Problem 14 - Basic Technical Mathematics with Calculus

Question

Find values of the trigonometric functions of the angle (in standard position) whose terminal side passes through the given points. For Exercises \(3-14,\) give answers in exact form. For Exercises 15 and \(16,\) the coordinates are approximate. $$\left(1, \frac{1}{2}\right)$$

Step-by-Step Solution

Answer

\( \sin(\theta) = \frac{\sqrt{5}}{5}, \cos(\theta) = \frac{2\sqrt{5}}{5}, \tan(\theta) = \frac{1}{2}; \csc(\theta) = \sqrt{5}, \sec(\theta) = \frac{\sqrt{5}}{2}, \cot(\theta) = 2 \).

Step 1: Identify the components

The given point is \( (1, \frac{1}{2}) \). Here, \( x = 1 \) and \( y = \frac{1}{2} \). These values will be used to calculate the trigonometric functions.

Step 2: Calculate the distance (r) from the origin

The distance, \(r\), from the origin to the point \((1, \frac{1}{2})\) can be calculated using the Pythagorean theorem: \(r = \sqrt{x^2 + y^2} = \sqrt{1^2 + \left(\frac{1}{2}\right)^2} = \sqrt{1 + \frac{1}{4}} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}\).

Step 3: Calculate \(\sin(\theta)\)

The sine of the angle \(\theta\) is calculated as \( \sin(\theta) = \frac{y}{r} = \frac{\frac{1}{2}}{\frac{\sqrt{5}}{2}} = \frac{1}{\sqrt{5}} = \frac{\sqrt{5}}{5} \) after rationalizing the denominator.

Step 4: Calculate \(\cos(\theta)\)

The cosine of the angle \(\theta\) is calculated as \( \cos(\theta) = \frac{x}{r} = \frac{1}{\frac{\sqrt{5}}{2}} = \frac{2}{\sqrt{5}} = \frac{2\sqrt{5}}{5} \) after rationalizing the denominator.

Step 5: Calculate \(\tan(\theta)\)

The tangent of the angle \(\theta\) is calculated as \( \tan(\theta) = \frac{y}{x} = \frac{\frac{1}{2}}{1} = \frac{1}{2} \).

Step 6: Calculate \(\csc(\theta)\)

The cosecant of the angle \(\theta\) is the reciprocal of sine: \( \csc(\theta) = \frac{1}{\sin(\theta)} = \sqrt{5} \).

Step 7: Calculate \(\sec(\theta)\)

The secant of the angle \(\theta\) is the reciprocal of cosine: \( \sec(\theta) = \frac{1}{\cos(\theta)} = \frac{\sqrt{5}}{2} \).

Step 8: Calculate \(\cot(\theta)\)

The cotangent of the angle \(\theta\) is the reciprocal of tangent: \( \cot(\theta) = \frac{1}{\tan(\theta)} = 2 \).

Key Concepts

Understanding Trigonometry

Trigonometry is a fascinating branch of mathematics that focuses on the relationships between the angles and sides of triangles. It is particularly useful when dealing with right triangles and can help in calculating distances and angles. In trigonometry, the three primary functions are sine, cosine, and tangent. These functions were utilized in our exercise to find the values of the trigonometric functions for the angle formed by the point \((1, \frac{1}{2})\).

• **Sine** is the ratio of the opposite side to the hypotenuse.
• **Cosine** is the ratio of the adjacent side to the hypotenuse.
• **Tangent** is the ratio of the opposite side to the adjacent side.

These ratios allow us to find angles and distances with remarkable precision. Besides sine, cosine, and tangent, we also calculate their reciprocals: cosecant, secant, and cotangent, which provide additional insights into the triangle's properties.

Pythagorean Theorem in Trigonometry

The Pythagorean Theorem is a fundamental pillar in geometry and trigonometry. Named after the Greek mathematician Pythagoras, this theorem provides a simple equation relating the lengths of the sides of a right triangle:
\[ a^2 + b^2 = c^2 \]
Here, \(a\) and \(b\) are the triangle's legs, and \(c\) is the hypotenuse. This theorem is invaluable when you're working with trigonometric problems because it allows you to solve for unknown distances like the hypotenuse (\(r\)), which was calculated as \(\frac{\sqrt{5}}{2}\) in our example.
When calculating any trigonometric function, knowing \(r\) allows you to measure how far the point is from the origin in a coordinate system, solidifying the bridge between algebraic expressions and geometric interpretations.

Rationalizing Denominators

In trigonometric exercises, you'll often encounter expressions like \(\frac{1}{\sqrt{5}}\). Though valid, such expressions are not in 'simplified' form according to mathematical standards. Rationalizing the denominator simplifies this expression by eliminating the radical from the bottom of the fraction. This involves multiplying both the numerator and the denominator by the radical.
For instance, the rationalization process turns \(\frac{1}{\sqrt{5}}\) into \(\frac{\sqrt{5}}{5}\). Here's how:

• Multiply the numerator and the denominator by \(\sqrt{5}\).
• Calculate \(\frac{\sqrt{5}}{5}\).

This practice not only simplifies the expression but also fits standard mathematical conventions, making calculations easier and cleaner. It was applied step-by-step in our solution to calculate both sine and cosine, ensuring the most exact form of the trigonometric ratios.