Maximizing the Value of sin(x)cos(x): A Comprehensive Guide

Introduction

In the field of trigonometry, understanding the behavior of various trigonometric functions is crucial for solving complex problems. One such function is the product of sine and cosine, sin(x)cos(x). This article delves into finding the maximum value of sin(x)cos(x), exploring different methods and providing insightful mathematical derivations.

Using Trigonometric Identities for Maximization

One of the most straightforward methods to find the maximum value of sin(x)cos(x) is by utilizing trigonometric identities. Specifically, we can use the identity that states:

sin(x)cos(x) frac{1}{2}sin(2x)

The maximum value of sin(2x) is 1, which occurs when 2x frac{pi}{2} 2kpi for any integer k. Therefore, the maximum value of sin(x)cos(x) is:

frac{1}{2} cdot 1 frac{1}{2}

Thus, the greatest value of sin(x)cos(x) is frac{1}{2}.

Using Calculus for Maximization

To find the maximum value of sin(x)cos(x) using calculus, we can start by defining the function:

F(x) sin(x)cos(x)

To find the critical points, we take the first derivative:

F'(x) cos^2(x) - sin^2(x) 0

From this equation, we get:

cos^2(x) sin^2(x)

Since cos^2(x) sin^2(x) 1, we have:

cos(2x) 0

Solving for x, we get:

2x frac{pi}{2} kpi

Solving for x, we find:

x frac{pi}{4} frac{kpi}{2}

Next, we use the second derivative test to confirm the nature of these critical points:

F''(x) -2cos(x)sin(x) - 2sin(x)cos(x) -4sin(x)cos(x)

For x frac{pi}{4}, we have:

F''left(frac{pi}{4}right) -4sinleft(frac{pi}{4}right)cosleft(frac{pi}{4}right) -4left(frac{sqrt{2}}{2}right)left(frac{sqrt{2}}{2}right) -2

This indicates that F(x) has a local maximum at x frac{3pi}{4} (since the second derivative is negative), and the maximum value is:

Fleft(frac{3pi}{4}right) frac{1}{2}

Using the Arithmetic Mean-Geometric Mean Inequality

Another method to find the maximum value of sin(x)cos(x) is by applying the Arithmetic Mean-Geometric Mean (AM-GM) Inequality. Given two non-negative numbers a and b, AM-GM states that:

frac{a b}{2} geq sqrt{ab}

Letting a sin(x) and b cos(x), we get:

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

Squaring both sides, we obtain:

sin(x)cos(x) leq left(frac{sin(x) cos(x)}{2}right)^2

Since sin(x)^2 cos(x)^2 1, we can simplify further:

sin(x)cos(x) leq frac{1}{2}

Equality holds when sin(x) cos(x), which occurs at x frac{pi}{4} kpi. Hence, the maximum value of sin(x)cos(x) is frac{1}{2}.

Conclusion

In conclusion, the maximum value of sin(x)cos(x) is frac{1}{2}. This value can be obtained using trigonometric identities, calculus, and the AM-GM inequality. These methods provide a comprehensive approach to understanding and solving trigonometric problems.

References

For further reading and detailed proofs, refer to the following resources:

Trigonometry: A Modern Approach by Thomas Koshy Calculus: Early Transcendentals by James Stewart Mathematics for the International Student: Trigonometry by David Bedford