7.4: Sum-to-Product and Product-to-Sum Formulas
Expressing Products as Sums
We have already learned a number of formulas useful for expanding or simplifying trigonometric expressions, but sometimes we may need to express the product of cosine and sine as a sum. We can use the product-to-sum formulas, which express products of trigonometric functions as sums. Let’s investigate the cosine identity first and then the sine identity.
Expressing Products as Sums for Cosine
We can derive the product-to-sum formula from the sum and difference identities for cosine . If we add the two equations, we get:
\[\begin{align*} \cos \alpha \cos \beta+\sin \alpha \sin \beta&= \cos(\alpha-\beta)\\[4pt] \underline{+ \cos \alpha \cos \beta-\sin \alpha \sin \beta}&= \underline{ \cos(\alpha+\beta) }\\[4pt] 2 \cos \alpha \cos \beta&= \cos(\alpha-\beta)+\cos(\alpha+\beta)\end{align*}\]
Then, we divide by 2 to isolate the product of cosines:
\[ \cos \alpha \cos \beta= \dfrac{1}{2}[\cos(\alpha-\beta)+\cos(\alpha+\beta)] \label{eq1}\]
- Write the formula for the product of cosines.
- Substitute the given angles into the formula.
- Simplify.
Write the following product of cosines as a sum: \(2\cos\left(\dfrac{7x}{2}\right) \cos\left(\dfrac{3x}{2}\right)\).
Solution
We begin by writing the formula for the product of cosines (Equation \ref{eq1}):
\[ \cos \alpha \cos \beta = \dfrac{1}{2}[ \cos(\alpha-\beta)+\cos(\alpha+\beta) ] \nonumber \]
We can then substitute the given angles into the formula and simplify.
\[\begin{align*} 2 \cos\left(\dfrac{7x}{2}\right)\cos\left(\dfrac{3x}{2}\right)&= 2\left(\dfrac{1}{2}\right)[ \cos\left(\dfrac{7x}{2}-\dfrac{3x}{2}\right)+\cos\left(\dfrac{7x}{2}+\dfrac{3x}{2}\right) ]\\[4pt] &= \cos\left(\dfrac{4x}{2}\right)+\cos\left(\dfrac{10x}{2}\right) \\[4pt] &= \cos 2x+\cos 5x \end{align*}\]
Use the product-to-sum formula (Equation \ref{eq1}) to write the product as a sum or difference: \(\cos(2\theta)\cos(4\theta)\).
- Answer
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\(\dfrac{1}{2}(\cos 6\theta+\cos 2\theta)\)
Expressing the Product of Sine and Cosine as a Sum
Next, we will derive the product-to-sum formula for sine and cosine from the sum and difference formulas for sine . If we add the sum and difference identities, we get:
\[\begin{align*} \cos \alpha \cos \beta+\sin \alpha \sin \beta&= \cos(\alpha-\beta)\\[4pt] \underline{+ \cos \alpha \cos \beta-\sin \alpha \sin \beta}&= \cos(\alpha+\beta)\\[4pt] 2 \cos \alpha \cos \beta&= \cos(\alpha-\beta)+\cos(\alpha+\beta)\\[4pt] \text{Then, we divide by 2 to isolate the product of cosines:}\\[4pt] \cos \alpha \cos \beta&= \dfrac{1}{2}\left[\cos(\alpha-\beta)+\cos(\alpha+\beta)\right] \end{align*}\]
Express the following product as a sum containing only sine or cosine and no products: \(\sin(4\theta)\cos(2\theta)\).
Solution
Write the formula for the product of sine and cosine. Then substitute the given values into the formula and simplify.
\[\begin{align*} \sin \alpha \cos \beta&= \dfrac{1}{2}[ \sin(\alpha+\beta)+\sin(\alpha-\beta) ]\\[4pt] \sin(4\theta)\cos(2\theta)&= \dfrac{1}{2}[\sin(4\theta+2\theta)+\sin(4\theta-2\theta)]\\[4pt] &= \dfrac{1}{2}[\sin(6\theta)+\sin(2\theta)] \end{align*}\]
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:
\[\begin{align*} \cos(\alpha-\beta)&= \cos \alpha \cos \beta+\sin \alpha \sin \beta\\[4pt] \underline{-\cos(\alpha+\beta)}&= -(\cos \alpha \cos \beta-\sin \alpha \sin \beta)\\[4pt] \cos(\alpha-\beta)-\cos(\alpha+\beta)&= 2 \sin \alpha \sin \beta\\[4pt] \text{Then, we divide by 2 to isolate the product of sines:}\\[4pt] \sin \alpha \sin \beta&= \dfrac{1}{2}[ \cos(\alpha-\beta)-\cos(\alpha+\beta) ] \end{align*}\]
Similarly we could express the product of cosines in terms of sine or derive other product-to-sum formulas.
The product-to-sum formulas are as follows:
\[\cos \alpha \cos \beta=\dfrac{1}{2}[\cos(\alpha−\beta)+\cos(\alpha+\beta)]\]
\[\sin \alpha \cos \beta=\dfrac{1}{2}[\sin(\alpha+\beta)+\sin(\alpha−\beta)]\]
\[\sin \alpha \sin \beta=\dfrac{1}{2}[\cos(\alpha−\beta)−\cos(\alpha+\beta)]\]
\[\cos \alpha \sin \beta=\dfrac{1}{2}[\sin(\alpha+\beta)−\sin(\alpha−\beta)]\]
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 \(\dfrac{u+v}{2}=\alpha\) and \(\dfrac{u−v}{2}=\beta\).
Then,
\[\begin{align*} \alpha+\beta&= \dfrac{u+v}{2}+\dfrac{u-v}{2}\\[4pt] &= \dfrac{2u}{2}\\[4pt] &= u \end{align*}\]
\[\begin{align*} \alpha-\beta&= \dfrac{u+v}{2}-\dfrac{u-v}{2}\\[4pt] &= \dfrac{2v}{2}\\[4pt] &= v \end{align*}\]
Thus, replacing \(\alpha\) and \(\beta\) in the product-to-sum formula with the substitute expressions, we have
\[\begin{align*} \sin \alpha \cos \beta&= \dfrac{1}{2}[\sin(\alpha+\beta)+\sin(\alpha-\beta)]\\[4pt] \sin \left ( \frac{u+v}{2} \right ) \cos \left ( \frac{u-v}{2} \right )&= \frac{1}{2}[\sin u + \sin v]\qquad \text{Substitute for } (\alpha+\beta) \text{ and } (\alpha\beta)\\[4pt] 2\sin\left(\dfrac{u+v}{2}\right) \cos\left(\dfrac{u-v}{2}\right)&= \sin u+\sin v \end{align*}\]
The other sum-to-product identities are derived similarly.
The sum-to-product formulas are as follows:
\[\sin \alpha+\sin \beta=2\sin\left(\dfrac{\alpha+\beta}{2}\right)\cos\left(\dfrac{\alpha−\beta}{2}\right)\]
\[\sin \alpha-\sin \beta=2\sin\left(\dfrac{\alpha-\beta}{2}\right)\cos\left(\dfrac{\alpha+\beta}{2}\right)\]
\[\cos \alpha−\cos \beta=−2\sin\left(\dfrac{\alpha+\beta}{2}\right)\sin\left(\dfrac{\alpha−\beta}{2}\right)\]
\[\cos \alpha+\cos \beta=2\sin\left(\dfrac{\alpha+\beta}{2}\right)\sin\left(\dfrac{\alpha−\beta}{2}\right)\]
Write the following difference of sines expression as a product: \(\sin(4\theta)−\sin(2\theta)\).
Solution
We begin by writing the formula for the difference of sines.
\[\begin{align*} \sin \alpha-\sin \beta&= 2\sin\left(\dfrac{\alpha-\beta}{2}\right)\cos\left(\dfrac{\alpha+\beta}{2}\right)\\[4pt] \text {Substitute the values into the formula, and simplify.}\\[4pt] \sin(4\theta)-\sin(2\theta)&= 2\sin\left(\dfrac{4\theta-2\theta}{2}\right) \cos\left(\dfrac{4\theta+2\theta}{2}\right)\\[4pt] &= 2\sin\left(\dfrac{2\theta}{2}\right) \cos\left(\dfrac{6\theta}{2}\right)\\[4pt] &= 2 \sin \theta \cos(3\theta) \end{align*}\]