How to Find the Minimum Value of sin x cos x

To find the minimum value of the function ( sin x cos x ), we can explore a few different methods, including a strategic substitution and the use of calculus to identify critical points. This detailed guide explores both analytic and numerical approaches.

Method 1: Substitution and Simplification

Welcome to the task of finding the minimum value of ( sin x cos x ). Let's begin by simplifying the function:

We can rewrite the function:

sin x cos x (frac{2}{2 sin x cos x}) (frac{2 sin x cos x}{2 sin x cos x})

This simplification can be further manipulated by substituting ( x frac{pi}{4} ), where (sin x cos x frac{sqrt{2}}{2}). This transforms the function to:

(sin x cos x sqrt{2} cos y)

where (2 sin x cos x sin 2x cos 2y). Therefore, the critical value can be found by analyzing the function:

(f(z) z frac{2}{z^2 - 1})

By introducing a substitution (z sqrt{2} cos y), we get (-sqrt{2} leq z leq sqrt{2}), excluding (z 1), as it makes the denominator zero. The function simplifies to:

g(z) z frac{2}{z^2 - 1} z frac{2}{z - 1})

Method 2: Analytical Approach with Derivatives

To identify the critical points, we take the derivative of (g(z)) and set it to zero:

(g'(z) frac{2(z - 1) - 2z}{(z - 1)^2} frac{-2}{(z - 1)^2})

Setting (g'(z) 0), we observe that there are no real solutions since the numerator is a constant. However, we can use the first derivative test to determine the behavior of the function around the critical points.

By analyzing the function (g(z)), we find that the minimum value occurs at (z 1 - sqrt{2}). Evaluating the function at this point:

(g(1 - sqrt{2}) 2sqrt{2} - 1)

Verification and Numerical Methods

Wolfram Alpha confirms the numerical result, providing an approximate answer of 1.828. If we do not have access to computational tools, we can employ numerical methods like the gradient descent algorithm to approximate the minimum value.

Behavior of Trigonometric Functions

Consider the function (f(x) sin x cos x tan xcos xsec xcsc x). This function is undefined at integer multiples of (frac{pi}{2}) and is continuous on intervals not containing these values.

Since these functions are periodic with a period of (2pi), we only need to consider the intervals: