Sum-to-Product Identities
Solved Problems
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 \(\sin \frac{\pi }{5}\) 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 \(\cos \alpha = \frac{1}{{\sqrt 3 }}.\)
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 \(\sin \alpha =\frac{2}{{\sqrt 5 }}.\)
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
\[E = 2\cos 4\alpha \sin \alpha = 2 \cdot \left( { - \frac{7}{{25}}} \right) \cdot \frac{2}{{\sqrt 5 }} = - \frac{{28}}{{25\sqrt 5 }}.\]
