Problem 116 - Algebra and Trigonometry Real Mathematics, Real People

Question

Rewrite each trigonometric function of \(\theta\) in terms of \(\cos \theta\)

Step-by-Step Solution

Answer

The trigonometric functions in terms of \(\cos \theta\) are as follows: \(\sin \theta = \sqrt{1 - \cos^2 \theta}\), \(\tan \theta = \frac{\sqrt{1 - \cos^2 \theta}}{\cos \theta}\), \(\csc \theta = \frac{1}{\sqrt{1 - \cos^2 \theta}}\), \(\sec \theta = \frac{1}{\cos \theta}\), \(\cot \theta = \frac{\cos \theta}{\sqrt{1 - \cos^2 \theta}}\).

Step 1: Rewrite \(\sin \theta\)

In order to re-write \(\sin \theta\) in terms of \(\cos \theta\), we need to use the Pythagorean identity in trigonometry which states that:\(\sin^2 \theta + \cos^2 \theta = 1\) Reformulate for \(\sin \theta\):\(\sin \theta = \sqrt{1 - \cos^2 \theta}\)

Step 2: Rewrite \(\tan \theta\)

\(\tan \theta\) is interpreted as \(\sin \theta\) divided by \(\cos \theta\). Therefore, you can substitute the prior obtained expression for \(\sin \theta\) to express \(\tan \theta\) in terms of \(\cos \theta\):\(\tan \theta = \frac{\sqrt{1 - \cos^2 \theta}}{\cos \theta}\)

Step 3: Rewrite \(\csc \theta\)

\(\csc \theta\) equals \(1\) divided by \(\sin \theta\), so by substituting \(\sin \theta\) from step 1, \(\csc \theta\) in terms of \(\cos \theta\) would be:\(\csc \theta = \frac{1}{\sqrt{1 - \cos^2 \theta}}\)

Step 4: Rewrite \(\sec \theta\)

\(\sec \theta\) is \(1\) divided by \(\cos \theta\), so \(\sec \theta\) in terms of \(\cos \theta\) is:\(\sec \theta = \frac{1}{\cos \theta}\)

Step 5: Rewrite \(\cot \theta\)

\(\cot \theta\) is \(1\) divided by \(\tan \theta\), so substituting \(\tan \theta\) from step 2, \(\cot \theta\) in terms of \(\cos \theta\) is:\(\cot \theta = \frac{\cos \theta}{\sqrt{1 - \cos^2 \theta}}\)

Key Concepts

Pythagorean Identity

The Pythagorean identity is a cornerstone in trigonometry, simplifying the understanding of relationships between trigonometric functions. This identity is expressed as:

• \( \sin^2 \theta + \cos^2 \theta = 1 \)

This equation shows that the square of the sine of an angle plus the square of the cosine of the same angle always equals one. It's derived from the Pythagorean theorem applied to a unit circle.
This identity is particularly useful for expressing one trigonometric function in terms of another. For example, to express \( \sin \theta \) in terms of \( \cos \theta \), rearrange the identity:

• \( \sin \theta = \sqrt{1 - \cos^2 \theta} \)

Utilizing this identity allows for solving trigonometric problems by focusing on a single trigonometric function. It's a powerful tool in algebraic manipulation of trigonometric functions.

Trigonometric Functions

Trigonometric functions are functions of an angle, crucial for understanding a wide range of mathematical concepts, from geometry to calculus. The primary trigonometric functions include:

• \( \sin \theta \) - represents the sine function
• \( \cos \theta \) - represents the cosine function
• \( \tan \theta \) - represents the tangent function, defined as \( \frac{\sin \theta}{\cos \theta} \)
• \( \csc \theta \) - represents the cosecant function, defined as \( \frac{1}{\sin \theta} \)
• \( \sec \theta \) - represents the secant function, defined as \( \frac{1}{\cos \theta} \)
• \( \cot \theta \) - represents the cotangent function, defined as \( \frac{1}{\tan \theta} \) or \( \frac{\cos \theta}{\sin \theta} \)

These functions help describe relationships in circular and oscillatory motion. Understanding them in terms of \( \cos \theta \) allows for simplifications and insights in many applications.

Algebraic Manipulation

Algebraic manipulation involves rearranging equations to isolate specific terms or simplify expressions. For trigonometric functions, this often means expressing one function in terms of another. In the given solution, algebraic manipulation helps us express multiple trigonometric functions using \( \cos \theta \):

• \( \sin \theta = \sqrt{1 - \cos^2 \theta} \)
• \( \tan \theta = \frac{\sqrt{1 - \cos^2 \theta}}{\cos \theta} \)
• \( \csc \theta = \frac{1}{\sqrt{1 - \cos^2 \theta}} \)
• \( \sec \theta = \frac{1}{\cos \theta} \)
• \( \cot \theta = \frac{\cos \theta}{\sqrt{1 - \cos^2 \theta}} \)

This process makes it easier to solve problems by focusing on a single trigonometric aspect like \( \cos \theta \). It also reveals deeper connections between the functions and can simplify calculations significantly.