Problem 105 - Precalculus

Question

In calculus, the value of \(F(b)-F(a)\) of a function \(F(x)\) at \(x=a\) and \(x=b\) plays an important role in the calculation of definite integrals. Find the exact value of \(F(b)-F(a)\). $$F(x)=\sec x, a=\frac{\pi}{6}, b=\frac{\pi}{3}$$

Step-by-Step Solution

Answer

The exact value of \(F(b)-F(a)\) is \(\frac{2(\sqrt{3} - 1)}{\sqrt{3}}\).

Step 1: Evaluate F(x) at x=b

First, substitute the value of \(x=b\) into the function \(F(x) = \sec x\). For \(b = \frac{\pi}{3}\), we have \(F\left(\frac{\pi}{3}\right) = \sec\left(\frac{\pi}{3}\right)\). Recall that \(\sec x = \frac{1}{\cos x}\), and \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\). Thus, \(\sec\left(\frac{\pi}{3}\right) = \frac{1}{\frac{1}{2}} = 2\).

Step 2: Evaluate F(x) at x=a

Next, substitute the value of \(x=a\) into the function \(F(x) = \sec x\). For \(a = \frac{\pi}{6}\), we have \(F\left(\frac{\pi}{6}\right) = \sec\left(\frac{\pi}{6}\right)\). Again, use the identity \(\sec x = \frac{1}{\cos x}\), and recall that \(\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\). Thus, \(\sec\left(\frac{\pi}{6}\right) = \frac{1}{\frac{\sqrt{3}}{2}} = \frac{2}{\sqrt{3}}\).

Step 3: Subtract F(a) from F(b)

Now, find the difference \(F(b) - F(a)\). We have \(F\left(\frac{\pi}{3}\right) = 2\) and \(F\left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}}\). Thus, \(F(b) - F(a) = 2 - \frac{2}{\sqrt{3}}\).

Step 4: Simplify the Expression

To simplify \(2 - \frac{2}{\sqrt{3}}\), we need a common denominator. Multiply \(2\) by \(\frac{\sqrt{3}}{\sqrt{3}}\) to get \(\frac{2\sqrt{3}}{\sqrt{3}}\). Now, the expression becomes \(\frac{2\sqrt{3} - 2}{\sqrt{3}}\). The final result is \(\frac{2(\sqrt{3} - 1)}{\sqrt{3}}\).

Key Concepts

Secant Function

The secant function, denoted as \(\sec x\), is a fundamental trigonometric function that is crucial in calculus and trigonometry.
It is related closely to the cosine function and is the reciprocal of the cosine function. The secant function is defined as follows:

• \(\sec x = \frac{1}{\cos x}\)

Understanding this definition helps you make connections between secants and cosines.
This function is undefined for values of \(x\) where \(\cos x = 0\), as this would lead to division by zero.
In the context of definite integrals, evaluating the secant function at particular points requires substituting into \(F(x) = \sec x\). For example:

• For \(x = \frac{\pi}{3}\), \(\sec\left(\frac{\pi}{3}\right) = 2\).
• For \(x = \frac{\pi}{6}\), \(\sec\left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}}\).

These evaluations play a pivotal role in calculating the value differences across intervals.

Trigonometric Identities

Trigonometric identities are equations that hold true for all values involved in the functions.
Using identities simplifies expressions and makes calculations much easier.
A crucial identity in the context of this exercise is \(\sec x = \frac{1}{\cos x}\). This identity is foundational for transforming a secant into a more manageable form.

• Knowing other identities like \(\cos^2 x + \sin^2 x = 1\) helps you solve related problems more efficiently.
• These identities are especially useful when you're working with trigonometric functions over different intervals.

With such identities, you can relate secant back to cosine, facilitating the computation of definite integrals.

Cosine Function

The cosine function is one of the primary trigonometric functions, denoted as \(\cos x\).
The cosine of an angle in a right triangle is the ratio of the adjacent side to the hypotenuse.
Mathematically, it can be expressed as:

• \(\cos x = \frac{\text{Adjacent}}{\text{Hypotenuse}}\)

In our example, computing the cosine of specific angles helps in easily finding their secant values.

• The angle \(\frac{\pi}{3}\) has \(\cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\).
• The angle \(\frac{\pi}{6}\) has \(\cos\left(\frac{\pi}{6}\right) = \frac{\sqrt{3}}{2}\).

Understanding these values is crucial, as they are directly used to compute related secant values in exercises involving definite integrals.