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

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Solution.

Using the product of cosines formula, we get:

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

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,

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

Solution.

We use the product of sine and cosine identity:

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

Solution.

Applying the product of cosines identity yields:

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

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,

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

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

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

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,

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

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,

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