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Write expression in terms of sine and cosine, and simplify it. (The final expression does not have to be in terms of sine and cosine.) $$\cot ^{2} \theta\left(1+\tan ^{2} \theta\right)$$

Short Answer

Expert verified
The simplified expression is \( \csc^2 \theta \).

Step by step solution

01

Express Logarithms in Trigonometric Functions

The given expression is \( \cot^2 \theta (1 + \tan^2 \theta) \). To express it in terms of sine and cosine, recall the definitions of cotangent and tangent.\( \cot \theta = \frac{\cos \theta}{\sin \theta} \) and \( \tan \theta = \frac{\sin \theta}{\cos \theta} \).
02

Substitute Definitions into Expression

Substitute the definitions of cotangent and tangent into the expression:\[ \cot^2 \theta = \left(\frac{\cos \theta}{\sin \theta}\right)^2 = \frac{\cos^2 \theta}{\sin^2 \theta} \]\[ 1 + \tan^2 \theta = 1 + \left(\frac{\sin \theta}{\cos \theta}\right)^2 = 1 + \frac{\sin^2 \theta}{\cos^2 \theta} \].
03

Simplify the Trigonometric Expression

The expression becomes:\[ \frac{\cos^2 \theta}{\sin^2 \theta} \left(1 + \frac{\sin^2 \theta}{\cos^2 \theta}\right) \]Now simplify the term inside the parentheses:\[ 1 + \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{\cos^2 \theta}{\cos^2 \theta} + \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{\cos^2 \theta + \sin^2 \theta}{\cos^2 \theta}\].
04

Use Pythagorean Identity

Apply the Pythagorean identity \( \cos^2 \theta + \sin^2 \theta = 1 \) to simplify:\[ \frac{\cos^2 \theta + \sin^2 \theta}{\cos^2 \theta} = \frac{1}{\cos^2 \theta} = \sec^2 \theta \].
05

Full Simplification of Expression

Therefore, the expression becomes:\[ \frac{\cos^2 \theta}{\sin^2 \theta} \cdot \sec^2 \theta = \frac{\cos^2 \theta}{\sin^2 \theta} \cdot \frac{1}{\cos^2 \theta} \]This simplifies to \( \frac{1}{\sin^2 \theta} = \csc^2 \theta \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Understanding Cotangent in Trigonometry
In trigonometry, the cotangent function is essential for expressing angles in terms of sine and cosine. It is the reciprocal of the tangent function. Specifically, for any angle \( \theta \), cotangent is represented as:
  • \( \cot \theta = \frac{\cos \theta}{\sin \theta} \)
By understanding cotangent in this way, you can rewrite expressions to focus on sine and cosine, which are the foundational trigonometric functions. This approach simplifies complex expressions and helps solve trigonometric exercises efficiently.
Consistent with these simplifications, you often encounter expressions involving \( \cot^2 \theta = \left(\frac{\cos \theta}{\sin \theta}\right)^2 = \frac{\cos^2 \theta}{\sin^2 \theta} \), which allow conversions between different trigonometric identities.
Tangent and Its Role in Simplifying Expressions
The tangent function is another critical trigonometric function and is defined as the ratio of sine to cosine for an angle \( \theta \). It is expressed as:
  • \( \tan \theta = \frac{\sin \theta}{\cos \theta} \)
In exercises, such as the one we're considering, expressing equations in terms of tangent can lead to elegant simplifications. When we have \( 1 + \tan^2 \theta \), this can be rephrased into:
  • \( 1 + \left(\frac{\sin \theta}{\cos \theta}\right)^2 = 1 + \frac{\sin^2 \theta}{\cos^2 \theta} \)
This set-up often calls for the Pythagorean identity in the following steps. Understanding tangent and its square, \( \tan^2 \theta \), makes it easy to manipulate expressions by aligning them with this foundational trigonometric formula.
The Crucial Role of the Pythagorean Identity
The Pythagorean Identity is a foundation of trigonometry. It states that for any angle \( \theta \):
  • \( \cos^2 \theta + \sin^2 \theta = 1 \)
This identity is crucial in simplifying trigonometric expressions, such as the one in our example. By substituting the identity into complex equations, you reduce their complexity significantly, dramatically simplifying solutions. In our problem, applying this identity alters:
  • \( 1 + \frac{\sin^2 \theta}{\cos^2 \theta} = \frac{\cos^2 \theta + \sin^2 \theta}{\cos^2 \theta} \)
Using the Pythagorean Identity, the equation further simplifies to \( \sec^2 \theta \), enabling straightforward continuation to final results. Recognizing and using this identity transforms complicated trigonometric expressions into manageable pieces. This is why it lies at the heart of many trigonometric simplifications.

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Most popular questions from this chapter

Solve each equation ( \(x\) in radians and \(\theta\) in degrees) for all exact solutions where appropriate. Round approximale values in radians to four decimal places and approximate values in degrees to the nearest tenth. Write answers using the least possible nonnegative angle measures. $$2 \sqrt{3} \sin \frac{x}{2}=3$$

Solve each problem. Nautical Mile The British nautical mile is defined as the length \(L\) of a minute of arc on any meridian. Since Earth is flatter at its poles, the nautical mile varies with latitude and, in feet, is $$ L=6077-31 \cos 2 \theta $$ where \(\theta\) is the latitude in degrees. (See the figure.) (Figure can't copy) (a) Find the latitude between \(0^{\circ}\) and \(90^{\circ}\) at which the nautical mile is 6074 feet. (b) At what latitude between \(0^{\circ}\) and \(90^{\circ}\) (inclusive) is the nautical mile 6108 feet? (c) In the United States, the nautical mile is defined everywhere as 6080.2 feet. At what latitude between \(0^{\circ}\) and \(90^{\circ}\) does this agree with the British nautical mile?

Verify that each equation is an identity. $$\frac{\sin (A-B)}{\sin B}+\frac{\cos (A-B)}{\cos B}=\frac{\sin A}{\sin B \cos B}$$

Write each expression as an algebraic expression in \(u, u>0\). $$\cos \left(\tan ^{-1} \frac{3}{u}\right)$$

Solve each equation in part (a) analytically over the interval \([0,2 \pi) .\) Then use a graph to solve each inequality in part (b). (a) \(\sin \frac{x}{2}=\sqrt{2}-\sin \frac{x}{2}\) (b) \(\sin \frac{x}{2}>\sqrt{2}-\sin \frac{x}{2}\)

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