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Ch. 1 - Functions
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 1, Problema 86

Prove the following identities.
sinθ1+cosθ=1−cosθsinθ\(\frac{\sin\theta}{1+\cos\theta}\)=\(\frac{1-\cos\theta}{\sin\theta}\)

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Start by examining the left-hand side of the identity: \( \frac{\sin\theta}{1+\cos\theta} \).
Multiply both the numerator and the denominator of the left-hand side by the conjugate of the denominator, which is \( 1-\cos\theta \). This gives: \( \frac{\sin\theta (1-\cos\theta)}{(1+\cos\theta)(1-\cos\theta)} \).
Simplify the denominator using the difference of squares formula: \((1+\cos\theta)(1-\cos\theta) = 1 - \cos^2\theta\).
Recognize that \(1 - \cos^2\theta\) is equal to \(\sin^2\theta\) by the Pythagorean identity.
Simplify the expression: \( \frac{\sin\theta (1-\cos\theta)}{\sin^2\theta} = \frac{1-\cos\theta}{\sin\theta} \), which matches the right-hand side of the identity.

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