# Product-to-Sum Identities

The product-to-sum formulas can be derived from the addition and subtraction formulas for sine and cosine.

### Product of Sines

Consider the cosine formulas:

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

$\cos \left( {\alpha – \beta } \right) = \cos \alpha \cos \beta + \sin \alpha \sin \beta .$

Subtract the second expression from the first one:

${\cos \left( {\alpha + \beta } \right) – \cos \left( {\alpha – \beta } \right) }={ – 2\sin \alpha \sin \beta ,}$

that is,

### Product of Cosines

If we add the sum and difference identities above, we get

${\cos \left( {\alpha – \beta } \right) + \cos \left( {\alpha + \beta } \right) }={ 2\cos \alpha \cos \beta .}$

Hence,

### Product of Sine and Cosine

Similarly we can express the product of sine and cosine as a sum of trigonometric functions. Adding the equations

${\sin \left( {\alpha + \beta } \right) }={ \sin \alpha \cos \beta + \cos \alpha \sin \beta ,}$

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

yields

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

So, the product of sine and cosine is given by

### Product of Tangents

To derive the product-to-sum identity for tangents we use the following formulas:

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

${\cot \alpha + \cot \beta }={ \frac{{\sin \left( {\alpha + \beta } \right)}}{{\sin \alpha \sin \beta }}.}$

If we divide the first expression by the second, we obtain

$\require{cancel}{\frac{{\tan \alpha + \tan \beta }}{{\cot \alpha + \cot \beta }} }={ \frac{{\frac{{\sin \left( {\alpha + \beta } \right)}}{{\cos \alpha \cos \beta }}}}{{\frac{{\sin \left( {\alpha + \beta } \right)}}{{\sin \alpha \sin \beta }}}} }={ \frac{{\cancel{\sin \left( {\alpha + \beta } \right)} \cdot \sin \alpha \sin \beta }}{{\cos \alpha \cos \beta \cdot \cancel{\sin \left( {\alpha + \beta } \right)}}} }={ \frac{{\sin \alpha \sin \beta }}{{\cos \alpha \cos \beta }} }={ \tan \alpha \tan \beta .}$

Thus,

### Product of Cotangents

Since $$\cot \theta = \large{\frac{1}{{\tan \theta }}}\normalsize,$$ we get

${\cot \alpha \cot \beta }={ \frac{1}{{\tan \alpha \tan \beta }} }={ \frac{{\cot \alpha + \cot \beta }}{{\tan \alpha + \tan \beta }},}$

that is,

### Product of Tangent and Cotangent

We take the previous formula and replace $$\beta \to \frac{\pi }{2} – \beta$$ in it. Using the cofunction identities

${\tan \left( {\frac{\pi }{2} – \beta } \right) = \cot \beta\;\;\text{and}\;\;}\kern0pt{ \cot \left( {\frac{\pi }{2} – \beta } \right) = \tan \beta ,}$

we have

${\tan \alpha \cot \beta }={ \tan \alpha \tan \left( {\frac{\pi }{2} – \beta } \right) }={ \frac{{\tan \alpha + \tan \left( {\frac{\pi }{2} – \beta } \right)}}{{\cot \alpha + \cot \left( {\frac{\pi }{2} – \beta } \right)}} }={ \frac{{\tan \alpha + \cot \beta }}{{\cot \alpha + \tan \beta }}.}$

Hence,

## Solved Problems

Click or tap a problem to see the solution.

### Example 1

Represent as a sum of trigonometric functions: $\cos 4\alpha \cos 6\alpha .$

### Example 2

Represent as a sum of trigonometric functions: $2\sin 10^\circ \cos 100^\circ.$

### Example 3

Transform the product into a sum: $\sin \left( {\alpha – \beta } \right)\cos \left( {\alpha + \beta } \right).$

### Example 4

Transform the product into a sum: $\cos \left( {\alpha + \beta } \right)\cos \left( {\alpha – \beta } \right).$

### Example 5

Write the expression as a sum of trigonometric functions: $\sin \alpha \sin 2\alpha \sin 3\alpha .$

### Example 6

Write the expression as a sum of trigonometric functions: $\cos 2\alpha \cos 4\alpha \cos 6\alpha .$

### Example 7

Calculate $\cos {10^\circ}\cos {50^\circ}\cos {70^\circ}.$

### Example 8

Calculate $\sin {20^\circ}\sin {40^\circ}\sin {80^\circ}.$

### Example 1.

Represent as a sum of trigonometric functions: $\cos 4\alpha \cos 6\alpha .$

Solution.

Using the product of cosines formula, we get:

${\cos 4\alpha \cos 6\alpha }={ \frac{1}{2}\left[ {\cos \left( {4\alpha – 6\alpha } \right) + \cos \left( {4\alpha + 6\alpha } \right)} \right] }={ \frac{1}{2}\left[ {\cos \left( { – 2\alpha } \right) + \cos 10\alpha } \right] }={ \frac{1}{2}\cos 2\alpha + \frac{1}{2}\cos 10\alpha .}$

### Example 2.

Represent as a sum of trigonometric functions: $2\sin 10^\circ \cos 100^\circ.$

Solution.

${2\sin {10^\circ}\cos {100^\circ} }={ 2 \cdot \frac{1}{2}\left[ {\sin \left( {{{10}^\circ} – {{100}^\circ}} \right) + \sin \left( {{{10}^\circ} + {{100}^\circ}} \right)} \right] }={ \sin \left( { – {{90}^\circ}} \right) + \sin {110^\circ}.}$

Note that

$\sin \left( { – {{90}^\circ}} \right) = – \sin {90^\circ} = – 1,$

${\sin {110^\circ} }={ \sin \left( {{{180}^\circ} – {{110}^\circ}} \right) }={ \sin {70^\circ}.}$

Therefore,

$2\sin {10^\circ}\cos {100^\circ} = \sin {70^\circ} – 1.$

### Example 3.

Transform the product into a sum: $\sin \left( {\alpha – \beta } \right)\cos \left( {\alpha + \beta } \right).$

Solution.

We use the product of sine and cosine identity:

${\sin \left( {\alpha – \beta } \right)\cos \left( {\alpha + \beta } \right) }={ \frac{1}{2}\sin \left[ {\left( {\alpha – \beta } \right) – \left( {\alpha + \beta } \right)} \right] }+{ \frac{1}{2}\sin \left[ {\left( {\alpha – \beta } \right) + \left( {\alpha + \beta } \right)} \right] }={ \frac{1}{2}\sin \left( {\cancel{\alpha} – \beta – \cancel{\alpha} – \beta } \right) + \frac{1}{2}\sin \left( {\alpha – \cancel{\beta} + \alpha + \cancel{\beta} } \right) }={ \frac{1}{2}\sin \left( { – 2\beta } \right) + \frac{1}{2}\sin 2\alpha }={ \frac{1}{2}\sin 2\alpha – \frac{1}{2}\sin 2\beta .}$

### Example 4.

Transform the product into a sum: $\cos \left( {\alpha + \beta } \right)\cos \left( {\alpha – \beta } \right).$

Solution.

Applying the product of cosines identity yields:

${\cos \left( {\alpha + \beta } \right)\cos \left( {\alpha – \beta } \right) }={ \frac{1}{2}\cos \left[ {\left( {\alpha – \beta } \right) – \left( {\alpha + \beta } \right)} \right] }+{ \frac{1}{2}\cos \left[ {\left( {\alpha – \beta } \right) + \left( {\alpha + \beta } \right)} \right] }={ \frac{1}{2}\cos \left( {\cancel{\alpha} – \beta – \cancel{\alpha} – \beta } \right) + \frac{1}{2}\cos \left( {\alpha – \cancel{\beta} + \alpha + \cancel{\beta} } \right) }={ \frac{1}{2}\cos \left( { – 2\beta } \right) + \frac{1}{2}\cos 2\alpha }={ \frac{1}{2}\cos 2\alpha + \frac{1}{2}\cos 2\beta .}$

### Example 5.

Write the expression as a sum of trigonometric functions: $\sin \alpha \sin 2\alpha \sin 3\alpha .$

Solution.

Let this expression be denoted by $$E.$$ First we convert the product $$\sin\alpha\sin 3\alpha$$ into a sum:

${\sin \alpha \sin 3\alpha }={ \frac{1}{2}\left[ {\cos \left( {\alpha – 3\alpha } \right) – \cos \left( {\alpha + 3\alpha } \right)} \right] }={ \frac{1}{2}\left[ {\cos \left( { – 2\alpha } \right) – \cos 4\alpha } \right] }={ \frac{1}{2}\left( {\cos 2\alpha – \cos 4\alpha } \right).}$

Then the original expression is given by

${E = \sin \alpha \sin 2\alpha \sin 3\alpha }={ \sin 2\alpha \cdot \frac{1}{2}\left( {\cos 2\alpha – \cos 4\alpha } \right) }={ \frac{1}{2}\sin 2\alpha \cos 2\alpha }-{ \frac{1}{2}\sin 2\alpha \cos 4\alpha .}$

Here

${\sin 2\alpha \cos 2\alpha = \frac{1}{2}\sin 4\alpha ,}$

${\sin 2\alpha \cos 4\alpha }={ \frac{1}{2}\left[ {\sin \left( {2\alpha – 4\alpha } \right) + \sin \left( {2\alpha + 4\alpha } \right)} \right] }={ \frac{1}{2}\left[ {\sin \left( { – 2\alpha } \right) + \sin 6\alpha } \right] }={ \frac{1}{2}\sin 6\alpha – \frac{1}{2}\sin 2\alpha .}$

Hence,

${E = \frac{1}{2} \cdot \frac{1}{2}\sin 4\alpha }-{ \frac{1}{2} \left( {\frac{1}{2}\sin 6\alpha – \frac{1}{2}\sin 2\alpha } \right) }={ \frac{1}{4}\sin 2\alpha }+{ \frac{1}{4}\sin 4\alpha }-{ \frac{1}{4}\sin 6\alpha .}$

### Example 6.

Write the expression as a sum of trigonometric functions: $\cos 2\alpha \cos 4\alpha \cos 6\alpha .$

Solution.

We denote this expression by $$F.$$ Since

${\cos 2\alpha \cos 6\alpha }={ \frac{1}{2}\left[ {\cos \left( {2\alpha – 6\alpha } \right) + \cos \left( {2\alpha + 6\alpha } \right)} \right] }={ \frac{1}{2}\left[ {\cos \left( { – 4\alpha } \right) + \cos 8\alpha } \right] }={ \frac{1}{2}\cos 4\alpha + \frac{1}{2}\cos 8\alpha ,}$

we get

${F = \cos 2\alpha \cos 4\alpha \cos 6\alpha }={ \cos 4\alpha \left( {\frac{1}{2}\cos 4\alpha + \frac{1}{2}\cos 8\alpha } \right) }={ \frac{1}{2}{\cos ^2}4\alpha + \frac{1}{2}\cos 4\alpha \cos 8\alpha .}$

By the half-angle identity,

${\cos ^2}4\alpha = \frac{1}{2} + \frac{1}{2}\cos 8\alpha .$

Convert the product $$\cos 4\alpha \cos 8\alpha$$ into a sum:

${\cos 4\alpha \cos 8\alpha }={ \frac{1}{2}\left[ {\cos \left( {4\alpha – 8\alpha } \right) + \cos \left( {4\alpha + 8\alpha } \right)} \right] }={ \frac{1}{2}\left[ {\cos \left( { – 4\alpha } \right) + \cos 12\alpha } \right] }={ \frac{1}{2}\cos 4\alpha + \frac{1}{2}\cos 12\alpha .}$

As a result, we have

${F = \frac{1}{2} \cdot \left( {\frac{1}{2} + \frac{1}{2}\cos 8\alpha } \right) }+{ \frac{1}{2} \left( {\frac{1}{2}\cos 4\alpha + \frac{1}{2}\cos 12\alpha } \right) }={ \frac{1}{4} + \frac{1}{4}\cos 4\alpha }+{ \frac{1}{4}\cos 8\alpha }+{ \frac{1}{4}\cos 12\alpha .}$

### Example 7.

Calculate $\cos {10^\circ}\cos {50^\circ}\cos {70^\circ}.$

Solution.

Let $$A$$ be the original expression. First we transform the product $$\cos {10^\circ}\cos {70^\circ}$$ into a sum:

${\cos {10^\circ}\cos {70^\circ} }={ \frac{1}{2}\left[ {\cos \left( {{{10}^\circ} – {{70}^\circ}} \right) + \cos \left( {{{10}^\circ} + {{70}^\circ}} \right)} \right] }={ \frac{1}{2}\left[ {\cos \left( { – {{60}^\circ}} \right) + \cos {{80}^\circ}} \right] }={ \frac{1}{2}\cos {60^\circ} + \frac{1}{2}\cos {80^\circ} }={ \frac{1}{4} + \frac{1}{2}\cos {80^\circ}.}$

Then

${A = \cos {10^\circ}\cos {50^\circ}\cos {70^\circ} }={ \cos {50^\circ} \left( {\frac{1}{4} + \frac{1}{2}\cos {{80}^\circ}} \right) }={ \frac{1}{4}\cos {50^\circ} + \frac{1}{2}\cos {50^\circ}\cos {80^\circ}.}$

Again, convert the product $$\cos {50^\circ}\cos {80^\circ}$$ into a sum of trig functions:

${\cos {50^\circ}\cos {80^\circ} }={ \frac{1}{2}\left[ {\cos \left( {{{50}^\circ} – {{80}^\circ}} \right) + \cos \left( {{{50}^\circ} + {{80}^\circ}} \right)} \right] }={ \frac{1}{2}\left[ {\cos \left( { – {{30}^\circ}} \right) + \cos {{130}^\circ}} \right] }={ \frac{1}{2}\cos {30^\circ} + \frac{1}{2}\cos {130^\circ} }={ \frac{{\sqrt 3 }}{4} + \frac{1}{2}\cos {130^\circ}.}$

By the reduction formula,

${\cos {130^\circ} = \cos \left( {{{180}^\circ} – {{50}^\circ}} \right) }={ – \cos {50^\circ}.}$

Therefore,

${A = \frac{1}{4}\cos {50^\circ} + \frac{1}{2}\cos {50^\circ}\cos {80^\circ} }={ \frac{1}{4}\cos {50^\circ} }+{ \frac{1}{2}\left( {\frac{{\sqrt 3 }}{4} + \frac{1}{2}\cos {{130}^\circ}} \right) }={ \frac{1}{4}\cos {50^\circ} }+{ \frac{1}{2}\left( {\frac{{\sqrt 3 }}{4} – \frac{1}{2}\cos {{50}^\circ}} \right) }={ \cancel{\frac{1}{4}\cos {50^\circ}} + \frac{{\sqrt 3 }}{8} – \cancel{\frac{1}{4}\cos {50^\circ}} }={ \frac{{\sqrt 3 }}{8}.}$

### Example 8.

Calculate $\sin {20^\circ}\sin {40^\circ}\sin {80^\circ}.$

Solution.

We denote this expression by $$B.$$ Transform the product $$\sin {20^\circ}\sin {80^\circ}:$$

${\sin {20^\circ}\sin {80^\circ} }={ \frac{1}{2}\left[ {\cos \left( {{{20}^\circ} – {{80}^\circ}} \right) – \cos \left( {{{20}^\circ} + {{80}^\circ}} \right)} \right] }={ \frac{1}{2}\left[ {\cos \left( { – {{60}^\circ}} \right) – \cos {{100}^\circ}} \right] }={ \frac{1}{2}\cos {60^\circ} – \frac{1}{2}\cos {100^\circ} }={ \frac{1}{4} – \frac{1}{2}\cos {100^\circ}.}$

Substitute this into the original triple product:

${B = \sin {20^\circ}\sin {40^\circ}\sin {80^\circ} }={ \sin {40^\circ}\left( {\frac{1}{4} – \frac{1}{2}\cos {{100}^\circ}} \right) }={ \frac{1}{4}\sin {40^\circ} – \frac{1}{2}\sin {40^\circ}\cos {100^\circ}.}$

Now we convert the product $$\sin {40^\circ}\cos {100^\circ}:$$

${\sin {40^\circ}\cos {100^\circ} }={ \frac{1}{2}\left[ {\sin \left( {{{40}^\circ} – {{100}^\circ}} \right) + \sin \left( {{{40}^\circ} + {{100}^\circ}} \right)} \right] }={ \frac{1}{2}\left[ {\sin \left( { – {{60}^\circ}} \right) + \sin {{140}^\circ}} \right] }={ \frac{1}{2}\sin {140^\circ} – \frac{1}{2}\sin {60^\circ} }={ \frac{1}{2}\sin {140^\circ} – \frac{{\sqrt 3 }}{4}.}$

Note that

${\sin {140^\circ} = \sin \left( {{{180}^\circ} – {{40}^\circ}} \right) }={ \sin {40^\circ}.}$

Hence,

${B = \frac{1}{4}\sin {40^\circ} – \frac{1}{2}\left( {\frac{1}{2}\sin {{140}^\circ} – \frac{{\sqrt 3 }}{4}} \right) }={ \frac{1}{4}\sin {40^\circ} }-{ \frac{1}{2}\left( {\frac{1}{2}\sin {{40}^\circ} – \frac{{\sqrt 3 }}{4}} \right) }={ \cancel{\frac{1}{4}\sin {40^\circ}} – \cancel{\frac{1}{4}\sin {40^\circ}} + \frac{{\sqrt 3 }}{8} }={ \frac{{\sqrt 3 }}{8}.}$