2.12b to yield
(2.450)
where the first term in the right‐hand side represents the length of the portion of [AB] closer to A, while the second term represents the length of the remainder of [AB] closer to B; c may then multiply both sides to produce
Ordered addition of Eqs. (2.447) and (2.449) gives rise to
(2.452)
or else
after condensing terms alike; ordered subtraction of Eq. (2.451) from Eq. (2.453) generates
(2.454)
which breaks down to
(2.455)
upon cancelation of symmetrical terms – and which recovers Eq. (2.443), after swapping 2ab cos γ and c2 between sides.
Another important relationship in trigonometry relates the lengths of the sides of a triangle of any shape to the sines of its angles, according to
it is classically referred to as law of sines, or sine formula – and a, b, and c denote lengths of the sides opposing angles α, β, and γ, respectively, while R denotes radius of the circumcircle to the triangle. To derive the first two equalities in Eq. (2.456), one should recall the rule of calculation of the area of a triangle, S – as one half of the product of its base by its height (to be proven later); inspection of Fig. 2.12b indicates that this can be done in three different ways, i.e.
using [BC] as base and [AC] as hypotenuse,
using [AC] as base and [AB] as hypotenuse, or
using [AB] as base and [BC] as hypotenuse – in all cases at the expense of the graphical interpretation of sine conveyed by Eq. (2.290). Since Eqs. (2.457)–(2.459) share their left‐hand side, one may lump them as
(2.460)
whereas a further multiplication of all sides by 2/abc permits simplification to
the first two equalities in Eq. (2.461) degenerate to the corresponding equalities in Eq. (2.456), after taking reciprocals of all three sides. To prove the last equality of Eq. (2.456), one should redraw triangle [ABC] together with the corresponding circumscribed circle – as done in Fig. 2.12 c; furthermore, an altitude may be drawn from the center of the circle, O, toward side [BC], with intercept denoted as D. Since sides [OB] and [OC] of auxiliary triangle [OBC] are identical for coinciding with radius R of the circle, the altitude [OD] of this isosceles triangle splits in half the angle formed by [OB] and [OC], as well as segment [BC] of length a; therefore, angle ∠BOD is equal to angle ∠COD, and to half the angle ∠BOC – and thus also equal to angle ∠BAC of amplitude α, because it subtends side [BC] with corner A lying on the circle. By definition of sine as ratio of length of opposite leg to that of hypotenuse, one realizes that
(2.462)
pertaining to ∠BOD, or equivalently
after revisiting Eq. (2.458) as
following isolation of sin α, one may eliminate sin α between Eqs. (2.463) and (2.464) to get
(2.465)
– which degenerates to
when solved for the reciprocal
