# Trig Addition Identities – Explanation and Examples

Trig addition identities provide a formula for the sum of two trigonometric functions.

These sum identities can be manipulated to provide trig difference identities. Sometimes, trig addition identities are also called sum-to-product and product-to-sum identities.

Trig addition identities, like all trig identities, are important in physical sciences. These include astronomy and engineering.

Before reading this topic, brush up on trig identities and the related angle sum formulas.

Trig addition formulas are different from angle sum formulas. Recall that angle sum formulas give a trig ratio for an angle equal to the sum of two other angles. On the other hand, trig addition formulas give the sum of two trig ratios.

These formulas are also called the sum-to-product and product-to-sum identities. As the names imply, the two identities convert a sum of trigonometric ratios into a product of trigonometric ratios and vice versa.

Now, there are trig addition (sum-to-product) identities for sine and cosine. From these, it is possible to derive trig subtraction (or difference-to-product) identities. Then, both trig addition and trig subtraction identities for the other functions follow.

Importantly, the sum of two sine ratios is not the sine of the sums. That is:

$sinx+siny \neq sin(x+y)$.

Similarly,

$cosx+cosy \neq cos(x+y)$.

For two angles $x$ and $y$, the sum of the sines of $x$ and $y$ is:

$sinx+siny = 2sin(\frac{x+y}{2})cos(\frac{x+y}{2})$.

This is the sum-to-product identity. It also works in reverse as the product-to-sum identity.

$2sinxcosy = sin(x+y) + sin(x-y)$.

This is sometimes written as $sinxcosy = \frac{1}{2}(sin(x+y)+sin(x-y))$.

For two angles $x$ and $y$, the sum of the cosines of $x$ and $y$ is:

$cosx+cosy = 2cos(\frac{x+y}{2})cos(\frac{x-y}{2})$.

This is the sum-to-product identity. It also works in reverse as the product-to-sum identity.

$2cosxcosy = cos(x+y) + cos(x-y)$.

Like the corresponding sine formula, this is sometimes written $cosxcosy = \frac{1}{2}(cos(x+y)+cos(x-y))$.

### Other Product-to-Sum Formulas

Take note that there are several other product-to-sum formulas:

• $2sinxsiny = cos(x-y)-cos(x+y)$
• $2cosxsiny = sin(x+y)-sin(x-y)$

Unlike other trig identities, there are not any other convenient formulas for other trig functions. That is, there is no formula for the sum of two tangents, two secants, or two cosecants. Even writing these in terms of sine and cosine is difficult.

For example, $secx+secy = \frac{1}{cosx}+\frac{1}{siny} \neq \frac{1}{cosx+cosy}$.

## Examples

This section goes over common problems involving trig addition identities and their step-by-step solutions.

### Example 1

Use the product-to-sum identity to prove that $cos(2x) = 2cos^2x-1$

### Solution

Begin with the product-to-sum identity for cosine. This is:

$2cosxcosy = cos(x+y) + cos(x-y)$.

In this case, let $x=y$. Then, rewrite the formula:

$2cosxcosx = cos(x+x) + cos(x-x)$.

This then simplifies to:

$2cos^2x = cos(2x) + cos(0) = cos(2x) +1$.

Then subtract $1$ from both sides to get:

$2cos^2x-1 = cos(2x)$.

### Example 2

Use the sum-to-product identity for sine to find a difference-to-product identity.

### Solution

Recall that the sum-to-product addition identity for sine is:

$sinx+siny = 2sin(\frac{x+y}{2})cos(\frac{x-y}{2})$.

Now, finding a difference-to-product identity requires finding a product formula for:

$sinx-siny$.

Since sine is odd, however, this is the same as:

$sinx+sin(-y)$.

Now, use the original sum-to-product identity with $-y$:

$sinx+sin(-y) = 2sin( \frac{x-y}{2} )cos( \frac{x+y}{2} )$.

This is, in fact, the difference-to-product identity for sine.

### Example 3

Use the sum-to-product identity to find $cos(\frac{\pi}{12}) + cos( \frac{5\pi}{12} )$.

### Solution

Recall that the sum-to-product identity for cosine is:

$cosx + cosy = 2cos( \frac{x+y}{2} )cos( \frac{x-y}{2} )$.

In this case, let $x=\frac{5\pi}{12}$. Then, let $y =\frac{\pi}{12}$. Now, plugging these values into the formula yields:

$cos(\frac{5\pi}{12}) + cos( \frac{\pi}{12} ) = 2cos( \frac{\frac{5\pi}{12} + \frac{\pi}{12}}{2} ) cos(\frac{\frac{5\pi}{12} – \frac{\pi}{12}}{2} )$.

Then, this is:

$2cos( \frac{\frac{6\pi }{12}}{2} ) cos(\frac{ \frac{4\pi }{12}}{2}) = 2cos( \frac{\frac{\pi}{2} }{2}) cos(\frac{ \frac{\pi}{3} }{2})$.

Then, this yields:

$2cos( \frac{\pi}{4} )cos( \frac{\pi}{6} )$.

But $\frac{\pi}{4}$ and $\frac{\pi}{6}$ are major angles. Recall that at these angles, the cosine ratios are $\frac{\sqrt2}}{2}$ and $\frac{\sqrt{3}}{2}$, respectively.

Therefore:

$cos(\frac{5\pi}{12}) + cos(\frac{\pi}{12}) = 2(\frac{ \sqrt{2}}{2}) (\frac{\sqrt{3} }{2})$.

Finally, this becomes:

$\frac{ \sqrt{2}\sqrt{3} }{2} = \frac{ \sqrt{6}}{2}$.