# 3.5.1: Sum to Product Formulas for Sine and Cosine

Relation of the sum or difference of two trigonometric functions to a product.

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.

### 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:

$$\sin \alpha +\sin \beta =2\sin \dfrac{\alpha+\beta}{2}\times \cos \dfrac{\alpha −\beta }{2}$$

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

$$\begin{array}{l} 2 \sin \dfrac{\alpha+\beta}{2} \cos \dfrac{\alpha-\beta}{2} \\ =2\left[\sin \left(\dfrac{\alpha}{2}+\dfrac{\beta}{2}\right) \cos \left(\dfrac{\alpha}{2}-\dfrac{\beta}{2}\right)\right] \\ =2\left[\left(\sin \dfrac{\alpha}{2} \cos \dfrac{\beta}{2}+\cos \dfrac{\alpha}{2} \sin \dfrac{\beta}{2}\right)\left(\cos \dfrac{\alpha}{2} \cos \dfrac{\beta}{2}+\sin \dfrac{\alpha}{2} \sin \dfrac{\beta}{2}\right)\right] \\ =2\left[\sin \dfrac{\alpha}{2} \cos \dfrac{\alpha}{2} \cos ^{2} \dfrac{\beta}{2}+\sin ^{2} \dfrac{\alpha}{2} \sin \dfrac{\beta}{2} \cos \dfrac{\beta}{2}+\sin \dfrac{\beta}{2} \cos ^{2} \dfrac{\alpha}{2} \cos \dfrac{\beta}{2}+\sin \dfrac{\alpha}{2} \sin ^{2} \dfrac{\beta}{2} \cos \dfrac{\alpha}{2}\right] \\ =2\left[\sin \dfrac{\alpha}{2} \cos \dfrac{\alpha}{2}\left(\sin ^{2} \dfrac{\beta}{2}+\cos ^{2} \dfrac{\beta}{2}\right)+\sin \dfrac{\beta}{2} \cos \dfrac{\beta}{2}\left(\sin ^{2} \dfrac{\alpha}{2}+\cos ^{2} \dfrac{\alpha}{2}\right)\right] \\ =2\left[\sin \dfrac{\alpha}{2} \cos \dfrac{\alpha}{2}+\sin \dfrac{\beta}{2} \cos \dfrac{\beta}{2}\right] \\ =2 \sin \dfrac{\alpha}{2} \cos \dfrac{\alpha}{2}+2 \sin \dfrac{\beta}{2} \cos \dfrac{\beta}{2} \\ =\sin \left(2 \cdot \dfrac{\alpha}{2}\right)+\sin \left(2 \cdot \dfrac{\beta}{2}\right) \\ =\sin \alpha+\sin \beta \end{array}$$

The following variations can be derived similarly:

\begin{aligned} \sin \alpha −\sin \beta &=2\sin \dfrac{\alpha −\beta }{2}\times \cos \dfrac{\alpha +\beta }{2}\\ \cos \alpha +\cos \beta &=2\cos \dfrac{\alpha +\beta }{2}\times \cos \dfrac{\alpha −\beta }{2} \\ \cos \alpha −\cos \beta &=−2\sin \dfrac{\alpha +\beta }{2}\times \sin \dfrac{\alpha −\beta }{2}\end{aligned}

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

1. Change $$\sin 5x−\sin 9x$$ into a product.

Use the formula $$\sin \alpha −\sin \beta =2\sin \dfrac{\alpha −\beta }{2}\times \cos \dfrac{\alpha +\beta }{2}$$.

\begin{aligned} \sin 5x−\sin 9x&=2\sin \dfrac{5x−9x}{2} \cos \dfrac{5x+9x}{2} \\&=2\sin (−2x)\cos 7x \\ &=−2\sin 2x\cos 7x \end{aligned}

2. Change $$\cos (−3x)+\cos 8x$$ into a product.

Use the formula $$\cos \alpha +\cos \beta =2\cos \dfrac{\alpha +\beta }{2} \times \cos \dfrac{\alpha −\beta }{2}$$

\begin{aligned} \cos (−3x)+\cos (8x)&=2\cos \dfrac{−3x+8x}{2} \cos \dfrac{−3x−8x}{2} \\&=2\cos (2.5x)\cos (−5.5x)\\&=2\cos (2.5x)\cos (5.5x) \end{aligned}

3. Change $$2\sin 7x\cos 4x$$ 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, $$\sin \alpha +\sin \beta =2\sin \dfrac{\alpha +\beta }{2} \times \cos \dfrac{\alpha −\beta }{2}$$. Therefore, $$7x=\dfrac{\alpha +\beta }{2}$$ and $$4x=\dfrac{\alpha −\beta }{2}$$.

\begin{aligned} 7x&=\dfrac{\alpha +\beta }{2} & 4x&=\dfrac{\alpha −\beta }{2} \\ & \qquad \qquad \qquad \text{and}& & \\ 14x&=\alpha +\beta & 8x&=\alpha −\beta \\ \alpha &=14x−\beta & 8x&=[14x−\beta ]−\beta \\ & \qquad \qquad \qquad \text{so}& &\\ \alpha &=14x−3x & −6x&=−2\beta \\ \alpha &=11x & 3x&=\beta \end{aligned}

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 $$3x$$.

Example $$\PageIndex{1}$$

Earlier, you were asked to solve

$$\cos 10t+\cos 3t$$

Solution

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

$$\cos \alpha +\cos \beta =2\cos \dfrac{\alpha +\beta }{2} \times \cos \dfrac{\alpha −\beta }{2}$$

Substituting the known quantities:

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

Example $$\PageIndex{2}$$

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

Solution

Using the sum-to-product formula:

\begin{aligned} &\sin 9x+\sin 5x \\ &2\left(\sin \left(\dfrac{9x+5x}{2}\right)\cos \left(\dfrac{9x−5x}{2}\right)\right) \\ & 2\sin 7x\cos 2x \end{aligned}

Example $$\PageIndex{3}$$

Express the difference as a product: $$\cos 4y−\cos 3y$$

Solution

Using the difference-to-product formula:

\begin{aligned} &\cos 4y−\cos 3y \\ &−2\sin \left(\dfrac{4y+3y}{2}\right) \sin \left(\dfrac{4y−3y}{2}\right) \\ &−2\sin \dfrac{7y}{2} \sin \dfrac{y}{2} \end{aligned}

Example $$\PageIndex{4}$$

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

Solution

Using the difference-to-product formulas:

\begin{aligned} \dfrac{\cos 3a−\cos 5a}{\sin 3a−\sin 5a}&=−\tan 4a \\ \dfrac{−2\sin \left(\dfrac{3a+5a}{2}\right)\sin \left(\dfrac{3a−5a}{2}\right) }{2\sin \left(\dfrac{3a−5a}{2}\right) \cos \left(\dfrac{3a+5a}{2}\right)}& \\ −\dfrac{\sin 4a}{\cos 4a }&\\ −\tan 4a & \end{aligned}

### Review

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. $$\sin x+\sin \dfrac{1}{2} x$$
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 $$\cos \dfrac{A+B}{2}\cos \dfrac{A−B}{2}=\dfrac{1}{2}(\cos A+\cos B)$$.
8. Let $$u=\dfrac{A+B}{2}$$ and $$v=\dfrac{A−B}{2}$$. Show that $$\cos u\cos v=\dfrac{1}{2}(\cos (u+v)+\cos (u−v))$$.