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Chapter 5: Problem 40

Verify the identity. $$\frac{\cos x-\cos y}{\sin x+\sin y}+\frac{\sin x-\sin y}{\cos x+\cos y}=0$$

Short Answer

Expert verified
The identity \(\frac{\cos x-\cos y}{\sin x+\sin y}+\frac{\sin x-\sin y}{\cos x+\cos y}=0\) holds true. It has been verified using algebraic manipulation and trigonometric identities.

Step by step solution

01

Multiply By Conjugate

We first multiply the terms by their conjugates to simplify them. Multiply \(\frac{\cos x-\cos y}{\sin x+\sin y}\) by \(\frac{\cos x+\cos y}{\cos x+\cos y}\) and \(\frac{\sin x-\sin y}{\cos x+\cos y}\) by \(\frac{\sin x+\sin y}{\sin x+\sin y}\). This gives us \(\frac{(\cos x - \cos y)(\cos x + \cos y)}{(\sin x + \sin y)(\cos x + \cos y)} + \frac{(\sin x - \sin y)(\sin x + \sin y)}{(\cos x + \cos y)(\sin x + \sin y)}\).
02

Expand the Numerators

Expand out using the difference of squares formula \(a^2 - b^2 = (a-b)(a+b)\). We then have \(\frac{\cos^2 x- \cos^2 y}{(\sin x + \sin y)(\cos x + \cos y)} + \frac{\sin^2 x - \sin^2 y}{(\cos x + \cos y)(\sin x + \sin y)}\).
03

Apply Pythagoras Identity

Since \(\cos^2 + \sin^2 = 1\), we can replace \(\cos^2 x\) with \(1 - \sin^2 x\), and \(\cos^2 y\) with \(1 - \sin^2 y\). We also replace \(\sin^2 x\) with \(1 - \cos^2 x\), and \(\sin^2 y\) with \(1 - \cos^2 y\). Now the equation becomes \(\frac{(1 - \sin^2 x) - (1 - \sin^2 y)}{(\sin x + \sin y)(\cos x + \cos y)} + \frac{(1- \cos^2 x) - (1 - \cos^2 y)}{(\cos x + \cos y)(\sin x + \sin y)}\).
04

Simplify

This simplifies to \(\frac{\sin^2 y - \sin^2 x}{(\sin x + \sin y)(\cos x + \cos y)} + \frac{\cos^2 y - \cos^2 x}{(\cos x + \cos y)(\sin x + \sin y)}\). Combine the two fractions since they have the same denominator. This gives the expression \(\frac{\sin^2 y - \sin^2 x + \cos^2 y - \cos^2 x}{(\sin x + \sin y)(\cos x + \cos y)}\). This simplifies to 0 because \(\sin^2 y - \sin^2 x + \cos^2 y - \cos^2 x = 0\) (Adding equal terms).
05

Verify the Identity

The initial expression simplifies to 0, which shows that the trigonometric identity has been verified.

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