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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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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:

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

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:

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

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

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\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*}

The following variations can be derived similarly:

sinαsinβcosα+cosβcosαcosβ=2sinαβ2×cosα+β2=2cosα+β2×cosαβ2=2sinα+β2×sinαβ2\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*}

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

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

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

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

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

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*}

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

3. Change 2sin7xcos4x\begin{align*}2 \sin 7x \cos 4x\end{align*} 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, sinα+sinβ=2sinα+β2×cosαβ2.\begin{align*}\sin \alpha + \sin \beta = 2 \sin \frac{\alpha + \beta}{2} \times \cos \frac{\alpha - \beta}{2}.\end{align*} Therefore, 7x=α+β2\begin{align*}7x = \frac{\alpha + \beta}{2}\end{align*} and 4x=αβ2\begin{align*}4x = \frac{\alpha - \beta}{2}\end{align*}.

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

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

### Examples

#### Example 1

Earlier, you were asked to solve

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

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

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*}

Substituting the known quantities:

cos10t+cos3t=2cos13t2×cos7t2=2cos(6.5t)cos(3.5t)\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*}

#### Example 2

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

Using the sum-to-product formula:

sin9x+sin5x2(sin(9x+5x2)cos(9x5x2))2sin7xcos2x\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*}

#### Example 3

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

Using the difference-to-product formula:

cos4ycos3y2sin(4y+3y2)sin(4y3y2)2sin7y2siny2\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*}

#### Example 4

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

Using the difference-to-product formulas:

cos3acos5asin3asin5a=tan4a2sin(3a+5a2)sin(3a5a2)2sin(3a5a2)cos(3a+5a2)  sin4acos4a  tan4a\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*}

### Review

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*}

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

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### Vocabulary Language: English

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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