7.4 Sum-to-product and product-to-sum formulas (Page 2/6)

Precalculus

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Expressing products of sines in terms of cosine

Expressing the product of sines in terms of cosine is also derived from the sum and difference identities for cosine. In this case, we will first subtract the two cosine formulas:

\begin{array}{l} \underset{____________________________________________________}{\begin{array}{l} \begin{array}{l} \cos (α - β) = \cos α \cos β + \sin α \sin β \end{array} \\ - \cos (α + β) = - (\cos α \cos β - \sin α \sin β) \end{array}} \\ \cos (α - β) - \cos (α + β) = 2 \sin α \sin β \end{array}

Then, we divide by 2 to isolate the product of sines:

\sin α \sin β = \frac{1}{2} [\cos (α - β) - \cos (α + β)]

Similarly we could express the product of cosines in terms of sine or derive other product-to-sum formulas.

A General Note

The product-to-sum formulas

The product-to-sum formulas are as follows:

\cos α \cos β = \frac{1}{2} [\cos (α - β) + \cos (α + β)]

\sin α \cos β = \frac{1}{2} [\sin (α + β) + \sin (α - β)]

\sin α \sin β = \frac{1}{2} [\cos (α - β) - \cos (α + β)]

\cos α \sin β = \frac{1}{2} [\sin (α + β) - \sin (α - β)]

Express the product as a sum or difference

Write $\cos (3 θ) \cos (5 θ)$ as a sum or difference.

We have the product of cosines, so we begin by writing the related formula. Then we substitute the given angles and simplify.

\begin{array}{l} \cos α \cos β = \frac{1}{2} [\cos (α - β) + \cos (α + β)] \\ \cos (3 θ) \cos (5 θ) = \frac{1}{2} [\cos (3 θ - 5 θ) + \cos (3 θ + 5 θ)] \\ = \frac{1}{2} [\cos (2 θ) + \cos (8 θ)] & Use even-odd identity . \end{array}

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Try It

Use the product-to-sum formula to evaluate $\cos \frac{11 π}{12} \cos \frac{π}{12} .$

$\frac{- 2 - \sqrt{3}}{4}$

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Expressing sums as products

Some problems require the reverse of the process we just used. The sum-to-product formulas allow us to express sums of sine or cosine as products. These formulas can be derived from the product-to-sum identities. For example, with a few substitutions, we can derive the sum-to-product identity for sine . Let $\frac{u + v}{2} = α$ and $\frac{u - v}{2} = β .$

Then,

\begin{array}{l} α + β = \frac{u + v}{2} + \frac{u - v}{2} \\ = \frac{2 u}{2} \\ = u \\ α - β = \frac{u + v}{2} - \frac{u - v}{2} \\ = \frac{2 v}{2} \\ = v \end{array}

Thus, replacing $α$ and $β$ in the product-to-sum formula with the substitute expressions, we have

\begin{array}{l} \sin α \cos β = \frac{1}{2} [\sin (α + β) + \sin (α - β)] \\ \sin (\frac{u + v}{2}) \cos (\frac{u - v}{2}) = \frac{1}{2} [\sin u + \sin v] & Substitute for (α + β) and (α - β) \\ 2 \sin (\frac{u + v}{2}) \cos (\frac{u - v}{2}) = \sin u + \sin v \end{array}

The other sum-to-product identities are derived similarly.

A General Note

Sum-to-product formulas

The sum-to-product formulas are as follows:

\sin α + \sin β = 2 \sin (\frac{α + β}{2}) \cos (\frac{α - β}{2})

\sin α - \sin β = 2 \sin (\frac{α - β}{2}) \cos (\frac{α + β}{2})

\cos α - \cos β = - 2 \sin (\frac{α + β}{2}) \sin (\frac{α - β}{2})

\cos α + \cos β = 2 \cos (\frac{α + β}{2}) \cos (\frac{α - β}{2})

Writing the difference of sines as a product

Write the following difference of sines expression as a product: $\sin (4 θ) - \sin (2 θ) .$

We begin by writing the formula for the difference of sines.

\sin α - \sin β = 2 \sin (\frac{α - β}{2}) \cos (\frac{α + β}{2})

Substitute the values into the formula, and simplify.

\begin{array}{l} \sin (4 θ) - \sin (2 θ) = 2 \sin (\frac{4 θ - 2 θ}{2}) \cos (\frac{4 θ + 2 θ}{2}) \\ = 2 \sin (\frac{2 θ}{2}) \cos (\frac{6 θ}{2}) \\ = 2 \sin θ \cos (3 θ) \end{array}

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Try It

Use the sum-to-product formula to write the sum as a product: $\sin (3 θ) + \sin (θ) .$

$2 \sin (2 θ) \cos (θ)$

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Evaluating using the sum-to-product formula

Evaluate $\cos (15^{\circ}) - \cos (75^{\circ}) .$

We begin by writing the formula for the difference of cosines.

\cos α - \cos β = - 2 \sin (\frac{α + β}{2}) \sin (\frac{α - β}{2})

Then we substitute the given angles and simplify.

\begin{array}{l} \cos (15^{\circ}) - \cos (75^{\circ}) = - 2 \sin (\frac{15^{\circ} + 75^{\circ}}{2}) \sin (\frac{15^{\circ} - 75^{\circ}}{2}) \\ = - 2 \sin (45^{\circ}) \sin (- 30^{\circ}) \\ = - 2 (\frac{\sqrt{2}}{2}) (- \frac{1}{2}) \\ = \frac{\sqrt{2}}{2} \end{array}

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Source: OpenStax, Precalculus. OpenStax CNX. Jan 19, 2016 Download for free at https://legacy.cnx.org/content/col11667/1.6

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