Math Problem Statement

Write the expression in terms of sine and​ cosine, and then simplify so that no quotients appear in the final expression.

negative sine squared theta left parenthesis negative 1 minus cotangent squared theta right parenthesis−sin2 θ−1−cot2 θ

Question content area bottom

Part 1

Choose the correct answer below.

A.StartFraction cosine squared theta Over sine squared theta EndFraction

StartFraction cosine squared theta Over sine squared theta EndFractioncos2 θsin2 θ

B.negative 1

negative 1−1

C.tangent squared theta

tangent squared thetatan2 θ

D.1

1

E.cotangent squared theta

cotangent squared thetacot2 θ

F.negative StartFraction cosine squared theta Over sine squared theta EndFraction

negative StartFraction cosine squared theta Over sine squared theta EndFraction

Solution

To rewrite the expression −sin⁡2θ(−1−cot⁡2θ)−sin⁡2θ−1−cot⁡2θ-\sin^2 \theta (-1 - \cot^2 \theta) - \sin^2 \theta - 1 - \cot^2 \theta in terms of sine and cosine, let's break it down step by step.

  1. Recall that cot⁡2θ=cos⁡2θsin⁡2θ\cot^2 \theta = \frac{\cos^2 \theta}{\sin^2 \theta}.

  2. Substitute cot⁡2θ\cot^2 \theta into the expression: −sin⁡2θ(−1−cos⁡2θsin⁡2θ)−sin⁡2θ−1−cos⁡2θsin⁡2θ-\sin^2 \theta \left(-1 - \frac{\cos^2 \theta}{\sin^2 \theta}\right) - \sin^2 \theta - 1 - \frac{\cos^2 \theta}{\sin^2 \theta}

  3. Distribute −sin⁡2θ-\sin^2 \theta: sin⁡2θ+sin⁡2θcos⁡2θsin⁡2θ−sin⁡2θ−1−cos⁡2θsin⁡2θ\sin^2 \theta + \frac{\sin^2 \theta \cos^2 \theta}{\sin^2 \theta} - \sin^2 \theta - 1 - \frac{\cos^2 \theta}{\sin^2 \theta} This simplifies to: sin⁡2θ+cos⁡2θ−sin⁡2θ−1−cos⁡2θsin⁡2θ\sin^2 \theta + \cos^2 \theta - \sin^2 \theta - 1 - \frac{\cos^2 \theta}{\sin^2 \theta}

  4. Since sin⁡2θ+cos⁡2θ=1\sin^2 \theta + \cos^2 \theta = 1, we can substitute: 1−sin⁡2θ−1−cos⁡2θsin⁡2θ1 - \sin^2 \theta - 1 - \frac{\cos^2 \theta}{\sin^2 \theta} Which simplifies to: −sin⁡2θ−cos⁡2θsin⁡2θ-\sin^2 \theta - \frac{\cos^2 \theta}{\sin^2 \theta}

  5. Factor out −1-1: −(sin⁡2θ+cos⁡2θsin⁡2θ)-\left(\sin^2 \theta + \frac{\cos^2 \theta}{\sin^2 \theta}\right) This is: −(sin⁡4θ+cos⁡2θsin⁡2θ)-\left(\frac{\sin^4 \theta + \cos^2 \theta}{\sin^2 \theta}\right)

  6. Since we are asked to remove quotients, let's express it all under the same denominator: −(sin⁡4θ+cos⁡2θsin⁡2θ)-\left(\frac{\sin^4 \theta + \cos^2 \theta}{\sin^2 \theta}\right)

The final expression is simplified to:

−cos⁡2θsin⁡2θ-\frac{\cos^2 \theta}{\sin^2 \theta}

So the answer is:

F. −cos⁡2θsin⁡2θ-\frac{\cos^2 \theta}{\sin^2 \theta}

If you have any questions or need further clarification, feel free to ask!

Related Questions:

  1. What are the Pythagorean identities related to sine and cosine?
  2. How can you derive cot⁡2θ\cot^2 \theta in terms of sine and cosine?
  3. What happens to the expression if you change the sign of θ\theta?
  4. Can you explain how to convert cotangent to sine and cosine?
  5. What are some applications of these trigonometric identities?

Tip: Remember that simplifying expressions can often involve substituting identities to reduce complexity.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Sine and Cosine Functions
Cotangent

Formulas

cot^2 θ = cos^2 θ / sin^2 θ
sin^2 θ + cos^2 θ = 1

Theorems

-

Suitable Grade Level

Grades 9-12

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