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3.13: Sum to Product Formulas for Sine and Cosine

Difficulty Level: At Grade Created by: CK-12
Practice Sum to Product Formulas for Sine and Cosine
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Can you solve problems that involve the sum of sines or cosines? For example, consider the equation:

\cos 10t + \cos 3t

You could just compute each expression separately and add their values at the end. However, there is an easier way to do this. You can simplify the equation first, and then solve.

Read this Concept, and at the end of it, you'll be able to simplify this equation and transform it into a product of trig functions instead of a sum!

Watch This

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

James Sousa: Sum to Product and Product to Sum Identities


In some problems, the product of two trigonometric functions is more conveniently found by the sum of two trigonometric functions by use of identities such as this one:

\sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \times \cos \frac{\alpha - \beta}{2}

This can be verified by using the sum and difference formulas:

2 \sin \frac{\alpha + \beta}{2} \cos \frac{\alpha - \beta}{2} &= 2 \begin{bmatrix} \sin \left( \frac{\alpha}{2} + \frac{\beta}{2} \right) \cos \left(\frac{\alpha}{2} - \frac{\beta}{2} \right) \end{bmatrix} \\ &= 2 \begin{bmatrix} \left( \sin \frac{\alpha}{2} \cos \frac{\beta}{2} + \cos \frac{\alpha}{2} \sin \frac{\beta}{2} \left) \right( \cos \frac{\alpha}{2} \cos \frac{\beta}{2} + \sin \frac{\alpha}{2} \sin \frac{\beta}{2} \right ) \end{bmatrix}\\ &= 2 \begin{bmatrix} \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \cos ^2 \frac{\beta}{2} + \sin ^2 \frac{\alpha}{2} \sin \frac{\beta}{2} \cos \frac{\beta}{2} + \sin \frac{\beta}{2} \cos ^2 \frac{\alpha}{2} \cos \frac{\beta}{2} + \sin \frac{\alpha}{2} \sin ^2 \frac{\beta}{2} \cos \frac{\alpha}{2} \end{bmatrix}\\ &= 2 \begin{bmatrix} \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} \left( \sin^2 \frac{\beta}{2} + \cos^2 \frac{\beta}{2} \right) + \sin \frac{\beta}{2} \cos \frac{\beta}{2} \left( \sin^2 \frac{\alpha}{2} + \cos^2 \frac{\alpha}{2} \right) \end{bmatrix}\\ &= 2 \begin{bmatrix} \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} + \sin \frac{\beta}{2} \cos \frac{\beta}{2} \end{bmatrix}\\ &= 2 \sin \frac{\alpha}{2} \cos \frac{\alpha}{2} + 2 \sin \frac{\beta}{2} \cos \frac{\beta}{2}\\ &= \sin \left( 2 \cdot \frac{\alpha}{2} \right) + \sin \left( 2 \cdot \frac{\beta}{2} \right)\\ &= \sin \alpha + \sin \beta

The following variations can be derived similarly:

 \sin \alpha - \sin \beta &= 2 \sin \frac{\alpha - \beta}{2} \times \cos \frac{\alpha + \beta}{2}\\ \cos \alpha + \cos \beta &= 2 \cos \frac{\alpha + \beta}{2} \times \cos \frac{\alpha - \beta}{2}\\ \cos \alpha - \cos \beta &= -2 \sin \frac{\alpha + \beta}{2} \times \sin \frac{\alpha - \beta}{2}\\

Here are some examples of this type of transformation from a sum of terms to a product of terms.

Example A

Change \sin 5x - \sin 9x into a product.

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

 \sin 5x - \sin 9x &= 2 \sin \frac{5x - 9x}{2} \cos \frac{5x + 9x}{2}\\ &= 2 \sin (-2x) \cos 7x\\ &= -2 \sin 2x  \cos 7x

Example B

Change \cos(-3x) + \cos 8x into a product.

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

\cos (-3x) + \cos (8x) &= 2 \cos \frac{-3x + 8x}{2} \cos \frac {-3x - 8x}{2}\\ &= 2 \cos (2.5x) \cos (-5.5x)\\ &= 2 \cos (2.5x) \cos (5.5x)

Example C

Change 2 \sin 7x \cos 4x to a sum.

Solution: This is the reverse of what was done in the previous two examples. Looking at the four formulas above, take the one that has sine and cosine as a product, \sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \times \cos \frac{\alpha - \beta}{2}. Therefore, 7x = \frac{\alpha + \beta}{2} and 4x = \frac{\alpha - \beta}{2} .

7x & = \frac{\alpha + \beta}{2} &&&& 4x = \frac{\alpha - \beta}{2} \\&&& \text{and} \\14x & = \alpha+\beta &&&& 8x= \alpha - \beta \\&\alpha = 14x - \beta &&&& 8x=[14x-\beta]-\beta \\&&& \text{so} \\&&&&&-6x = -2\beta\\&&&&&3x=\beta\\\alpha=14x-3x\\\alpha=11x

So, this translates to \sin(11x) + \sin(3x) . A shortcut for this problem, would be to notice that the sum of 7x and 4x is 11x and the difference is 3x .

Guided Practice

1. Express the sum as a product: \sin 9x + \sin 5x

2. Express the difference as a product: \cos 4y - \cos 3y

3. Verify the identity (using sum-to-product formula): \frac{\cos 3a - \cos 5a}{\sin 3a - \sin 5a} = - \tan 4a


1. Using the sum-to-product formula:

& \sin 9x + \sin 5x\\& 2 \left(\sin \left(\frac{9x + 5x}{2} \right) \cos \left(\frac{9x - 5x}{2} \right) \right)\\& 2 \sin 7x \cos 2x


Using the difference-to-product formula:

& \cos 4y - \cos 3y\\& -2 \sin \left(\frac{4y + 3y}{2} \right) \sin \left(\frac{4y - 3y}{2} \right)\\& -2 \sin \frac{7y}{2} \sin \frac{y}{2}


Using the difference-to-product formulas:

& \qquad \qquad \frac{\cos 3a - \cos 5a}{\sin 3a - \sin 5a} = - \tan 4a\\& \frac{-2 \sin \left (\frac{3a + 5a}{2} \right ) \sin \left (\frac {3a - 5a}{2} \right )}{2 \sin \left (\frac{3a - 5a}{2} \right ) \cos \left(\frac{3a + 5a}{2} \right)}\\&  \qquad \qquad \qquad \ \ - \frac{\sin 4a}{\cos 4a}\\& \qquad \qquad \qquad  \  \ - \tan 4a

Concept Problem Solution

Prior to learning the sum to product formulas for sine and cosine, evaluating a sum of trig functions, such as

\cos 10t + \cos 3t

might have been considered difficult. But you can easily transform this equation into a product of two trig functions using:

\cos \alpha + \cos \beta = 2 \cos \frac{\alpha + \beta}{2} \times \cos \frac{\alpha - \beta}{2}

Substituting the known quantities:

\cos 10t + \cos 3t = 2 \cos \frac{13t}{2} \times \cos \frac{7t}{2} = 2\cos(6.5t) \cos(3.5t)

Explore More

Change each sum or difference into a product.

  1. \sin 3x + \sin 2x
  2. \cos 2x + \cos 5x
  3. \sin (-x) - \sin 4x
  4. \cos 12x + \cos 3x
  5. \sin 8x - \sin 4x
  6. \sin x + \sin \frac{1}{2}x
  7. \cos 3x - \cos (-3x)

Change each product into a sum or difference.

  1. -2\sin 3.5x \sin 2.5x
  2. 2\cos 3.5x \sin 0.5x
  3. 2\cos 3.5x \cos 5.5x
  4. 2\sin 6x \cos 2x
  5. -2\sin 3x \sin x
  6. 2\sin 4x \cos x
  7. Show that \cos\frac{A+B}{2}\cos\frac{A-B}{2}=\frac{1}{2}(\cos A + \cos B) .
  8. Let u=\frac{A+B}{2} and v=\frac{A-B}{2} . Show that \cos u\cos v =\frac{1}{2}(\cos (u+v)+\cos(u-v)).

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Difficulty Level:

At Grade



Date Created:

Sep 26, 2012

Last Modified:

Feb 26, 2015
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