# 3.5.2: Product to Sum Formulas for Sine and Cosine

- Page ID
- 4240

Relation of the product of two trigonometric functions to a sum or difference.

Let's say you are in class one day, working on calculating the values of trig functions, when your instructor gives you an equation like this:

\(\sin 75^{\circ}\sin 15^{\circ}\)

Can you solve this sort of equation? You might want to just calculate each term separately and then compute the result. However, there is another way. You can transform this product of trig functions into a sum of trig functions.

Read on, and by the end of this lesson, you'll know how to solve this problem by changing it into a sum of trig functions.

**Product to Sum Formulas for sine and cosine**

Here we'll begin by deriving formulas for how to convert the product of two trig functions into a sum or difference of trig functions.

There are two formulas for transforming a product of sine or cosine into a sum or difference. First, let’s look at the product of the sine of two angles. To do this, we need to start with the cosine of the difference of two angles.

\(\begin{array}{l}

\cos (a-b)=\cos a \cos b+\sin a \sin b \text { and } \cos (a+b)=\cos a \cos b-\sin a \sin b \\

\cos (a-b)-\cos (a+b)=\cos a \cos b+\sin a \sin b-(\cos a \cos b-\sin a \sin b) \\

\cos (a-b)-\cos (a+b)=\cos a \cos b+\sin a \sin b-\cos a \cos b+\sin a \sin b \\

\cos (a-b)-\cos (a+b)=2 \sin a \sin b \\

\dfrac{1}{2}[\cos (a-b)-\cos (a+b)]=\sin a \sin b

\end{array}\)

The following product to sum formulas can be derived using the same method:

\(\begin{aligned} \cos \alpha \cos \beta=\dfrac{1}{2}[\cos (\alpha −\beta )+\cos (\alpha +\beta )] \\ \sin \alpha \cos \beta=\dfrac{1}{2}[\sin (\alpha +\beta )+\sin (\alpha −\beta )] \\ \cos \alpha \sin \beta =\dfrac{1}{2}[\sin (\alpha +\beta )−\sin (\alpha −\beta )] \end{aligned}\)

**Using the ****Product to Sum Formula**

**Product to Sum Formula**

1. Change \(\cos 2x\cos 5y\) to a sum.

Use the formula \(\cos \alpha \cos \beta =\dfrac{1}{2}[\cos (\alpha −\beta )+\cos (\alpha +\beta )]\). Set \(\alpha =2x\) and \(\beta =5y\).

\(\cos 2x \cos 5y=\dfrac{1}{2}[\cos (2x−5y)+\cos (2x+5y)]\)

2. Change \(\dfrac{\sin 11z+\sin z}{2}\) to a product.

Use the formula \(\sin \alpha \cos \beta =\dfrac{1}{2}[\sin (\alpha +\beta )+\sin (\alpha −\beta )]\). Therefore, \(\alpha +\beta =11z\) and \(\alpha −\beta =z\). Solve the second equation for \alpha and plug that into the first.

\(\begin{aligned} \alpha =z+\beta \rightarrow (z+\beta )+\beta &=11z \qquad \text{ and }\quad \alpha =z+5z=6z\\ z+2\beta &=11z \\ 2\beta &=10z \\ \beta &=5z \end{aligned}\)

\(\dfrac{\sin 11z+\sin z}{2}=\sin 6z\cos 5z\). Again, the sum of \(6z\) and \(5z\) is \(11z\) and the difference is \(z\).

3. Solve \(\cos 5x+\cos x=\cos 2x\).

Use the formula \(\cos \alpha +\cos \beta =2\cos \dfrac{\alpha +\beta }{2}\times \cos \dfrac{\alpha −\beta }{2}\).

\(\begin{aligned}

\cos 5 x+\cos x&=\cos 2 x \\

2 \cos 3 x \cos 2 x&=\cos 2 x \\

2 \cos 3 x \cos 2 x-\cos 2 x&=0 \\

\cos 2 x(2 \cos 3 x-1)&=0 \end{aligned} \\ \begin{aligned}

\swarrow &&\searrow \qquad \qquad &\\

\cos 2 x&=0 & 2 \cos 3 x-1&=0 \\

& & 2 \cos 3 x&=1 \\

2 x&=\dfrac{\pi}{2}, \dfrac{3 \pi}{2} &\text { and } \quad \cos 3 x=& \dfrac{1}{2} \\

x&=\dfrac{\pi}{4}, \dfrac{3 \pi}{4} & 3 x&=\dfrac{\pi}{3}, \dfrac{5 \pi}{3}, \dfrac{7 \pi}{3}, \dfrac{11 \pi}{3}, \dfrac{13 \pi}{3}, \dfrac{17 \pi}{3} \\

& & x&=\dfrac{\pi}{9}, \dfrac{5 \pi}{9}, \dfrac{7 \pi}{9}, \dfrac{11 \pi}{9}, \dfrac{13 \pi}{9}, \dfrac{17 \pi}{9}

\end{aligned}\)

Example \(\PageIndex{1}\)

Earlier, you were asked to solve sin \(75^{\circ} \sin 15^{\circ}\).

**Solution**

Changing \(\sin 75^{\circ}\sin 15^{\circ}\) to a product of trig functions can be accomplished using

\(\sin a\sin b=\dfrac{1}{2} [\cos (a−b)−\cos (a+b)]\)

Substituting in known values gives:

\(\sin 75^{\circ}\sin 15^{\circ}=\dfrac{1}{2} [\cos (60^{\circ})−\cos (90^{\circ})]=\dfrac{1}{2}\left[\dfrac{1}{2}−0\right]=\dfrac{1}{4}\)

Example \(\PageIndex{2}\)

Express the product as a sum: \(\sin (6\theta )\sin (4\theta )\)

**Solution**

Using the product-to-sum formula:

\(\sin 6\theta \sin 4\theta \\ \dfrac{1}{2} (\cos (6\theta −4\theta )−\cos (6\theta +4\theta )) \\ \dfrac{1}{2} (\cos 2\theta −\cos 10\theta )\)

Example \(\PageIndex{3}\)

Express the product as a sum: \(\sin (5\theta )\cos (2\theta )\)

**Solution**

Using the product-to-sum formula:

\(\sin 5\theta \cos 2\theta \\ \dfrac{1}{2}(\sin (5\theta +2\theta )−\sin (5\theta −2\theta )) \\ \dfrac{1}{2} (\sin 7\theta −\sin 3\theta )\)

Example \(\PageIndex{4}\)

Express the product as a sum: \(\cos (10\theta )\sin (3\theta )\)

**Solution**

Using the product-to-sum formula:

\(\cos 10\theta \sin 3\theta \\ \dfrac{1}{2}(\sin (10\theta +3\theta )−\sin (10\theta −3\theta )) \\ \dfrac{1}{2} (\sin 13\theta −\sin 7\theta ) \)

**Review**

Express each product as a sum or difference.

- \(\sin (5\theta )\sin (3\theta )\)
- \(\sin (6\theta )\cos (\theta )\)
- \(\cos (4\theta )\sin (3\theta )\)
- \(\cos (\theta )\cos (4\theta )\)
- \(\sin (2\theta )\sin (2\theta )\)
- \(\cos (6\theta )\sin (8\theta )\)
- \(\sin (7\theta )\cos (4\theta )\)
- \(\cos (11\theta )\cos (2\theta )\)

Express each sum or difference as a product.

- \(\dfrac{\sin 8\theta +\sin 6\theta }{2}\)
- \(\dfrac{\sin 6\theta −\sin 2\theta }{2}\)
- \(\dfrac{\cos 12\theta +\cos 6\theta }{2}\)
- \(\dfrac{\cos 12\theta −\cos 4\theta }{2}\)
- \(\dfrac{\sin 10\theta +\sin 4\theta }{2}\)
- \(\dfrac{\sin 8\theta −\sin 2\theta }{2}\)
- \(\dfrac{\cos 8\theta −\cos 4\theta }{2}\)

**Review (Answers)**

To see the Review answers, open this PDF file and look for section 3.14.

## Vocabulary

Term | Definition |
---|---|

Product to Sum Formula |
A product to sum formula relates the product of two trigonometric functions to the sum of two trigonometric functions. |