7.3. (B) Product-Sum Formulas (Not on the Quiz and Not on the Exam)

Product-Sum Formulas

$sin u cos v = \frac{1}{2} [sin (u + v) + sin (u - v)$ ]

$cos u sin v = \frac{1}{2} [sin (u + v) - sin (u - v)]$

$cos u cos v = \frac{1}{2} [cos (u + v) + cos (u - v)]$

$sin u sin v = \frac{1}{2} [cos (u - v) - cos (u + v)]$

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Example

Write the product as a sum.

$sin (3 x) cos (4 x)$

Solution

From product-sum formula $sin u cos v = \frac{1}{2} [sin (u + v) + sin (u - v)$ ], we have

$sin (3 x) cos (4 x) = \frac{1}{2} [sin (3 x + 4 x) + sin (3 x - 4 x)] = \frac{1}{2} [sin 7 x + sin (- x)] = \frac{1}{2} (sin 7 x - sin x)$

because sine function is an odd function.

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$cos x sin 7 x$

Solution

From product-sum formula $cos u sin v = \frac{1}{2} [sin (u + v) - sin (u - v)]$ , we have

$cos x sin 7 x = \frac{1}{2} [sin (x + 7 x) - s i n (x - 7 x)]$

$= \frac{1}{2} [sin (8 x) - s i n (- 6 x)]$

$= \frac{1}{2} [sin (8 x) + s i n (6 x)]$

because sine is an odd function, that is, $LaTeX: \sin\left(-x\right)=-\sin x$

Remark

$cos x sin 7 x$ can be written as $sin 7 x cos x$ . Thus, you can use $sin u cos v = \frac{1}{2} [sin (u + v) + sin (u - v)$ ].

In fact, two formulas for $sin u cos v$ and $cos u sin v$ are equivalent formulas.

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$cos 7 x cos 4 x$

Solution

From the product-sum formula $cos u cos v = \frac{1}{2} [cos (u + v) + cos (u - v)]$ , we have

$cos 7 x cos 4 x = \frac{1}{2} [cos (7 x + 4 x) + cos (7 x - 4 x)] = \frac{1}{2} [cos 11 x + cos 3 x]$

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$sin x sin 9 x$

Solution

From the product-sum formula $sin u sin v = \frac{1}{2} [cos (u - v) - cos (u + v)]$ , we have

$sin x sin 9 x = \frac{1}{2} [cos (x - 9 x) - cos (x + 9 x)]$

$= \frac{1}{2} [cos (- 8 x) - cos (10 x)]$

$= \frac{1}{2} (cos 8 x - cos 10 x)$

because cosine function is an even function, that is, $LaTeX: \cos\left(-x\right)=\cos x$

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Sum-Product Formulas

$sin x + sin y = 2 sin \frac{x + y}{2} cos \frac{x - y}{2}$

$sin x - sin y = 2 cos \frac{x + y}{2} sin \frac{x - y}{2}$

$cos x + cos y = 2 cos \frac{x + y}{2} cos \frac{x - y}{2}$

$cos x - cos y = - 2 sin \frac{x + y}{2} sin \frac{x - y}{2}$

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Example

Write a sum as a product.

$sin 4 x + sin 2 x$

Solution

From sum-product formula $sin x + sin y = 2 sin \frac{x + y}{2} cos \frac{x - y}{2}$ , we have

$sin 4 x + sin 2 x = 2 sin \frac{4 x + 2 x}{2} cos \frac{4 x - 2 x}{2} = 2 sin \frac{6 x}{2} cos \frac{2 x}{2} = 2 sin 3 x cos x$

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$sin x - sin 4 x$

Solution

From sum-product formula $sin x - sin y = 2 cos \frac{x + y}{2} sin \frac{x - y}{2}$ , we have

$sin x - sin 4 x = 2 cos \frac{x + 4 x}{2} sin \frac{x - 4 x}{2} = 2 cos \frac{5 x}{2} sin (- \frac{3 x}{2}) = - 2 cos \frac{5 x}{2} sin \frac{3 x}{2}$

Because sine function is an odd function.

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$cos 5 x + cos 2 x$

Solution

From sum-product formula $cos x + cos y = 2 cos \frac{x + y}{2} cos \frac{x - y}{2}$ , we have

$cos 5 x + cos 2 x = 2 cos \frac{5 x + 2 x}{2} cos \frac{5 x - 2 x}{2} = 2 cos \frac{7 x}{2} cos \frac{3 x}{2}$

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$cos 6 x - cos 8 x$

Solution

From sum-product formula $cos x - cos y = - 2 sin \frac{x + y}{2} sin \frac{x - y}{2}$ , we have

$cos 6 x - cos 8 x = - 2 sin \frac{6 x + 8 x}{2} sin \frac{6 x - 8 x}{2} = - 2 sin \frac{14 x}{2} sin (\frac{- 2 x}{2}) = - 2 sin 7 x sin (- x) = 2 sin 7 x sin x$

because sine function is an odd function, that is, $sin (- x) = - sin x .$