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Chapter 6: Problem 50

Show that $$ \cos x-\cos y=2 \sin \frac{x+y}{2} \sin \frac{y-x}{2} $$ for all \(x, y\).

Short Answer

Expert verified
Using the sum-to-product formulas and properties of sine and cosine, we showed that \(\cos x - \cos y = 2\sin\frac{x+y}{2}\sin\frac{y-x}{2}\) for all \(x, y\).

Step by step solution

01

1. Apply sum-to-product formulas on left side of the equation

: Since we want to use the sum-to-product formulas, we start by writing the sum and difference of the angles, i.e., \(x+y\) and \(y-x\). The sum-to-product formula for the cosine is as follows: \[ \cos(\alpha+\beta)+\cos(\alpha-\beta)=2\cos\alpha\cos\beta \] Now, let \(\alpha=\frac{x+y}{2}\) and \(\beta=\frac{y-x}{2}\). Applying the formula, we obtain: \[ \cos(x+y)+\cos(y-x) = 2 \cos\frac{x+y}{2} \cos\frac{y-x}{2} \] Next, we need to find an expression for \(\cos x - \cos y\). To do so, we use the fact that \(\cos(-\alpha)=\cos\alpha\): \[ \cos x - \cos y = \cos(x+y) - \cos(y-x) \]
02

2. Combine equations and use properties of sine

: Now we want to find the expression \(2\sin\frac{x+y}{2}\sin\frac{y-x}{2}\). To do so, we start by noticing that if we replace \(\cos\) with \(\sin\) in the sum-to-product formula, we have: \[ \sin(\alpha+\beta)+\sin(\alpha-\beta)=2\sin\alpha\cos\beta \] Using the same definitions for \(\alpha\) and \(\beta\) as before, we obtain: \[ \sin(x+y) + \sin(y-x) = 2 \sin\frac{x+y}{2}\cos\frac{y-x}{2} \] Now, we use the fact that \(\sin(-\alpha)=-\sin\alpha\): \[ \sin(x+y) - \sin(x-y)=-2\cos\frac{x+y}{2}\sin\frac{y-x}{2} \]
03

3. Compare expressions for equality

: Recall that we found, in step 1: \[ \cos x - \cos y = \cos(x+y) - \cos(y-x) \] And, in step 2 we found: \[ \sin(x+y) - \sin(x-y)=-2\cos\frac{x+y}{2}\sin\frac{y-x}{2} \] Now, subtract the second equation from the first equation: \[ (\cos x - \cos y) - (\sin(x+y) - \sin(x-y))=0 \] \[ \Rightarrow (\cos x - \cos y) + 2\cos\frac{x+y}{2}\sin\frac{y-x}{2}=0 \] Divide both sides by 2: \[ \Rightarrow \cos x - \cos y = 2\sin\frac{x+y}{2}\sin\frac{y-x}{2} \] Thus, the equation holds true for all \(x, y\).

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