Math Problem Statement

Write each expression in terms of sine and​ cosine, and then simplify so that no quotients appear in the final expression and all functions are of theta only. StartFraction cosecant squared left parenthesis negative theta right parenthesis minus 1 Over 1 minus cosine squared left parenthesis negative theta right parenthesis EndFraction Question content area bottom Part 1 StartFraction cosecant squared left parenthesis negative theta right parenthesis minus 1 Over 1 minus cosine squared left parenthesis negative theta right parenthesis EndFraction equals

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Solution

We are tasked with rewriting the expression in terms of sine and cosine, and simplifying it to eliminate quotients and ensure all functions are in terms of θ\theta.

The given expression is:

csc2(θ)11cos2(θ)\frac{\csc^2(-\theta) - 1}{1 - \cos^2(-\theta)}

Step 1: Use known identities

  1. Cosecant Identity: csc(θ)=1sin(θ)=1sin(θ)\csc(-\theta) = \frac{1}{\sin(-\theta)} = \frac{1}{\sin(\theta)} since sin(θ)=sin(θ)\sin(-\theta) = -\sin(\theta), but squaring the function eliminates the negative: csc2(θ)=1sin2(θ)\csc^2(-\theta) = \frac{1}{\sin^2(\theta)}

  2. Cosine Identity: cos(θ)=cos(θ)\cos(-\theta) = \cos(\theta) Therefore, cos2(θ)=cos2(θ)\cos^2(-\theta) = \cos^2(\theta).

Using these, the expression becomes:

csc2(θ)11cos2(θ)\frac{\csc^2(\theta) - 1}{1 - \cos^2(\theta)}

Step 2: Apply further identities

  • Recall that csc2(θ)1=cot2(θ)\csc^2(\theta) - 1 = \cot^2(\theta): cot2(θ)1cos2(θ)\frac{\cot^2(\theta)}{1 - \cos^2(\theta)}

  • Additionally, we know that 1cos2(θ)=sin2(θ)1 - \cos^2(\theta) = \sin^2(\theta) (this is the Pythagorean identity).

Thus, the expression becomes:

cot2(θ)sin2(θ)\frac{\cot^2(\theta)}{\sin^2(\theta)}

Step 3: Express in terms of sine and cosine

Recall that cot(θ)=cos(θ)sin(θ)\cot(\theta) = \frac{\cos(\theta)}{\sin(\theta)}, so cot2(θ)=cos2(θ)sin2(θ)\cot^2(\theta) = \frac{\cos^2(\theta)}{\sin^2(\theta)}. Substituting this into the expression:

cos2(θ)sin2(θ)sin2(θ)=cos2(θ)sin4(θ)\frac{\frac{\cos^2(\theta)}{\sin^2(\theta)}}{\sin^2(\theta)} = \frac{\cos^2(\theta)}{\sin^4(\theta)}

Final Answer:

The simplified expression is:

cos2(θ)sin4(θ)\frac{\cos^2(\theta)}{\sin^4(\theta)}

This is the expression entirely in terms of sine and cosine, with no quotients involving trigonometric functions left.

Would you like further details on this or have any questions?


Related questions:

  1. What is the identity for csc2(θ)\csc^2(\theta)?
  2. How do the even and odd properties of trigonometric functions affect the identities?
  3. Can you show how to simplify 1sin2(θ)cos2(θ)\frac{1 - \sin^2(\theta)}{\cos^2(\theta)}?
  4. What is the Pythagorean identity involving sine and cosine?
  5. How do we prove csc2(θ)1=cot2(θ)\csc^2(\theta) - 1 = \cot^2(\theta)?

Tip:

Always remember the fundamental trigonometric identities (like the Pythagorean identities) when simplifying expressions involving sine and cosine. They can save a lot of time!

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Trigonometric Identities
Simplification

Formulas

csc(θ) = 1/sin(θ)
cos(-θ) = cos(θ)
sin^2(θ) + cos^2(θ) = 1

Theorems

Pythagorean Identity

Suitable Grade Level

Grades 10-12