Precalculus

Trigonometry

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Product-to-Sum Identities

Solved Problems

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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,

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