How Can We Prove the Derivative of Sine at π/6?
This article aims to explore the process of proving the derivative of the sine function at a specific point, namely π/6. Understanding this involves delving into the concepts of calculus, particularly differentiation, and the fundamental properties of trigonometric functions. By the end of this discussion, readers will have a clear understanding of the steps involved in proving the derivative of the sine function at π/6, enhancing their knowledge in calculus and trigonometry.
Introduction to Differentiation and Trigonometric Functions
Differentiation is a core concept in calculus, which allows us to find the rate at which a function changes at any given point. Trigonometric functions, such as sine and cosine, play a significant role in various applications in mathematics and physics. The process of finding the derivative of the sine function at a specific point requires a solid understanding of these functions and their properties. Let's break down the steps to find the derivative of sin(x) at x π/6.
Step 1: Understanding the Definition of a Derivative
The derivative of a function f(x) is defined as the limit of the difference quotient as the change in x approaches zero:
f'(x) limh→0 (f(x h) - f(x)) / h
For the function f(x) sin(x), we can write the derivative at a specific point x a as:
sin'(a) limh→0 (sin(a h) - sin(a)) / h
Step 2: Applying the Sum Formula for Sine
The sum formula for sine is given by:
sin(a h) cos(a)sin(h) sin(a)cos(h)
Using this formula, we can rewrite the expression for the derivative of sin(x) at x π/6:
sin'(π/6) limh→0 (cos(π/6)sin(h) sin(π/6)cos(h) - sin(π/6)) / h
Simplifying further, we get:
sin'(π/6) limh→0 (sin(π/6)(cos(h) - 1) cos(π/6)sin(h)) / h
Since sin(π/6) 1/2 and cos(π/6) √3/2, we can substitute these values:
sin'(π/6) limh→0 (1/2(cos(h) - 1) (√3/2)sin(h)) / h
Step 3: Evaluating the Limit Using Fundamental Trigonometric Limits
To evaluate the above limit, we utilize the fundamental trigonometric limits:
limh→0 sin(h) / h 1
and
limh→0 (1 - cos(h)) / h 0
Substituting these limits, we get:
sin'(π/6) (1/2)(1 - 1) (√3/2)(1) 0 (√3/2) √3/2
Therefore, the derivative of sin(x) at x π/6 is:
sin'(π/6) √3/2
Conclusion
In conclusion, we have demonstrated how to prove the derivative of the sine function at x π/6 using the principles of calculus and trigonometry. By applying the definition of a derivative, utilizing the sum formula for sine, and evaluating fundamental limits, we were able to determine that sin'(π/6) √3/2. This process not only enhances our understanding of calculus and trigonometry but also provides a valuable problem-solving technique for more complex functions.
These concepts form the foundation for more advanced topics in mathematics and physics, such as differential equations and Fourier analysis. By mastering the techniques discussed here, students and professionals can approach a wide range of mathematical and scientific problems with confidence and precision.