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Chapter 6: Problem 51

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

Short Answer

Expert verified
Using the angle addition formula, we find: \(\sin x - \sin y = (\sin(x-y)\cos y + \cos(x-y)\sin y) - (\sin(y-x)\cos x + \cos(y-x)\sin x)\) Applying sum-to-product identities, we have: \(\sin x - \sin y = 2\sin\frac{x-y}{2}(\cos\frac{x+y}{2} + \sin\frac{x+y}{2})\) Substituting the angle addition formula for the sine function in the last term, we obtain: \(\sin x - \sin y = 2 \cos \frac{x+y}{2} \sin \frac{x-y}{2}\) Thus, the given trigonometric equation holds true for all values of \(x\) and \(y\).

Step by step solution

01

Apply Angle Addition Formula to both sides

First, apply the angle addition formula to both \(\sin x\) and \(\sin y\): \(\sin x = \sin((x-y) + y) = \sin(x-y)\cos y + \cos(x-y)\sin y\) \(\sin y = \sin(y-x) + x) = \sin(y-x)\cos x + \cos(y-x)\sin x\) Subtracting the second equation from the first equation gives: \(\sin x - \sin y = (\sin(x-y)\cos y + \cos(x-y)\sin y) - (\sin(y-x)\cos x + \cos(y-x)\sin x)\)
02

Apply Sum-to-Product Identities

Notice that \(\sin(x-y)\cos y - \sin(y-x)\cos x = 2\sin\frac{x-y}{2}\cos\frac{x+y}{2}\) and \(\cos(x-y)\sin y - \cos(y-x)\sin x = 2\sin\frac{x-y}{2}\sin\frac{x+y}{2}\). Substitute these identities into the equation obtained in Step 1: \(2\sin\frac{x-y}{2}\cos\frac{x+y}{2} + 2\sin\frac{x-y}{2}\sin\frac{x+y}{2} = 2\sin\frac{x-y}{2}(\cos\frac{x+y}{2} + \sin\frac{x+y}{2})\)
03

Substitute the angle addition formula for the sine

Finally, substitute back the angle addition formula for the sine function in the last term: \(\sin\frac{x-y}{2}(\cos\frac{x+y}{2} + \sin\frac{x+y}{2}) = \sin\frac{x-y}{2}(2\cos\frac{x+y}{2}\sin\frac{x-y}{2})\) We have arrived at the given trigonometric equation: \(\sin x - \sin y = 2 \cos \frac{x+y}{2} \sin \frac{x-y}{2}\) Since we performed the operations and substitutions without any constraints, the equation is true for all values of \(x\) and \(y\).

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