Problem 9 - 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. $$(1, \sqrt{15})$$

Step-by-Step Solution

Answer

The trigonometric functions are: \(\sin(\theta) = \frac{\sqrt{15}}{4}\), \(\cos(\theta) = \frac{1}{4}\), \(\tan(\theta) = \sqrt{15}\), \(\csc(\theta) = \frac{4\sqrt{15}}{15}\), \(\sec(\theta) = 4\), \(\cot(\theta) = \frac{\sqrt{15}}{15}\).

Step 1: Identify the point

The point given is \(1, \sqrt{15}\). This point lies on the terminal side of an angle in standard position, where \ x = 1 \ and \ y = \sqrt{15} \.

Step 2: Calculate the hypotenuse

The hypotenuse \(r\) is calculated using the Pythagorean theorem: \\[ r = \sqrt{x^2 + y^2} = \sqrt{1^2 + (\sqrt{15})^2} = \sqrt{1 + 15} = \sqrt{16} = 4. \]

Step 3: Compute the sine function

The sine of the angle \(\theta\) is given by \(\sin(\theta) = \frac{y}{r}\). Substitute the known values: \\[ \sin(\theta) = \frac{\sqrt{15}}{4}. \]

Step 4: Compute the cosine function

The cosine of the angle is \(\cos(\theta) = \frac{x}{r}\). Substitute the known values: \\[ \cos(\theta) = \frac{1}{4}. \]

Step 5: Compute the tangent function

The tangent of the angle is \(\tan(\theta) = \frac{y}{x}\). Substitute the known values: \\[ \tan(\theta) = \frac{\sqrt{15}}{1} = \sqrt{15}. \]

Step 6: Compute the cosecant function

The cosecant of the angle is the reciprocal of the sine function: \(\csc(\theta) = \frac{r}{y}\). Substitute the known values: \\[ \csc(\theta) = \frac{4}{\sqrt{15}} = \frac{4\sqrt{15}}{15}. \]

Step 7: Compute the secant function

The secant of the angle is the reciprocal of the cosine function: \(\sec(\theta) = \frac{r}{x}\). Substitute the known values: \\[ \sec(\theta) = \frac{4}{1} = 4. \]

Step 8: Compute the cotangent function

The cotangent of the angle is the reciprocal of the tangent function: \(\cot(\theta) = \frac{x}{y}\). Substitute the known values: \\[ \cot(\theta) = \frac{1}{\sqrt{15}} = \frac{\sqrt{15}}{15}. \]

Key Concepts

Pythagorean Theorem

The Pythagorean theorem is a fundamental concept in trigonometry and geometry. It is crucial when determining the distance of a point from the origin on a coordinate plane. The theorem states that in a right-angled triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the other two sides. This can be expressed with the equation:

• \(a^2 + b^2 = c^2\)

where:

• \(a\) and \(b\) are the lengths of the triangle's legs,
• and \(c\) is the length of the hypotenuse.

In our problem, the coordinates (1, \(\sqrt{15}\)) are used to form a right triangle where \(x = 1\) and \(y = \sqrt{15}\). Using the Pythagorean theorem:

• \[r = \sqrt{1^2 + (\sqrt{15})^2} = \sqrt{16} = 4\].

This resulting \(r\) represents the hypotenuse or the distance of the point from the origin. The hypotenuse is essential for calculating other trigonometric function values.

Sine Function

The sine function is a fundamental trigonometric function, representing the ratio of the length of the side opposite to the angle to the hypotenuse in a right triangle. For an angle \(\theta\) in standard position, with a terminal side passing through the point \((1, \sqrt{15})\), the sine can be calculated using the following expression:

• \(\sin(\theta) = \frac{y}{r}\)

Here, \(y = \sqrt{15}\) and \(r = 4\). Thus:

• \[\sin(\theta) = \frac{\sqrt{15}}{4}\].

This ratio helps in understanding how the angle \(\theta\) relates to the height of a vertical line drawn from the x-axis to the point \((1, \sqrt{15})\). The sine is crucial in various real-world applications such as waves, vibrations, and circular motion.

Cosine Function

The cosine function is another key trigonometric function that measures the ratio of the adjacent side to the hypotenuse in a right triangle. When positioning an angle in the standard position, you can compute the cosine using:

• \( \cos(\theta) = \frac{x}{r} \)

For our example, with \(x = 1\) and \(r = 4\), the formula gives:

• \[ \cos(\theta) = \frac{1}{4} \].

This measure provides insight into the horizontal distance from the origin to the position where the terminal side meets the circle with radius \(r\). Understanding the cosine is vital for contexts where the length along the ground is a factor, such as in physics and engineering problems involving force and motion.

Tangent Function

The tangent function links the sine and cosine functions by expressing their ratio. It is defined as the ratio of the opposite side to the adjacent side in a right triangle. For our angle \(\theta\) in standard position, tangent is computed by:

• \( \tan(\theta) = \frac{y}{x} \)

With \(y = \sqrt{15}\) and \(x = 1\), this gives:

• \[ \tan(\theta) = \frac{\sqrt{15}}{1} = \sqrt{15} \].

The tangent function provides a way to calculate the "steepness" or "slope" of the line connecting the origin to the point \((1, \sqrt{15})\). It is widely used in calculations involving angles and slopes, such as in navigation, architecture, and even computer graphics.