# 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.