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Chapter 3: Problem 48

Prove that each equation is an identity. \(\cos 2 y=\frac{1-\tan ^{2} y}{1+\tan ^{2} y}\)

Short Answer

Expert verified
Since we derived \( \cos 2y = \frac{1 - \tan^2 y}{1 + \tan^2 y} \), the equation is an identity.

Step by step solution

01

Recall the double-angle identity for cosine

The double-angle identity for cosine is: \[ \cos 2y = \cos^2 y - \sin^2 y \]
02

Convert to tangent form using sine and cosine

Recall that \( \tan y = \frac{\sin y}{\cos y} \), so \( \sin y = \tan y \cdot \cos y \). Substitute this into the double-angle identity: \[ \cos 2y = \cos^2 y - (\tan y \cdot \cos y)^2 \]
03

Simplify the expression

Simplify the equation: \[ \cos 2y = \cos^2 y - \tan^2 y \cdot \cos^2 y \] Factor out \( \cos^2 y \): \[ \cos 2y = \cos^2 y (1 - \tan^2 y) \]
04

Use the Pythagorean identity

Recall that \( 1 + \tan^2 y = \frac{1}{\cos^2 y} \). Rewrite the denominator with this identity: \[ \frac{1}{\cos^2 y} = 1 + \tan^2 y \] Then multiply both numerator and denominator by \( \cos^2 y \): \[ \cos 2y = \frac{(1 - \tan^2 y) \cos^2 y}{(1 + \tan^2 y) \cos^2 y} \]
05

Cancel out the common terms

Cancel \( \cos^2 y \) from the numerator and denominator: \[ \cos 2y = \frac{1 - \tan^2 y}{1 + \tan^2 y} \]

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Key Concepts

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

double-angle identity
The double-angle identity is a key concept in trigonometry. This identity lets you express trigonometric functions of double angles (like \(2y\)) in terms of single angles (like \(y\)). The double-angle identity for cosine states that: \[ \cos 2y = \cos^2 y - \sin^2 y \] This formula is incredibly useful for transforming and simplifying trigonometric expressions. For instance, you can convert everything into sine and cosine forms. Then, apply these transformations to further simplify the expressions.
tangent identity
The tangent identity is fundamental when working with trigonometric functions involving tangents. Recall that: \[\tan y = \frac{\sin y}{\cos y} \] This means you can express \$\sin y\$ as \$\tan y \cdot \cos y\$. In our exercise, we used this to rewrite sine in terms of tangent and cosine: \[\sin y = \tan y \cdot \cos y\] When simplifying the expression later, we also dealt with \tan^2 y\. Using these identities, one can easily switch between tangent and sine/cosine forms, which is helpful for proving complex identities.
Pythagorean identity
The Pythagorean identity is another essential trigonometric identity that relates sine and cosine: \[ \sin^2 y + \cos^2 y = 1 \] This formula is immensely useful for converting between different trigonometric forms. Additionally, when combined with tangent identities, you get: \[ 1 + \tan^2 y = \frac{1}{\cos^2 y} \] This conversion plays a crucial role in our exercise. It allows us to manipulate the expressions into a simpler form that can be easily simplified and solved. By recalling and applying these identities correctly, you'll be able to solve complex trigonometric problems efficiently.

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Most popular questions from this chapter

Simplify each expression by applying the odd/even identities, cofunction identities, and cosine of a sum or difference identities. Do not use a calculator $$ \cos \left(10^{\circ}\right) \cos \left(20^{\circ}\right)+\sin \left(-10^{\circ}\right) \cos \left(70^{\circ}\right) $$

Use an identity to simplify each expression. a. \(\sin 3.5 \cos 2.1+\cos 3.5 \sin 2.1\) b. \(\sin (2 x) \cos (x)-\cos (2 x) \sin (x)\) c. \(2 \sin (4.8) \cos (4.8)\)

Verify that each equation is an identity. \(\frac{\sin (x-y)}{\sin x \sin y}=\cot y-\cot x\)

Prove that each equation is an identity. \((\sin \alpha-\cos \alpha)^{2}=1-\sin 2 \alpha\)

Prove that each equation is an identity. \((\sin \alpha-\cos \alpha)^{2}=1-\sin 2 \alpha\)
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