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3.13: Sum to Product Formulas for Sine and Cosine

Difficulty Level: At Grade Created by: CK-12
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Can you solve problems that involve the sum of sines or cosines? For example, consider the equation:

\begin{align*}\cos 10t + \cos 3t\end{align*}cos10t+cos3t

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.

Sine and Cosine Sum to Product Formulas

In some problems, the product of two trigonometric functions is more conveniently found by the sum of two trigonometric functions by use of identities.


Here is an example: 

\begin{align*}\sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \times \cos \frac{\alpha - \beta}{2}\end{align*}sinα+sinβ=2sinα+β2×cosαβ2

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

\begin{align*}& 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\end{align*}2sinα+β2cosαβ2=2[sin(α2+β2)cos(α2β2)]=2[(sinα2cosβ2+cosα2sinβ2)(cosα2cosβ2+sinα2sinβ2)]=2[sinα2cosα2cos2β2+sin2α2sinβ2cosβ2+sinβ2cos2α2cosβ2+sinα2sin2β2cosα2]=2[sinα2cosα2(sin2β2+cos2β2)+sinβ2cosβ2(sin2α2+cos2α2)]=2[sinα2cosα2+sinβ2cosβ2]=2sinα2cosα2+2sinβ2cosβ2=sin(2α2)+sin(2β2)=sinα+sinβ

The following variations can be derived similarly:

\begin{align*} \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}\\\end{align*}sinαsinβcosα+cosβcosαcosβ=2sinαβ2×cosα+β2=2cosα+β2×cosαβ2=2sinα+β2×sinαβ2

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

1. Change \begin{align*}\sin 5x - \sin 9x\end{align*}sin5xsin9x into a product.

Use the formula \begin{align*}\sin \alpha - \sin \beta = 2 \sin \frac{\alpha - \beta}{2} \times \cos \frac{\alpha + \beta}{2}.\end{align*}sinαsinβ=2sinαβ2×cosα+β2.

\begin{align*} \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\end{align*}sin5xsin9x=2sin5x9x2cos5x+9x2=2sin(2x)cos7x=2sin2xcos7x

2. Change \begin{align*}\cos(-3x) + \cos 8x\end{align*}cos(3x)+cos8x into a product.

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*}\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)\end{align*}cos(3x)+cos(8x)=2cos3x+8x2cos3x8x2=2cos(2.5x)cos(5.5x)=2cos(2.5x)cos(5.5x)

3. Change \begin{align*}2 \sin 7x \cos 4x\end{align*}2sin7xcos4x to a sum.

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, \begin{align*}\sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \times \cos \frac{\alpha - \beta}{2}.\end{align*}sinα+sinβ=2sinα+β2×cosαβ2. Therefore, \begin{align*}7x = \frac{\alpha + \beta}{2}\end{align*}7x=α+β2 and \begin{align*}4x = \frac{\alpha - \beta}{2}\end{align*}4x=αβ2.

\begin{align*}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\end{align*}7x14xα=14x3xα=11x=α+β2=α+βα=14xβandso4x=αβ28x=αβ8x=[14xβ]β6x=2β3x=β

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


Example 1

Earlier, you were asked to solve 

\begin{align*}\cos 10t + \cos 3t\end{align*}cos10t+cos3t

You can easily transform this equation into a product of two trig functions using:

\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

Substituting the known quantities:

\begin{align*}\cos 10t + \cos 3t = 2 \cos \frac{13t}{2} \times \cos \frac{7t}{2} = 2\cos(6.5t) \cos(3.5t)\end{align*}cos10t+cos3t=2cos13t2×cos7t2=2cos(6.5t)cos(3.5t)

Example 2

Express the sum as a product: \begin{align*}\sin 9x + \sin 5x\end{align*}sin9x+sin5x

Using the sum-to-product formula:

\begin{align*}& \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\end{align*}sin9x+sin5x2(sin(9x+5x2)cos(9x5x2))2sin7xcos2x

Example 3

 Express the difference as a product: \begin{align*}\cos 4y - \cos 3y \end{align*}cos4ycos3y

Using the difference-to-product formula:

\begin{align*}& \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}\end{align*}cos4ycos3y2sin(4y+3y2)sin(4y3y2)2sin7y2siny2

Example 4

Verify the identity (using sum-to-product formula): \begin{align*}\frac{\cos 3a - \cos 5a}{\sin 3a - \sin 5a} = - \tan 4a\end{align*}cos3acos5asin3asin5a=tan4a

Using the difference-to-product formulas:

\begin{align*}& \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\end{align*}cos3acos5asin3asin5a=tan4a2sin(3a+5a2)sin(3a5a2)2sin(3a5a2)cos(3a+5a2)  sin4acos4a  tan4a


Change each sum or difference into a product.

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

Change each product into a sum or difference.

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

Review (Answers)

To see the Review answers, open this PDF file and look for section 3.13. 

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Sum to Product Formula

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

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Difficulty Level:
At Grade
Date Created:
Sep 26, 2012
Last Modified:
Mar 25, 2016
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