trigonometric identity



A trigonometric identity is (formally) a commutative diagram in the category of cartesian spaces and partial functions whose edges are labelled by rational functions (or sometimes algebraic functions) and trigonometric functions.

Slightly more precisely: each rational function p(x 1,,x n)(x 1,,x n)p(x_1, \ldots, x_n) \in \mathbb{R}(x_1, \ldots, x_n) is interpreted as a partial function nDf n\mathbb{R}^n \hookleftarrow D \stackrel{f}{\to} \mathbb{R}^n where DD is the “natural domain” of pp (see rational function for more discussion); these are partial analytic functions. The basic trigonometric functions sine (sin\sin) and cosine (cos\cos) are (total) analytic functions \mathbb{R} \to \mathbb{R}. All of these may be interpreted as partial functions n n\mathbb{R}^n \rightharpoonup \mathbb{R}^n, and generate a class of functions by applying the monoidal category structure on the category of partial functions that is induced by the cartesian product on cartesian spaces. A trigonometric identity is then (formally) an equality of morphisms in the monoidal category thus generated.

Of course, this is complete overkill; category theorists are not oblivious to the fact that this is exactly the kind of description lampooned in Linderholm’s Mathematics Made Difficult. It’s just a formal way of covering bases. So let us add that in practice, a trigonometric identity usually involves functions obtained by substituting trigonometric functions into rational functions, or substituting rational linear (affine) functions into trigonometric functions: the class of functions considered is usually fairly limited in scope.


Virtually all trigonometric identities can be seen as arising from suitable exponential function identities on complex numbers such as

These imply relations between the sine function and the cosine function as follows:

sin(α+β)=sin(α)cos(β)+cos(α)sin(β) \sin\left( \alpha + \beta \right) \;=\; \sin(\alpha) \cos(\beta) + \cos(\alpha) \sin(\beta)
cos(α+β)=cos(α)cos(β)sin(α)sin(β) \cos(\alpha + \beta) \;=\; \cos(\alpha) \cos(\beta) - \sin(\alpha) \sin(\beta)

which in turn implies

(1)sin(α)cos(β)=12(sin(α+β)+sin(αβ)) \sin(\alpha) \cos(\beta) \;=\; \tfrac{1}{2} \left( \sin(\alpha + \beta) + \sin(\alpha - \beta) \right)



See also