(2.388) and (2.407). For consistency between the linear expression on ι, the coefficients of ι‐dependent and ‐independent terms in both sides of Eq. (2.425) must match – so one may write
(2.427)
complemented with
(2.428)
– based specifically on Eq. (2.425), after expressing 2n − (2j + 1) as 2(n − j) − 1; and similarly
(2.429)
coupled with
(2.430)
– stemming from Eq. (2.426), with convenient factoring out of 2 in both exponents. Note, in either case, the need for a linear combination of cross powers of sine and cosine – with cosine always requiring n + 1 terms, while sine requires n terms for an even multiple but n + 1 terms for an odd multiple; and the (preferred) use of x as argument rather than θ, thus calling for rad as units.
2.3.3 Fundamental Theorem of Trigonometry
Consider a right triangle, i.e. a triangle containing a right angle – as depicted in Fig. 2.11a. In Euclidean geometry, a fundamental relationship exists encompassing the three sides of any right triangle – namely, the square of the hypotenuse (or side opposite to the right angle), of length c, equals the sum of the squares of the other two sides, of lengths a and b. In other words,
which is classically known as Pythagorean equation – in honor to ancient Greek mathematician Pythagoras (570–495 BCE), historically credited for its first (recorded) proof.
Figure 2.11 Illustration of Pythagoras’ theorem as (a) graphical statement, and graphical proof based on (b) combination of simple polygons or (c) relationships between similar triangles.
Despite the 400+ distinct proofs available, one may to advantage take four copies of a right triangle with sides a, b, and c – arranged inside a square with side c, as outlined in Fig. 2.11 b; the triangles share their area, ab/2 (a formula to be derived in due course), and the smaller square has side b − a. The area c2 (also to be derived) of the larger square may thus be given by
(2.432)
where (b − a)2 represents the area of the smaller square; expansion of the square of the binomial in the right‐hand side as per Eq. (2.238), followed by lumping of constants between numerator and denominator in the last term unfold
(2.433)
– whereas cancelation of symmetrical terms immediately retrieves Eq. (2.431).
One may alternatively resort to the proportionality of the sides of two similar triangles; let [ABC] accordingly represent a right triangle, with the right angle located at C – as per Fig. 2.11c. After drawing an altitude from point C toward side [AB], and denoting its intersection with said side as D, one ends up with point D dividing the hypotenuse [AB] into segments [AD] and [BD]. The new triangle [ACD] is similar to triangle [ABC], because they both have a right angle – at C in [ABC] (by hypothesis) and at D in [ACD] (by definition of altitude), and they share angle θ at A; therefore, the remaining angles must be identical, since the angles of a triangle always add up to π rad (see also proof at a later stage). By the same token, triangle [BCD] is similar to [ABC] – since they share the angle at B, and they both have a right angle, i.e. ∠BDC and ∠ACB, respectively. Remember that similarity of triangles enforces equality of ratios of the corresponding sides, i.e.
relating length of sides in [BCD] and [ABC] opposed to right angle as left‐hand side, and relating length of sides also in [BCD] and [ABC] opposed to angle of amplitude θ (note the mutually perpendicular sides of said angles) as right‐hand side; coupled with
relating length of sides in triangles [ACD] and [ABC] opposed to the right angle as left‐hand side, and relating length of sides also in [ACD] and [ABC] opposed to angle of amplitude π − π/2 − θ = π/2 − θ as right‐hand side. Equation (2.434) may be rewritten as
upon elimination of denominators – and a similar result, viz.
can be produced from Eq. (2.435); ordered addition of Eqs. (2.436) and (2.437) unfolds
(2.438)
where
