Problem 12 - Precalculus

Question

Suppose that \(\triangle A B C\) is a right triangle with \(\angle C=90^{\circ}\) If \(A B=3\) and \(A C=1,\) compute the values of the six trigonometric functions of angle \(A\) If \(A B=1\) and \(B C=\sqrt{3} / 2,\) compute the following. (a) \(\cos A, \sin B\) (b) \(\tan A, \cot B\) (c) \(\sec A, \csc B\)

Step-by-Step Solution

Answer

(a) \( \cos A = \frac{\sqrt{3}}{2}, \sin B = \frac{\sqrt{3}}{2} \); (b) \( \tan A = \frac{1}{2}, \cot B = \frac{1}{2} \); (c) \( \sec A = \frac{2}{\sqrt{3}}, \csc B = \frac{2}{\sqrt{3}} \).

Step 1: Understanding the Triangle Configuration

In a right triangle \( \triangle ABC \), \( \angle C = 90^\circ \) is given. We need to find trigonometric functions with different given lengths for the sides \( AB \), \( AC \), and \( BC \) for \( \angle A \) and \( \angle B \).

Step 2: Calculating Values for Given \( AB = 3, AC = 1 \)

For this configuration, determine the hypotenuse \( BC \) using the Pythagorean theorem:\[BC = \sqrt{AB^2 - AC^2} = \sqrt{3^2 - 1^2} = \sqrt{8} = 2\sqrt{2} \approx 2.83.\] Calculate the trigonometric functions:- \( \sin A = \frac{AC}{BC} = \frac{1}{2\sqrt{2}} \)- \( \cos A = \frac{AB}{BC} = \frac{3}{2\sqrt{2}} \)- \( \tan A = \frac{AC}{AB} = \frac{1}{3} \)- \( \sec A = \frac{1}{\cos A} \approx 0.94 \)- \( \csc A = \frac{1}{\sin A} = 2\sqrt{2} \)- \( \cot A = \frac{1}{\tan A} = 3 \)

Step 3: Calculating Values for Given \( AB = 1, BC = \sqrt{3}/2 \)

We have \( AC \) by using the Pythagorean theorem:\[AC = \sqrt{AB^2 - BC^2} = \sqrt{1^2 - \left(\frac{\sqrt{3}}{2}\right)^2} = \sqrt{1 - \frac{3}{4}} = \sqrt{\frac{1}{4}} = \frac{1}{2}.\] Using these side lengths, calculate the functions:

Step 3a: Calculate \( \cos A, \sin B \)

- \( \cos A = \frac{BC}{AB} = \frac{\sqrt{3}/2}{1} = \frac{\sqrt{3}}{2} \)- \( \sin B = \cos A = \frac{\sqrt{3}}{2} \) (in a right triangle, \( \sin B = \cos A \))

Step 3b: Calculate \( \tan A, \cot B \)

- \( \tan A = \frac{AC}{AB} = \frac{1/2}{1} = \frac{1}{2} \)- \( \cot B = \tan A = \frac{1}{2} \) (in a right triangle, \( \cot B = \tan A \))

Step 3c: Calculate \( \sec A, \csc B \)

- \( \sec A = \frac{1}{\cos A} = \frac{2}{\sqrt{3}} \)- \( \csc B = \sec A = \frac{2}{\sqrt{3}} \) (in a right triangle, \( \csc B = \sec A \))

Key Concepts

Right Triangle

A right triangle is a type of triangle in which one of the angles is exactly 90 degrees, making it a special and very common form in geometry. Understanding right triangles is crucial as they form the basis of trigonometry. In a right triangle, the longest side is called the hypotenuse, and it is opposite the right angle. The other two sides are referred to as legs. Right triangles adhere to specific properties, allowing us to use trigonometric functions, such as sine, cosine, and tangent, to find side lengths and angle measures.
These functions are defined based on the ratios between the sides of the triangle. For example:

• **Sine** (\(\sin \theta = \frac{{\text{{opposite}}}}{{\text{{hypotenuse}}}}\)): This function relates the angle to the ratio of the length of the side opposite the angle to the hypotenuse.
• **Cosine** (\(\cos \theta = \frac{{\text{{adjacent}}}}{{\text{{hypotenuse}}}}\)): This function relates the angle to the ratio of the length of the side adjacent to the angle to the hypotenuse.
• **Tangent** (\(\tan \theta = \frac{{\text{{opposite}}}}{{\text{{adjacent}}}}\)): This function relates the angle to the ratio of the length of the side opposite the angle to the side adjacent to it.

These relationships make right triangles an essential part of understanding trigonometry.

Pythagorean Theorem

The Pythagorean Theorem is a fundamental principle in geometry that connects the side lengths of a right triangle. This theorem states that, in a right triangle, the square of the length of the hypotenuse is equal to the sum of the squares of the lengths of the other two sides. This can be expressed with the formula:
\[c^2 = a^2 + b^2\] where:

• \(c\) is the hypotenuse.
• \(a\) and \(b\) are the lengths of the other two sides.

Using this theorem allows us to solve for unknown side lengths. For instance, if one knows the lengths of the two legs, the hypotenuse can be calculated. Or, if the hypotenuse and one leg's length are known, the other leg can be determined. This theorem is not just limited to theoretical exercises; it applies to real-world problems, such as construction, navigation, and more.
In the context of our exercise, knowing the Pythagorean Theorem is crucial as we used it to determine unknown side lengths, which then enabled us to calculate trigonometric functions like sine and cosine.

Angle Ratios

Angle ratios in trigonometry are invaluable for understanding the relationships between different parts of a triangle. These ratios help define the six fundamental trigonometric functions: sine, cosine, tangent, cotangent, secant, and cosecant. Understanding these ratios is crucial because they enable us to solve complex problems involving triangles by relating angles to side lengths. Here's a breakdown:

• **Sine** (\(\sin\)): Ratio of the length of the opposite side to the hypotenuse.
• **Cosine** (\(\cos\)): Ratio of the length of the adjacent side to the hypotenuse.
• **Tangent** (\(\tan\)): Ratio of the length of the opposite side to the adjacent side.
• **Cotangent** (\(\cot\)): The reciprocal of tangent (\(\cot \theta = \frac{1}{\tan \theta}\)).
• **Secant** (\(\sec\)): The reciprocal of cosine (\(\sec \theta = \frac{1}{\cos \theta}\)).
• **Cosecant** (\(\csc\)): The reciprocal of sine (\(\csc \theta = \frac{1}{\sin \theta}\)).

These functions and their corresponding angle ratios are not just theoretical. They are used in various applications, including physics, engineering, and computer graphics, to solve problems involving waves, oscillations, and rotational calculations. In our specific exercise, we used these ratios to compute the exact values of trigonometric functions for given angles.