Precalculus

Trigonometry

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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,

Product of sines

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 cosines

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 sine and cosine

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 tangents

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,

Product of cotangents

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,

Product of tangent and cotangent

See solved problems on Page 2.

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