Problem 35
Question
In Exercises \(30-36,\) perform the indicated operations, then simplify your answers by using appropriate definitions and identities. $$(\cot t-\tan t)\left(\cot ^{2} t+1+\tan ^{2} t\right)$$
Step-by-Step Solution
Verified Answer
Answer: The simplified form of the expression is \(\cot^3t - \tan^3t\).
1Step 1: Apply the distributive property
Distribute the terms in the parentheses, like this:
\((\cot t - \tan t)(\cot^2t + 1 + \tan^2 t) = \cot^3t + \cot t - \cot^2 t \tan t - \tan^3 t - \tan t\)
2Step 2: Use the Pythagorean identities
Applying the Pythagorean identities, substitute \(\cot^2t + 1 = \csc^2t\) and \(1 + \tan^2t = \sec^2t\):
\(= \cot^3t + \cot t\csc^2t - \cot^2 t \tan t - \tan^3 t - \tan t\sec^2t\)
3Step 3: Apply the identity \(\cot t \tan t = 1\)
Utilize the \(\cot t \tan t = 1\) identity to simplify further:
\(= \cot^3t + \cot t\csc^2t - \cot t - \tan^3 t - \tan t\sec^2t\)
4Step 4: Factor out common terms
Factor out common terms \(\cot t\) and \(\tan t\):
\(= \cot t (\cot^2t + \csc^2t - 1) - \tan t (\tan^2t + \sec^2t - 1)\)
Remembering the Pythagorean identities used before, we see that \((\cot^2t + \csc^2t - 1)\) can be written as \(\cot^2t\) and \((\tan^2t + \sec^2t - 1)\) can be written as \(\tan^2t\). Replace them in the expression:
\(= \cot t (\cot^2t) - \tan t (\tan^2t)\)
5Step 5: Simplify the expression
The final step is to simplify the expression:
\(= \cot^3t - \tan^3t\)
Thus, the simplified expression is \(\cot^3t - \tan^3t\).
Key Concepts
Pythagorean IdentitiesDistributive PropertyTrigonometric Functions
Pythagorean Identities
When simplifying trigonometric expressions, one essential tool is the set of Pythagorean identities. These stem from the Pythagorean theorem applied to a right triangle with a unit hypotenuse. Let’s recall the basic identities:
These identities allow us to transform trigonometric expressions that seem complex into simpler forms. For example, if you encounter \(\cot^2t\) in an expression, knowing that \(\cot^2t + 1\) equals \(\csc^2t\), you can replace \(\cot^2t\) by \(\csc^2t - 1\) to possibly simplify further. This step is often critical in solving trigonometric equations or proving identities. As illustrated in the provided exercise, recognizing where to apply these identities can greatly simplify the problem.
- \(\sin^2t + \cos^2t = 1\)
- \(1 + \tan^2t = \sec^2t\)
- \(\cot^2t + 1 = \csc^2t\)
These identities allow us to transform trigonometric expressions that seem complex into simpler forms. For example, if you encounter \(\cot^2t\) in an expression, knowing that \(\cot^2t + 1\) equals \(\csc^2t\), you can replace \(\cot^2t\) by \(\csc^2t - 1\) to possibly simplify further. This step is often critical in solving trigonometric equations or proving identities. As illustrated in the provided exercise, recognizing where to apply these identities can greatly simplify the problem.
Distributive Property
The distributive property is a cornerstone of algebra and is also incredibly useful in trigonometry. It states that for any three numbers, calls them a, b, and c, the equation \(a(b + c) = ab + ac\) holds true. This means when you multiply a sum by a number, you can multiply each addend by the number and then sum the products.
In the context of trigonometric expressions, the distributive property allows us to expand multiplication between two sums of trigonometric functions. In the provided problem, \(\cot t - \tan t\) is distributed over \(\cot^2 t + 1 + \tan^2 t\), resulting in a new expression where each term of the first polynomial is multiplied by every term of the second polynomial. This is the first, vital step in simplifying the expression because it breaks down a complex problem into smaller, more manageable pieces that can be individually simplified.
In the context of trigonometric expressions, the distributive property allows us to expand multiplication between two sums of trigonometric functions. In the provided problem, \(\cot t - \tan t\) is distributed over \(\cot^2 t + 1 + \tan^2 t\), resulting in a new expression where each term of the first polynomial is multiplied by every term of the second polynomial. This is the first, vital step in simplifying the expression because it breaks down a complex problem into smaller, more manageable pieces that can be individually simplified.
Trigonometric Functions
In the study of trigonometry, the trigonometric functions, like sine (sin), cosine (cos), tangent (tan), cotangent (cot), secant (sec), and cosecant (csc), play critical roles. Each function relates to the angles of a right triangle and the ratios of two of its sides.
Understanding the relationships between these functions is crucial when simplifying trigonometric expressions. For instance, \(\cot t\) is the reciprocal of \(\tan t\), and \(\tan t\) is the ratio of \(\sin t\) to \(\cos t\). Knowing these relationships can help you navigate through simplifying processes like those required in the provided problem. Specifically, the identity \(\cot t \tan t = 1\), which signifies one such reciprocal relationship, is used to eliminate certain terms and thus further simplify the expression. These functions and their relationships form the backbone of trigonometric problem-solving, whether you’re simplifying expressions, solving equations, or proving identities.
Understanding the relationships between these functions is crucial when simplifying trigonometric expressions. For instance, \(\cot t\) is the reciprocal of \(\tan t\), and \(\tan t\) is the ratio of \(\sin t\) to \(\cos t\). Knowing these relationships can help you navigate through simplifying processes like those required in the provided problem. Specifically, the identity \(\cot t \tan t = 1\), which signifies one such reciprocal relationship, is used to eliminate certain terms and thus further simplify the expression. These functions and their relationships form the backbone of trigonometric problem-solving, whether you’re simplifying expressions, solving equations, or proving identities.
Other exercises in this chapter
Problem 35
Graph the function over the interval \([0,2 \pi)\) and determine the location of all local maxima and minima. [This can be done either graphically or algebraica
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In Exercises \(31-36\), write the expression as a single real mum. ber. Do not use decimal approximations. IHint: Exercises \(15-21 \text { may be helpful. }]\)
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