Precalculus

Trigonometry

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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 of sines
Difference of sines

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 of cosines
Difference of cosines

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,

Sum of tangents

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,

Difference of tangents

Sum and Difference of Cotangents

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

Sum of cotangents
Difference of cotangents

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,

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

Use the double-angle identities for sine and cosine:

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

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

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