Problem 108 - 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 \(F(b)-F(a)\). $$F(x)=\frac{\cot x-4 \sin x}{\cos x}, a=\frac{\pi}{4}, b=\frac{\pi}{3}$$

Step-by-Step Solution

Answer

The exact value of \(F(b) - F(a)\) is \(\frac{-10}{\sqrt{3}} + \sqrt{2}\).

Step 1: Evaluate F(a)

First, we substitute \(a = \frac{\pi}{4}\) into the function \(F(x)\). \[F\left(\frac{\pi}{4}\right) = \frac{\cot\left(\frac{\pi}{4}\right) - 4 \sin\left(\frac{\pi}{4}\right)}{\cos\left(\frac{\pi}{4}\right)}\] To find the exact value, calculate each trigonometric function at \(\frac{\pi}{4}\): \[\cot\left(\frac{\pi}{4}\right) = 1, \ \sin\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}, \ \cos\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2}\] Substituting these, we get: \[F\left(\frac{\pi}{4}\right) = \frac{1 - 4 \cdot \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = \frac{1 - 2\sqrt{2}}{\frac{\sqrt{2}}{2}} = \frac{2\sqrt{2} - 4}{\sqrt{2}/2} = \sqrt{2} - 2\sqrt{2} = -\sqrt{2}\]

Step 2: Evaluate F(b)

Next, we substitute \(b = \frac{\pi}{3}\) into the function \(F(x)\). \[F\left(\frac{\pi}{3}\right) = \frac{\cot\left(\frac{\pi}{3}\right) - 4 \sin\left(\frac{\pi}{3}\right)}{\cos\left(\frac{\pi}{3}\right)}\] Calculate each trigonometric function at \(\frac{\pi}{3}\): \[\cot\left(\frac{\pi}{3}\right) = \frac{1}{\sqrt{3}}, \ \sin\left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2}, \ \cos\left(\frac{\pi}{3}\right) = \frac{1}{2}\] Substituting these, we get: \[F\left(\frac{\pi}{3}\right) = \frac{\frac{1}{\sqrt{3}} - 4 \cdot \frac{\sqrt{3}}{2}}{\frac{1}{2}} = \frac{\frac{1}{\sqrt{3}} - 2\sqrt{3}}{\frac{1}{2}} = 2\left(\frac{1 - 6}{\sqrt{3}}\right) = \frac{-10}{\sqrt{3}}\]

Step 3: Calculate F(b) - F(a)

Finally, calculate \(F(b) - F(a)\) using the evaluated values from Steps 1 and 2. \[F(b) - F(a) = \frac{-10}{\sqrt{3}} - (-\sqrt{2})\] Simplifying the expression, we have:\[F(b) - F(a) = \frac{-10}{\sqrt{3}} + \sqrt{2}\]This is the exact value of \(F(b) - F(a)\).

Key Concepts

Understanding Trigonometric Functions

Trigonometric functions are essential tools in mathematics, particularly when dealing with angles and periodic phenomena. They originate from the angles of triangles and relationships between the lengths of their sides. The primary trigonometric functions include sine (\( \sin \)), cosine (\( \cos \)), and tangent (\( \tan \)). Each function has a distinct role in trigonometry:

• Sine is the ratio of the length of the side opposite the angle to the hypotenuse in a right triangle.
• Cosine is the ratio of the length of the adjacent side to the hypotenuse.
• Cotangent (\( \cot \)) is the reciprocal of tangent, which itself is the ratio of sine to cosine. It can be defined as the cosine divided by sine.

Understanding each function makes it easier to tackle problems involving angles, such as calculating definite integrals with trigonometric elements. For example:

• At \( \frac{\pi}{4} \), \( \cot \left(\frac{\pi}{4}\right) = 1 \), \( \sin \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \), and \( \cos \left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} \)
• At \( \frac{\pi}{3} \), \( \cot \left(\frac{\pi}{3}\right) = \frac{1}{\sqrt{3}} \), \( \sin \left(\frac{\pi}{3}\right) = \frac{\sqrt{3}}{2} \), and \( \cos \left(\frac{\pi}{3}\right) = \frac{1}{2} \)

Evaluating Functions at Specific Points

To evaluate a function at a given point, you simply substitute the given point into the function formula. Functions like \( F(x) = \frac{\cot x - 4 \sin x}{\cos x} \)involve more complex substitutions due to trigonometric components. Make sure to accurately substitute each part and simplify step-by-step.
When evaluating at specific angles:

• Substitute the angle into the function, and replace each trigonometric function with its exact value at that angle.
• Simplify the expression by performing algebraic operations.

For instance:

• At \( a = \frac{\pi}{4} \), the function becomes:\[ F\left(\frac{\pi}{4}\right) = \frac{1 - 4 \cdot \frac{\sqrt{2}}{2}}{\frac{\sqrt{2}}{2}} = -\sqrt{2} \]
• At \( b = \frac{\pi}{3} \), it is:\[ F\left(\frac{\pi}{3}\right) = \frac{\frac{1}{\sqrt{3}} - 4 \cdot \frac{\sqrt{3}}{2}}{\frac{1}{2}} = \frac{-10}{\sqrt{3}} \]

This process requires an ability to handle both trigonometric identities and basic algebra, ensuring whatever simplification steps taken still correctly reflect the original function.

Finding Exact Values in Calculations

Exact values arise often in calculus and algebra. These values are precise, whole-number or simplified forms of numbers. They are particularly valuable in situations demanding high accuracy, like engineering or physics applications.
When working with functions and trigonometry:

• Ensure you are familiar with common angles and their trigonometric values. These include angles like \( \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \) and \( \frac{\pi}{2} \).
• Use known values to substitute back into functions to obtain an exact, rather than estimated, answer.

For the given function values:

• At \( \frac{\pi}{4} \), we substituted exact values: \( \cot \left( \frac{\pi}{4} \right) = 1 \), simplifying to: \( -\sqrt{2} \).
• At \( \frac{\pi}{3} \), similar steps led to: \( \frac{-10}{\sqrt{3}} \).

Finally, the difference \( F(b) - F(a) \)became: \( \frac{-10}{\sqrt{3}} + \sqrt{2} \). This value is exact without approximation, crucial in accurately solving analytical problems.