# Product-to-Sum Identities

## 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,

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