimum values of functions is crucial in both theoretical and practical contexts. The minimum value of cos x is -1, while the minimum value of sec x (which equals 1/cos x) can go to negative infinity because cos x can approach zero.

Steps to Find the Minimum Value

Step 1: Derive the Function

The first step is to find the derivative of the given function. We start with:

f(x) 2cos x 8sec^2 x

The derivative of this function is:

f'(x) -2sin x 16sec^2 x tan x

Step 2: Set the Derivative Equal to Zero

Next, we set the derivative equal to zero to find the critical points:

-2sin x 16sec^2 x tan x 0

Manipulating the equation:

-2sin x 16sec^3 x sin x 0 sin x - 8sec^3 x 0

This simplifies to:

sin x 0

Solving for x, we get:

x nπ, n is an integer

Step 3: Check Each Critical Point

Now, we must check each critical point to determine the minimum or maximum value:

fx 2cos x 8sec^2 x

Evaluating at x nπ (where n is even):

Let's test x 0:

fx(0) 2cos 0 8sec^2 0 2(1) 8(1) 2 8 10

Evaluating at x 2π:

fx(2π) 2cos 2π 8sec^2 2π 2(1) 8(1) 2 8 10

Step 4: Verify the Minimum Value Using Arithmetic Mean - Geometric Mean Theorem

To confirm the minimum value, we can use the Arithmetic Mean - Geometric Mean inequality (AM-GM inequality). According to the theorem:

f(x) / 3 ≥ (cos x)(cos x)(8sec^2 x)^{1/3}

Given that:

(cos x)(cos x)(8sec^2 x)^{1/3} 8^{1/3} 2

Therefore:

f(x) / 3 ≥ 2 f(x) ≥ 6

Hence, the minimum value of 2cos x 8sec^2 x is 6.

Conclusion

By applying calculus optimization techniques, we have demonstrated that the minimum value of the function 2cos x 8sec^2 x is 6. This process involves finding the derivative, setting it equal to zero, and verifying the result through the AM-GM inequality.