Problem 40 - Algebra and Trigonometry

Question

In Exercises \(33-42,\) let $$ \sin t=a, \cos t=b, \text { and } \tan t=c $$ Write each expression in terms of \(a, b,\) and \(c .\) $$ \sin (-t-2 \pi)+\cos (-t-4 \pi)-\tan (-t-\pi) $$

Step-by-Step Solution

Answer

The given expression \(\sin (-t-2 \pi) + \cos (-t-4 \pi) - \tan (-t-\pi)\) will turn into \(-a + b - c\).

Step 1: Understand the periodicity of sin, cos, and tan

The given expression has three components: \(\sin (-t-2 \pi)\), \(\cos (-t-4 \pi)\), and \(-\tan (-t-\pi)\). The sine, cosine, and tangent functions are periodic, so the values for these functions repeat every \(2\pi\) for \(\sin\) and \(\cos\), and every \(\pi\) for \(\tan\). Therefore, \(\sin (-t-2 \pi) = \sin (-t)\), \(\cos (-t-4 \pi) = \cos (-t)\), and \(-\tan (-t-\pi) = -\tan (-t)\).

Step 2: Apply the sine, cosine and tangent of negative angles identities

\(\sin (-t) = -\sin t\), \(\cos (-t) = \cos t\) and \(\tan (-t) = -\tan t\) because sine function is odd and cosine and tangent functions are even.

Step 3: Replace \(\sin t\), \(\cos t\), \(\tan t\) with their given correspondences

\(\sin (-t-2 \pi) = -\sin t = -a\), \(\cos (-t-4 \pi) = \cos t = b\), and \(-\tan (-t-\pi) = \tan t = -c\). Therefore, the given expression \(\sin (-t-2 \pi) + \cos (-t-4 \pi) - \tan (-t-\pi)\) will turn into \(-a + b - c\).

Key Concepts

Periodicity of Trigonometric Functions

To have a solid foundation in trigonometry, recognizing the periodicity of trigonometric functions is essential. Simply put, periodicity refers to the nature of these functions to repeat their values at regular intervals. For the sine and cosine functions, this interval is \(2\pi\), meaning that \(\sin(\theta + 2\pi) = \sin(\theta)\) and \(\cos(\theta + 2\pi) = \cos(\theta)\). On the other hand, the tangent function has a shorter period of \(\pi\), so \(\tan(\theta + \pi) = \tan(\theta)\).

This concept is particularly useful when dealing with complicated expressions. For instance, if you have \(\sin(-t-2\pi)\), you can simply recognize that the subtraction of \(2\pi\) keeps the sine function's value unchanged, simplifying the expression to \(\sin(-t)\). Understanding periodicity allows students to break down and simplify complex trigonometric expressions, making problem-solving much more manageable.

Trigonometric Functions of Negative Angles

When exploring trigonometry, you'll come across functions that include negative angles. A crucial point to learn is that trigonometric functions respond differently to negative inputs. Specifically, the sine function is odd, meaning that \(\sin(-\theta) = -\sin(\theta)\), whereas the cosine and tangent functions are even, so \(\cos(-\theta) = \cos(\theta)\) and \(\tan(-\theta) = -\tan(\theta)\).

Why does this matter? When you face an expression like \(\sin(-t)\) or \(\cos(-t)\), you can immediately rewrite it to \(\sin(-t) = -\sin(t)\) and \(\cos(-t) = \cos(t)\), thanks to these properties. This knowledge not only simplifies calculations but also helps you to better understand the symmetry and behavior of trigonometric functions along the coordinate plane.

Expressing Trigonometric Functions in Terms of Variables

A common task in trigonometry is to express trigonometric functions in terms of variables. In our example, we're asked to express trigonometric functions in terms of the variables \(a, b,\) and \(c\), which correspond to \(\sin t, \cos t,\) and \(\tan t\), respectively. Making these substitutions can provide a clearer understanding of trigonometric relationships and simplify the manipulation of expressions.

To make these changes, recognize the initial definitions: \(\sin t = a\), \(\cos t = b\), \(\tan t = c\). Then, apply the properties of trigonometric functions accordingly. For instance, if an expression includes \(\sin(-t)\) after considering periodicity and negation properties, it can be replaced by \(\sin(-t) = -a\). This approach makes it easier to handle trigonometric expressions algebraically, especially when solving equations or proving identities.