Product-to-Sum Identities
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,
\[\sin\alpha\sin\beta = \frac{1}{2}\left[{\cos\left({\alpha - \beta}\right) - \cos\left({\alpha + \beta}\right)}\right]\]
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,
\[\cos\alpha\cos\beta = \frac{1}{2}\left[{\cos\left({\alpha - \beta}\right) + \cos\left({\alpha + \beta}\right)}\right]\]
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
\[\sin\alpha\cos\beta = \frac{1}{2}\left[{\sin\left({\alpha - \beta}\right) + \sin\left({\alpha + \beta}\right)}\right]\]
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,
\[\tan\alpha\tan\beta = \frac{\tan\alpha + \tan\beta}{\cot\alpha + \cot\beta}\]
Product of Cotangents
Since \(\cot \theta = \frac{1}{{\tan \theta }},\) we get
\[\cot \alpha \cot \beta = \frac{1}{{\tan \alpha \tan \beta }} = \frac{{\cot \alpha + \cot \beta }}{{\tan \alpha + \tan \beta }},\]
that is,
\[\cot\alpha\cot\beta = \frac{\cot\alpha + \cot\beta}{\tan\alpha + \tan\beta}\]
Product of Tangent and Cotangent
We take the previous formula and replace \(\beta \to \frac{\pi }{2} - \beta \) in it. Using the cofunction identities
\[\tan \left( {\frac{\pi }{2} - \beta } \right) = \cot \beta\;\;\text{and}\;\;\cot \left( {\frac{\pi }{2} - \beta } \right) = \tan \beta ,\]
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,
\[\tan\alpha\cot\beta = \frac{\tan\alpha + \cot\beta}{\cot\alpha + \tan\beta}\]
See solved problems on Page 2.
