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Chapter 3: Problem 7

Prove the given identity. $$ \frac{\csc \theta}{\sin \theta}=\csc ^{2} \theta $$

Short Answer

Expert verified
The identity \( \frac{\text{csc} \theta}{\text{sin} \theta} = \text{csc}^{2} \theta \) is true because \( \frac{1}{\text{sin} \theta} \times \frac{1}{\text{sin} \theta} = \frac{1}{\text{sin}^{2} \theta} \).

Step by step solution

01

Understand the given identity

The given identity is \( \frac{\text{csc} \theta}{\text{sin} \theta} = \text{csc}^{2} \theta \). We need to prove that this identity is true.
02

Recall the definition of cosecant

Recall that \( \text{csc} \theta \) is the reciprocal of \(\text{sin} \theta \). Therefore, \( \text{csc} \theta = \frac{1}{\text{sin} \theta} \).
03

Substitute the definition of csc 胃 into the left side of the equation

Substitute \( \frac{1}{\text{sin} \theta} \) for \( \text{csc} \theta \):\[ \frac{\text{csc} \theta}{\text{sin} \theta} = \frac{\frac{1}{\text{sin} \theta}}{\text{sin} \theta} \].
04

Simplify the expression

Simplify \( \frac{\frac{1}{\text{sin} \theta}}{\text{sin} \theta} \) by multiplying the numerator and the denominator:\[ \frac{\frac{1}{\text{sin} \theta}}{\text{sin} \theta} = \frac{1}{\text{sin} \theta} \times \frac{1}{\text{sin} \theta} = \frac{1}{\text{sin}^{2} \theta} \].
05

Recognize the equivalent form

Notice that \( \frac{1}{\text{sin}^{2} \theta} \) is exactly \(\text{csc}^{2} \theta \). Therefore, we have proved the identity:\[ \frac{\text{csc} \theta}{\text{sin} \theta} = \text{csc}^{2} \theta. \]

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Key Concepts

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

Cosecant
In trigonometry, 'cosecant' is one of the six main trigonometric functions. It is denoted as \( \text{csc} \theta \). The cosecant function is the reciprocal of the sine function. This means that for any angle \( \theta \), \( \text{csc} \theta = \frac{1}{\text{sin} \theta} \). Knowing the definition of cosecant is crucial as it often appears in various trigonometric identities and proofs. When solving problems that involve cosecant, always remember its relationship to sine. This can help simplify equations and identities. Simply put, anytime you see \( \text{csc} \theta \), think about how it relates back to \( \text{sin} \theta \).
Reciprocal Identities
Reciprocal identities are fundamental in trigonometry and relate each trigonometric function to one of its 'inverse' functions. These identities help in transforming complicated trigonometric expressions into simpler ones.

For important reciprocal identities, consider:
  • \( \text{csc} \theta = \frac{1}{\text{sin} \theta} \)
  • \( \text{sec} \theta = \frac{1}{\text{cos} \theta} \)
  • \( \text{cot} \theta = \frac{1}{\text{tan} \theta} \)
Understanding these identities allows you to switch between the primary function (like sine, cosine, and tangent) and their reciprocals effectively. Many trigonometric proofs, including the given exercise, use reciprocal identities to simplify and solve equations.
Trigonometric Simplifications
Trigonometric simplifications are techniques to make complex trigonometric expressions easier to work with. In our exercise, we used a simplification process to prove the identity \( \frac{\text{csc} \theta}{\text{sin} \theta} = \text{csc}^{2} \theta \).
Let's simplify step-by-step:
  • Start by identifying known identities. Here, \( \text{csc} \theta = \frac{1}{\text{sin} \theta} \).
  • Substitute this identity into the given expression: \( \frac{\text{csc} \theta}{\text{sin} \theta} = \frac{\frac{1}{\text{sin} \theta}}{\text{sin} \theta} \).
  • Simplify the fraction: \( \frac{\frac{1}{\text{sin} \theta}}{\text{sin} \theta} = \frac{1}{\text{sin} \theta} \times \frac{1}{\text{sin} \theta} = \frac{1}{\text{sin}^{2} \theta} \).
  • Recognize that \( \frac{1}{\text{sin}^{2} \theta} = \text{csc}^{2} \theta \).
Hence, we prove the identity.
By mastering these simplifications and understanding each step, complex equations become more manageable, making trigonometry easier to navigate.

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Most popular questions from this chapter

We showed that \(\sin \theta=\pm \sqrt{1-\cos ^{2} \theta}\) for all \(\theta\). Give an example of an angle \(\theta\) such that \(\sin \theta=-\sqrt{1-\cos ^{2} \theta}\).

Prove the given identity. $$ \tan \frac{1}{2} \theta=\csc \theta-\cot \theta $$

Show that each trigonometric function can be put in terms of the sine function.

Prove Mollweide's second equation: For any triangle \(\triangle A B C, \frac{a+b}{c}=\frac{\cos \frac{1}{2}(A-B)}{\sin \frac{1}{2} C}\).

Prove the identity \(\sin \theta+\cos \theta=\sqrt{2} \sin \left(\theta+45^{\circ}\right)\). Explain why this shows that $$ -\sqrt{2} \leq \sin \theta+\cos \theta \leq \sqrt{2} $$ for all angles \(\theta\). For which \(\theta\) between \(0^{\circ}\) and \(360^{\circ}\) would \(\sin \theta+\cos \theta\) be the largest?
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