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3.14: Product to Sum Formulas for Sine and Cosine

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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:

\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 Concept, you'll know how to solve this problem by changing it into a sum of trig functions.

Watch This

In the second portion of this video you'll learn about Product to Sum formulas.

James Sousa: Sum to Product and Product to Sum Identities


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(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

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

\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 ]

Armed with these four formulas, we can work some examples.

Example A

Change \cos 2x \cos 5y to a sum.

Solution: Use the formula \cos \alpha \cos \beta = \frac{1}{2} \left [\cos (\alpha - \beta) +  \cos (\alpha + \beta) \right ] . Set \alpha = 2x and \beta = 5y .

\cos 2x \cos 5y = \frac{1}{2} \left [\cos (2x - 5y) + \cos (2x + 5y) \right ]

Example B

Change \frac{\sin11z + \sin z}{2} to a product.

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

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

\frac{\sin11z + \sin z}{2} = \sin 6z \cos 5z . Again, the sum of 6z and 5z is 11z and the difference is z .

Example C

Solve \cos 5x + \cos x = \cos 2x .

Solution: Use the formula \cos \alpha + \cos \beta = 2 \cos \frac{\alpha +\beta}{2} \times \cos \frac{\alpha - \beta}{2} .

& \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}


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

Guided Practice

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

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

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


1. Using the product-to-sum formula:

& \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)

2. Using the product-to-sum formula:

& \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)

3. Using the product-to-sum formula:

& \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)

Concept Problem Solution

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

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

Substituting in known values gives:

\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}


Express each product as a sum or difference.

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

Express each sum or difference as a product.

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

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Difficulty Level:

At Grade



Date Created:

Sep 26, 2012

Last Modified:

May 27, 2014
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