<img src="https://d5nxst8fruw4z.cloudfront.net/atrk.gif?account=iA1Pi1a8Dy00ym" style="display:none" height="1" width="1" alt="" />

# 3.14: Product to Sum Formulas for Sine and Cosine

Difficulty Level: At Grade Created by: CK-12
Estimated10 minsto complete
%
Progress
Practice Product to Sum Formulas for Sine and Cosine
Progress
Estimated10 minsto complete
%

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:

sin75sin15\begin{align*}\sin 75^\circ \sin 15^\circ\end{align*}

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.

cos(ab)=cosacosb+sinasinb and cos(a+b)=cosacosbsinasinbcos(ab)cos(a+b)=cosacosb+sinasinb(cosacosbsinasinb)cos(ab)cos(a+b)=cosacosb+sinasinbcosacosb+sinasinbcos(ab)cos(a+b)=2sinasinb12[cos(ab)cos(a+b)]=sinasinb\begin{align*}& \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\\ & \qquad \qquad \qquad \qquad \cos (a - b) - \cos (a + b) = 2 \sin a \sin b\\ & \qquad \qquad \qquad \qquad \frac{1}{2}\left[ \cos (a - b) - \cos (a+ b)\right] = \sin a \sin b\end{align*}

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

cosαcosβsinαcosβcosαsinβ=12[cos(αβ)+cos(α+β)]=12[sin(α+β)+sin(αβ)]=12[sin(α+β)sin(αβ)]\begin{align*}\cos \alpha \cos \beta &= \frac {1}{2} \left [ \cos (\alpha - \beta) + \cos (\alpha + \beta) \right ]\\ \sin \alpha \cos \beta &= \frac {1}{2} \left [\sin (\alpha + \beta) + \sin (\alpha - \beta) \right ]\\ \cos \alpha \sin \beta &= \frac {1}{2} \left[\sin (\alpha + \beta) - \sin (\alpha - \beta) \right ]\end{align*}

#### Using the Product to Sum Formula

1. Change cos2xcos5y\begin{align*}\cos 2x \cos 5y\end{align*} to a sum.

Use the formula cosαcosβ=12[cos(αβ)+cos(α+β)]\begin{align*}\cos \alpha \cos \beta = \frac{1}{2} \left [\cos (\alpha - \beta) + \cos (\alpha + \beta) \right ]\end{align*}. Set α=2x\begin{align*}\alpha = 2x\end{align*} and β=5y\begin{align*}\beta = 5y\end{align*}.

cos2xcos5y=12[cos(2x5y)+cos(2x+5y)]\begin{align*}\cos 2x \cos 5y = \frac{1}{2} \left [\cos (2x - 5y) + \cos (2x + 5y) \right ]\end{align*}

2. Change sin11z+sinz2\begin{align*}\frac{\sin11z + \sin z}{2}\end{align*} to a product.

Use the formula sinαcosβ=12[sin(α+β)+sin(αβ)]\begin{align*}\sin \alpha \cos \beta = \frac{1}{2} \left [\sin (\alpha + \beta) + \sin (\alpha - \beta) \right ]\end{align*}. Therefore, α+β=11z\begin{align*}\alpha + \beta = 11z\end{align*} and αβ=z\begin{align*}\alpha - \beta = z\end{align*}. Solve the second equation for α\begin{align*}\alpha\end{align*} and plug that into the first.

α=z+β(z+β)+β=11zz+2β=11z2β=10zβ=5zandα=z+5z=6z\begin{align*}\alpha = z + \beta \rightarrow (z + \beta) + \beta = 11z && \text {and} \quad \alpha = z + 5z = 6z\\ z + 2 \beta = 11z\\ 2 \beta = 10z\\ \beta = 5z\end{align*}

sin11z+sinz2=sin6zcos5z\begin{align*}\frac{\sin11z + \sin z}{2} = \sin 6z \cos 5z\end{align*}. Again, the sum of 6z\begin{align*}6z\end{align*} and 5z\begin{align*}5z\end{align*} is 11z\begin{align*}11z\end{align*} and the difference is z\begin{align*}z\end{align*}.

3. Solve cos5x+cosx=cos2x\begin{align*}\cos 5x + \cos x = \cos 2x\end{align*}.

Use the formula cosα+cosβ=2cosα+β2×cosαβ2\begin{align*}\cos \alpha + \cos \beta = 2 \cos \frac{\alpha +\beta}{2} \times \cos \frac{\alpha - \beta}{2}\end{align*}.

cos5x+cosx=cos2x2cos3xcos2x=cos2x2cos3xcos2xcos2x=0 cos2x(2cos3x1)=0   cos2x=02cos3x1=0 2cos3x=1  2x=π2,3π2and cos3x=12 x=π4,3π43x=π3,5π3,7π3,11π3,13π3,17π3 x=π9,5π9,7π9,11π9,13π9,17π9\begin{align*}& \qquad \qquad \cos 5x + \cos x = \cos 2x\\ & \qquad \qquad 2 \cos 3x \cos 2x = \cos 2x\\ & 2 \cos 3x \cos 2x - \cos 2x = 0\\ & \quad \ \cos 2x(2 \cos 3x-1)=0\\ & \quad \ \ \swarrow \qquad \qquad \ \searrow\\ & \cos 2x = 0 \qquad \qquad 2 \cos 3x-1=0\\ & \qquad \qquad \qquad \qquad \qquad \ 2\cos 3x = 1\\ & \quad \ \ 2x=\frac{\pi}{2}, \frac{3\pi}{2} \quad \text{and} \quad \ \cos 3x=\frac{1}{2}\\ & \qquad \ x = \frac{\pi}{4}, \frac{3 \pi}{4} \qquad \qquad \qquad 3x = \frac{\pi}{3}, \frac{5 \pi}{3}, \frac{7 \pi}{3}, \frac{11 \pi}{3}, \frac{13 \pi}{3}, \frac{17 \pi}{3}\\ & \qquad \qquad \qquad \qquad \qquad \qquad \quad \ x = \frac{\pi}{9}, \frac{5\pi}{9}, \frac{7 \pi}{9}, \frac{11 \pi}{9}, \frac{13 \pi}{9}, \frac{17 \pi}{9}\end{align*}

### Examples

#### Example 1

Earlier, you were asked to solve sin 75°sin15°.

Changing sin75sin15\begin{align*}\sin 75^\circ \sin 15^\circ\end{align*} to a product of trig functions can be accomplished using

sinasinb=12[cos(ab)cos(a+b)]\begin{align*}\sin a \sin b = \frac{1}{2}\left[ \cos (a - b) - \cos (a+ b)\right]\end{align*}

Substituting in known values gives:

sin75sin15=12[cos(60)cos(90)]=12[120]=14\begin{align*}\sin 75^\circ \sin 15^\circ = \frac{1}{2}\left[ \cos (60^\circ) - \cos (90^\circ)\right] = \frac{1}{2}[\frac{1}{2} - 0] = \frac{1}{4}\end{align*}

#### Example 2

Express the product as a sum: \begin{align*}\sin(6 \theta) \sin(4 \theta) \end{align*}

Using the product-to-sum formula:

\begin{align*}& \qquad \quad \sin 6 \theta \sin 4 \theta\\ & \frac{1}{2} \left( \cos (6 \theta - 4 \theta) - \cos (6 \theta + 4 \theta) \right )\\ & \qquad \frac{1}{2}(\cos 2 \theta - \cos 10 \theta)\end{align*}

#### Example 3

Express the product as a sum: \begin{align*}\sin(5 \theta) \cos(2 \theta)\end{align*}

Using the product-to-sum formula:

#### Example 4

Express the product as a sum: \begin{align*}\cos(10 \theta) \sin(3 \theta)\end{align*}

Using the product-to-sum formula:

\begin{align*} & \qquad \quad \cos 10 \theta \sin 3 \theta\\ & \frac{1}{2} \left( \sin ( 10 \theta + 3 \theta) - \sin ( 10 \theta - 3 \theta) \right )\\ & \qquad \frac{1}{2}(\sin 13 \theta - \sin 7 \theta)\end{align*}

### Review

Express each product as a sum or difference.

1. \begin{align*}\sin(5 \theta) \sin(3 \theta) \end{align*}
2. \begin{align*}\sin(6 \theta) \cos( \theta) \end{align*}
3. \begin{align*}\cos(4 \theta) \sin(3 \theta) \end{align*}
4. \begin{align*}\cos(\theta) \cos(4 \theta) \end{align*}
5. \begin{align*}\sin(2 \theta) \sin(2 \theta) \end{align*}
6. \begin{align*}\cos(6 \theta) \sin(8 \theta) \end{align*}
7. \begin{align*}\sin(7 \theta) \cos(4 \theta) \end{align*}
8. \begin{align*}\cos(11 \theta) \cos(2 \theta) \end{align*}

Express each sum or difference as a product.

1. \begin{align*}\frac{\sin8\theta + \sin6\theta}{2}\end{align*}
2. \begin{align*}\frac{\sin6\theta - \sin2\theta}{2}\end{align*}
3. \begin{align*}\frac{\cos12\theta + \cos6\theta}{2}\end{align*}
4. \begin{align*}\frac{\cos12\theta - \cos4\theta}{2}\end{align*}
5. \begin{align*}\frac{\sin10\theta + \sin4\theta}{2}\end{align*}
6. \begin{align*}\frac{\sin8\theta - \sin2\theta}{2}\end{align*}
7. \begin{align*}\frac{\cos8\theta - \cos4\theta}{2}\end{align*}

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

### Notes/Highlights Having trouble? Report an issue.

Color Highlighted Text Notes

### Vocabulary Language: English

Product to Sum Formula

A product to sum formula relates the product of two trigonometric functions to the sum of two trigonometric functions.

Show Hide Details
Description
Difficulty Level:
Tags:
Subjects: