# Sum-to-Product Identities

Sometimes we may need to simplify a trigonometric expression like $$\sin \alpha \pm \sin \beta$$ or $$\cos \alpha \pm \cos \beta$$ by converting the sum or difference of trigonometric functions into a product. These formulas are known as sum-to-product identities.

### Sum and Difference of Sines

To derive the sum-to-product identities for sine we use the sine addition and subtraction formulas:

${\sin \left( {x + y} \right) }={ \sin x \cos y + \cos x \sin y ,}$

${\sin \left( {x – y} \right) }={ \sin x \cos y – \cos x \sin y .}$

If we add and subtract the two equations, we get

${\sin \left( {x + y} \right) + \sin \left( {x – y} \right) }={ 2\sin x \cos y,}$

${\sin \left( {x + y} \right) – \sin \left( {x – y} \right) }={ 2\cos x \sin y.}$

Let now $$x = y = \alpha$$ and $$x – y = \beta.$$ Solving this system for $$x$$ and $$y,$$ we obtain

${x = \frac{{\alpha + \beta }}{2},\;\;}\kern0pt{y = \frac{{\alpha – \beta }}{2}.}$

Plugging this into the equations above yields:

### Sum and Difference of Cosines

Similarly, using the cosine addition and subtraction formulas

$\cos \left( {x + y} \right) = \cos x\cos y – \sin x\sin y,$

$\cos \left( {x – y} \right) = \cos x\cos y + \sin x\sin y,$

we can derive the sum-to-product identities for cosine:

### Sum and Difference of Tangents

We have

${\tan \alpha + \tan \beta }={ \frac{{\sin \alpha }}{{\cos \alpha }} + \frac{{\sin \beta }}{{\cos \beta }} }={ \frac{{\sin \alpha \cos \beta + \cos \alpha \sin \beta }}{{\cos \alpha \cos \beta}}.}$

The numerator in the last expression is equal to $$\sin \left( {\alpha + \beta } \right).$$ Hence,

This formula is valid if $$\cos\alpha \ne 0$$ and $$\cos\beta \ne 0,$$ or $$\alpha ,\beta \ne \large{\frac{\pi }{2}}\normalsize + \pi n,$$ $$n \in \mathbb{Z}.$$

Consider the difference of tangents:

${\tan \alpha – \tan \beta }={ \frac{{\sin \alpha }}{{\cos \alpha }} – \frac{{\sin \beta }}{{\cos \beta }} }={ \frac{{\sin \alpha \cos \beta – \cos \alpha \sin \beta }}{{\cos \alpha \cos \beta }} }={ \frac{{\sin \left( {\alpha – \beta } \right)}}{{\cos \alpha \cos \beta }},}$

that is,

### Sum and Difference of Cotangents

The sum-to-product formulas for cotangent can de derived in the same way. They are given by

where $$\sin\alpha \ne 0$$ and $$\sin\beta \ne 0,$$ or $$\alpha ,\beta \ne \pi n,$$ $$n \in \mathbb{Z}.$$

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Verify the identity $\cos {47^\circ} + \cos {73^\circ} = \cos {13^\circ}.$

### Example 2

Verify the identity $\sin {87^\circ} – \sin {27^\circ} = \cos {57^\circ}.$

### Example 3

Transform the sum into a product: $1 + \cos 2\alpha .$

### Example 4

Transform the sum into a product: $\sin \beta + \cos 2\beta – \sin 3\beta .$

### Example 5

Simplify the expression $\frac{{\cos \alpha – \cos \beta }}{{\sin \alpha + \sin \beta }}.$

### Example 6

Simplify the expression $\frac{{\sin 3\alpha – \sin 7\alpha }}{{\cos 4\alpha + \cos 6\alpha }}.$

### Example 7

Prove the identity $\cos \frac{\pi }{5} + \cos \frac{{3\pi }}{5} = \frac{1}{2}.$

### Example 8

Prove the identity $\frac{{\cos \alpha + \sin \alpha }}{{\cos \alpha – \sin \alpha }} = \tan \left( {\frac{\pi }{4} + \alpha } \right).$

### Example 9

Calculate the value of the expression $\cos 2\alpha – \cos 6\alpha$ if $$\cos \alpha = \large{\frac{1}{{\sqrt 3 }}}\normalsize.$$

### Example 10

Calculate the value of the expression $\sin 5\alpha – \sin 3\alpha$ if $$\sin \alpha = \large{\frac{2}{{\sqrt 5 }}}\normalsize.$$

### Example 1.

Verify the identity $\cos {47^\circ} + \cos {73^\circ} = \cos {13^\circ}.$

Solution.

Using the sum formula for cosine, we have

${LHS }={ \cos {47^\circ} + \cos {73^\circ} }={ 2\cos \frac{{{{47}^\circ} + {{73}^\circ}}}{2}\cos \frac{{{{47}^\circ} – {{73}^\circ}}}{2} }={ 2\cos \frac{{{{120}^\circ}}}{2}\cos \frac{{ – {{26}^\circ}}}{2} }={ 2\cos {60^\circ}\cos \left( { – {{13}^\circ}} \right) }={ 2 \cdot \frac{1}{2} \cdot \cos \left( { – {{13}^\circ}} \right) }={ \cos \left( { – {{13}^\circ}} \right).}$

The cosine function is even. Therefore,

${LHS = \cos \left( { – {{13}^\circ}} \right) }={ \cos {13^\circ} }={ RHS.}$

### Example 2.

Verify the identity $\sin {87^\circ} – \sin {27^\circ} = \cos {57^\circ}.$

Solution.

Using the difference identity for sine, we can write

${LHS = \sin {87^\circ} – \sin {27^\circ} }={ 2\cos \frac{{{{87}^\circ} + {{27}^\circ}}}{2}\sin \frac{{{{87}^\circ} – {{27}^\circ}}}{2} }={ 2\cos \frac{{{{114}^\circ}}}{2}\sin \frac{{{{60}^\circ}}}{2} }={ 2\cos {57^\circ}\sin {30^\circ} }={ 2 \cdot \cos {57^\circ} \cdot \frac{1}{2} }={ \cos {57^\circ} }={ RHS.}$

### Example 3.

Transform the sum into a product: $1 + \cos 2\alpha .$

Solution.

Since $$1=\cos 0,$$ we can write:

${1 + \cos 2\alpha }={ \cos 0 + \cos 2\alpha }={ 2\cos \frac{{0 + 2\alpha }}{2}\cos \frac{{0 – 2\alpha }}{2} }={ 2\cos \alpha \cos \left( { – \alpha } \right) }={ 2\,{\cos ^2}\alpha .}$

### Example 4.

Transform the sum into a product: $\sin \beta + \cos 2\beta – \sin 3\beta .$

Solution.

Using the difference formula for sine, we get

${\sin \beta + \cos 2\beta – \sin 3\beta }={ \left( {\sin \beta – \sin 3\beta } \right) + \cos 2\beta }={ 2\cos \frac{{\beta + 3\beta }}{2}\sin \frac{{\beta – 3\beta }}{2} + \cos 2\beta }={ 2\cos 2\beta \sin \left( { – \beta } \right) + \cos 2\beta }={ \cos 2\beta \left( {1 – 2\sin \beta } \right).}$

### Example 5.

Simplify the expression $\frac{{\cos \alpha – \cos \beta }}{{\sin \alpha + \sin \beta }}.$

Solution.

Using the sum-to-product identities, we obtain

$\require{cancel}{\frac{{\cos \alpha – \cos \beta }}{{\sin \alpha + \sin \beta }} }={ \frac{{ – \cancel{2}\cancel{\sin \frac{{\alpha + \beta }}{2}}\sin \frac{{\alpha – \beta }}{2}}}{{\cancel{2}\cancel{\sin \frac{{\alpha + \beta }}{2}}\cos \frac{{\alpha – \beta }}{2}}} }={ – \frac{{\sin \frac{{\alpha – \beta }}{2}}}{{\cos \frac{{\alpha – \beta }}{2}}} }={ – \tan \frac{{\alpha – \beta }}{2}.}$

### Example 6.

Simplify the expression $\frac{{\sin 3\alpha – \sin 7\alpha }}{{\cos 4\alpha + \cos 6\alpha }}.$

Solution.

We denote this expression by $$E$$ and apply the sum-to-product formulas to rewrite the difference of sines and sum of cosines:

${E }={ \frac{{\sin 3\alpha – \sin 7\alpha }}{{\cos 4\alpha + \cos 6\alpha }} }={ \frac{{\cancel{2}\cos \frac{{3\alpha + 7\alpha }}{2}\sin \frac{{3\alpha – 7\alpha }}{2}}}{{\cancel{2}\cos \frac{{4\alpha + 6\alpha }}{2}\cos \frac{{4\alpha – 6\alpha }}{2}}} }={ \frac{{\cancel{\cos 5\alpha} \sin \left( { – 2\alpha } \right)}}{{\cancel{\cos 5\alpha} \cos \left( { – \alpha } \right)}} }={ \frac{{\sin \left( { – 2\alpha } \right)}}{{\cos \left( { – \alpha } \right)}}.}$

The sine function is odd, and the cosine function is even. Hence,

${E = \frac{{\sin \left( { – 2\alpha } \right)}}{{\cos \left( { – \alpha } \right)}} }={ – \frac{{\sin 2\alpha }}{{\cos \alpha }}.}$

Using the double-angle identity for sine, we have

${E = – \frac{{\sin 2\alpha }}{{\cos \alpha }} }={ – \frac{{2\sin \alpha \cancel{\cos \alpha} }}{{\cancel{\cos \alpha} }} }={ – 2\sin \alpha .}$

### Example 7.

Prove the identity $\cos \frac{\pi }{5} + \cos \frac{{3\pi }}{5} = \frac{1}{2}.$

Solution.

First we transform the sum of cosines into a product:

${LHS = \cos \frac{\pi }{5} + \cos \frac{{3\pi }}{5} }={ 2\cos \frac{{\frac{\pi }{5} + \frac{{3\pi }}{5}}}{2}\cos \frac{{\frac{\pi }{5} – \frac{{3\pi }}{5}}}{2} }={ 2\cos \frac{{4\pi }}{{10}}\cos \left( { – \frac{{2\pi }}{{10}}} \right) }={ 2\cos \frac{{2\pi }}{5}\cos \frac{\pi }{5}.}$

Now we multiply and divide the left-hand side by $$\sin \large{\frac{\pi }{5}}\normalsize$$ and then apply the double-angle identity for sine twice:

${LHS }={ \frac{{2\cos \frac{{2\pi }}{5}\cos \frac{\pi }{5}\sin \frac{\pi }{5}}}{{\sin \frac{\pi }{5}}} }={ \frac{{\sin \frac{{2\pi }}{5}\cos \frac{{2\pi }}{5}}}{{\sin \frac{\pi }{5}}} }={ \frac{{2\sin \frac{{2\pi }}{5}\cos \frac{{2\pi }}{5}}}{{2\sin \frac{\pi }{5}}} }={ \frac{{\sin \frac{{4\pi }}{5}}}{{2\sin \frac{\pi }{5}}}.}$

Recall the reduction formula:

$\sin \left( {\pi – \alpha } \right) = \sin \alpha .$

Hence,

${LHS = \frac{{\sin \frac{{4\pi }}{5}}}{{2\sin \frac{\pi }{5}}} }={ \frac{{\sin \left( {\pi – \frac{\pi }{5}} \right)}}{{2\sin \frac{\pi }{5}}} }={ \frac{\cancel{\sin \frac{\pi }{5}}}{{2\cancel{\sin \frac{\pi }{5}}}} }={ \frac{1}{2} }={ RHS.}$

### Example 8.

Prove the identity $\frac{{\cos \alpha + \sin \alpha }}{{\cos \alpha – \sin \alpha }} = \tan \left( {\frac{\pi }{4} + \alpha } \right).$

Solution.

Using the cofunction identity and sum-to-product formulas, we represent the left-hand side $$\left( {LHS} \right)$$ in the form

${LHS }={ \frac{{\sin \left( {\frac{\pi }{2} – \alpha } \right) + \sin \alpha }}{{\sin \left( {\frac{\pi }{2} – \alpha } \right) – \sin \alpha }} }={ \frac{{\cancel{2}\sin \frac{{\frac{\pi }{2} – \cancel{\alpha} + \cancel{\alpha} }}{2}\cos \frac{{\frac{\pi }{2} – \alpha – \alpha }}{2}}}{{\cancel{2}\cos \frac{{\frac{\pi }{2} – \cancel{\alpha} + \cancel{\alpha} }}{2}\sin \frac{{\frac{\pi }{2} – \alpha – \alpha }}{2}}} }={ \frac{{\sin \frac{\pi }{4}\cos \left( {\frac{\pi }{4} – \alpha } \right)}}{{\cos \frac{\pi }{4}\sin \left( {\frac{\pi }{4} – \alpha } \right)}} }={ \frac{{\cancel{\frac{{\sqrt 2 }}{2}}\cos \left( {\frac{\pi }{4} – \alpha } \right)}}{{\cancel{\frac{{\sqrt 2 }}{2}}\sin \left( {\frac{\pi }{4} – \alpha } \right)}} }={ \cot \left( {\frac{\pi }{4} – \alpha } \right) }={ \tan \left( {\frac{\pi }{2} – \left( {\frac{\pi }{4} – \alpha } \right)} \right) }={ \tan \left( {\frac{\pi }{2} – \frac{\pi }{4} + \alpha } \right) }={ \tan \left( {\frac{\pi }{4} + \alpha } \right) }={ RHS.}$

### Example 9.

Calculate the value of the expression $\cos 2\alpha – \cos 6\alpha$ if $$\cos \alpha = \large{\frac{1}{{\sqrt 3 }}}\normalsize.$$

Solution.

We denote this expression by $$E.$$ By the cosine difference formula,

${E = \cos 2\alpha – \cos 6\alpha }={ – 2\sin \frac{{2\alpha + 6\alpha }}{2}\sin \frac{{2\alpha – 6\alpha }}{2} }={ – 2\sin 4\alpha \sin \left( { – 2\alpha } \right) }={ 2\sin 4\alpha \sin 2\alpha .}$

${E = 2\sin 4\alpha \sin 2\alpha }={ 2 \cdot 2\sin 2\alpha \cos 2\alpha \cdot \sin 2\alpha }={ 4\,{\sin ^2}2\alpha \cos 2\alpha }={ 4 \cdot {\left( {2\sin \alpha \cos \alpha } \right)^2} \cdot \left( {2{{\cos }^2}\alpha – 1} \right) }={ 4 \cdot 4\,{\sin ^2}\alpha \,{\cos ^2}\alpha \cdot \left( {2{{\cos }^2}\alpha – 1} \right) }={ 16\left( {{{\cos }^2}\alpha – {{\cos }^4}\alpha } \right)\left( {2{{\cos }^2}\alpha – 1} \right) }={ 16\left( {2{{\cos }^4}\alpha – 2{{\cos }^6}\alpha – {{\cos }^2}\alpha + {{\cos }^4}\alpha } \right) }={ 16\left( {3{{\cos }^4}\alpha – 2{{\cos }^6}\alpha – {{\cos }^2}\alpha } \right) }={ 16\left[ {3 \cdot \frac{1}{9} – 2 \cdot \frac{1}{{27}} – \frac{1}{3}} \right] }={ – \frac{{32}}{{27}}.}$

### Example 10.

Calculate the value of the expression $\sin 5\alpha – \sin 3\alpha$ if $$\sin \alpha = \large{\frac{2}{{\sqrt 5 }}}\normalsize.$$

Solution.

Let the expression be denoted by $$E.$$ By the sine difference formula,

${E = \sin 5\alpha – \sin 3\alpha }={ 2\cos \frac{{5\alpha + 3\alpha }}{2}\sin \frac{{5\alpha – 3\alpha }}{2} }={ 2\cos 4\alpha \sin \alpha .}$

Express $$\cos 4\alpha$$ in terms of the single angle $$\alpha$$ and find its value:

${\cos 4\alpha }={ 1 – 2\,{\sin ^2}2\alpha }={ 1 – 2 \cdot {\left( {2\sin \alpha \cos \alpha } \right)^2} }={ 1 – 2 \cdot 4\,{\sin ^2}\alpha \,{\cos ^2}\alpha }={ 1 – 8\,{\sin ^2}\alpha \left( {1 – {{\sin }^2}\alpha } \right) }={ 1 – 8\,{\sin ^2}\alpha + 8\,{\sin ^4}\alpha }={ 1 – 8 \cdot {\left( {\frac{2}{{\sqrt 5 }}} \right)^2} + 8 \cdot {\left( {\frac{2}{{\sqrt 5 }}} \right)^4} }={ 1 – \frac{{32}}{5} + \frac{{128}}{{25}} }={ \frac{{25 – 160 + 128}}{{25}} }={ – \frac{7}{{25}}.}$

Then

${E = 2\cos 4\alpha \sin \alpha }={ 2 \cdot \left( { – \frac{7}{{25}}} \right) \cdot \frac{2}{{\sqrt 5 }} }={ – \frac{{28}}{{25\sqrt 5 }}.}$