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Chapter 8: Problem 85

Establish each identity. $$\frac{\cos ^{2} \theta-\sin ^{2} \theta}{1-\tan ^{2} \theta}=\cos ^{2} \theta$$

Short Answer

Expert verified
The identity is established as \( \frac{\text {cos} ^{2} \theta-\text {sin} ^{2} \theta}{1-\tan ^{2} \theta} = \text{cos}^2(\theta) \) simplifies to \( \text{cos}^2(\theta) \).

Step by step solution

01

- Start with the left-hand side

The left-hand side of the equation is \(\frac{\text {cos} ^{2} \theta-\text {sin} ^{2} \theta}{1-\text {tan} ^{2} \theta}\). Remember, \( \text{tan}(\theta) = \frac{\text{sin}(\theta)}{\text{cos}(\theta)} \).
02

- Substitute \( \text{tan}(\theta) \)

Substitute \( \text{tan}(\theta) = \frac{\text{sin}(\theta)}{\text{cos}(\theta)} \) in the denominator: \(\frac{\text {cos} ^{2} \theta-\text {sin} ^{2} \theta}{1-\frac{\text{sin}^2(\theta)}{\text{cos}^2(\theta)}}\).
03

- Simplify the denominator

Simplify the denominator as follows: \(\frac{1-\frac{\text{sin}^2(\theta)}{\text{cos}^2(\theta)}}{1} = \frac{\text {cos}^2 (\theta) - \text {sin}^2 (\theta)}{\frac{\text{cos}^2 (\theta)-\text{sin}^2 (\theta)}{\text{cos}^2(\theta)}}\).
04

- Invert the complex fraction

Simplify the complex fraction by inverting the denominator and multiplying: \(\frac{\text {cos}^2 (\theta) - \text {sin}^2 (\theta)}{\frac{\text {cos}^2 (\theta)-\text{sin}^2 (\theta)}{\text {cos} ^2 (\theta)}} = (\text {cos}^2 (\theta) - \text {sin}^2(\theta)) \times \frac{\text {cos}^2 (\theta)}{\text {cos}^2 (\theta) - \text {sin}^2(\theta)}\).
05

- Simplify the expression

Notice that \( \text {cos} ^2 (\theta) - \text {sin} ^2 (\theta) \) will cancel out: \( \text{cos}^2(\theta) \).
06

- Verify both sides match

After canceling terms, both sides of the original equation are equal: \( \text{cos}^2(\theta) \).

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Key Concepts

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

Cosine
Cosine, or \(\text{cos}\), is a fundamental trigonometric function. It measures the adjacent side over the hypotenuse in a right triangle. For any angle \( \theta \), its cosine value is given by \(\text{cos}(\theta)\).
Some key properties include:
  • \(\text{cos}(0) = 1\)
  • \(\text{cos}(90^\text{o}) = 0\)
The cosine function is an even function, meaning \( \text{cos}(-\theta) = \text{cos}(\theta) \). This property is useful in simplifying expressions. Understanding cosine is critical in solving trigonometric identities.
Sine
Sine, represented as \(\text{sin}\), measures the opposite side over the hypotenuse in a right triangle. For an angle \( \theta \), \(\text{sin}(\theta)\) describes this relationship.
Important values:
  • \(\text{sin}(0) = 0\)
  • \(\text{sin}(90^\text{o}) = 1\)
The sine function is an odd function: \( \text{sin}(-\theta) = -\text{sin}(\theta) \). Recognizing this property helps in manipulating trigonometric identities correctly. Sine's interplay with cosine ([Pythagorean identity](https://en.wikipedia.org/wiki/Pythagorean_identity)) frequently comes into play when simplifying complex trigonometric expressions.
Tangent
Tangent, or \(\text{tan}\), represents the ratio of sine to cosine. \( \text{tan}(\theta) = \frac{\text{sin}(\theta)}{\text{cos}(\theta)} \). It can be thought of as the measure of the slope of the angle鈥檚 opposite over adjacent sides.
Critical values include:
  • \(\text{tan}(0) = 0\)
  • \(\text{tan}(45^\text{o}) = 1\)
Tangent identities, such as \( \text{tan}^2(\theta) = \text{sec}^2(\theta) - 1 \), are often employed in simplifying expressions and solving equations involving both sine and cosine functions.
Complex Fractions
Complex fractions, or fractions within fractions, occur frequently in trigonometry. Simplifying them involves:
  • Rewriting as a single fraction
  • Inverting the denominator and multiplying
Steps in the exercise showed how this was done: The initial fraction \( \frac{\text{cos}^2(\theta) - \text{sin}^2(\theta)}{1 - \text{tan}^2(\theta)} \) was broken down and simplified by recognizing the underlying trigonometric identities and properties. Using basic operations of algebra, we can simplify even formidable-looking expressions, making them manageable.

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