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3.14: Product to Sum Formulas for Sine and Cosine

Difficulty Level: At Grade Created by: CK-12
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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:

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

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{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*}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

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

\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*}cosαcosβsinαcosβcosαsinβ=12[cos(αβ)+cos(α+β)]=12[sin(α+β)+sin(αβ)]=12[sin(α+β)sin(αβ)]

Using the Product to Sum Formula

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

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

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

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

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

\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*}α=z+β(z+β)+β=11zz+2β=11z2β=10zβ=5zandα=z+5z=6z

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

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

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

\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*}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


Example 1

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

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

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

Substituting in known values gives:

\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*}sin75sin15=12[cos(60)cos(90)]=12[120]=14

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:

\begin{align*} & \qquad \quad \sin 5 \theta \cos 2 \theta\\ & \frac{1}{2} \left( \sin ( 5 \theta + 2 \theta) - \sin ( 5 \theta - 2 \theta) \right )\\ & \qquad \frac{1}{2}(\sin 7 \theta - \sin 3 \theta)\end{align*}

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*}


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*}

Review (Answers)

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

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Product to Sum Formula

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

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Difficulty Level:
At Grade
Date Created:
Sep 26, 2012
Last Modified:
Mar 23, 2016
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