That they contain some elements of simplicity I am convinced by the fact that several of my students worked out the proof of the formula for sin(x+y) with no other help than the mere suggestion to use polar coördinates. §§6, 7, 8 are based on memoranda given me by Professor Moore of work intended for his elementary calculus course. The proofs are made only for positive angles less than. 2. If an angle is inscribed in a circle of unit diameter its sine is the chord of the arc subtended. = If one side OB of the angle AOB is a diameter of the circle (see Fig. 2), AB AB then since OBA is a right angle, is the sine of AOX. If the angle is ОА 1 inscribed in any other way, by a familiar theorem, it subtends the same chord as AOB. The theorem is also true in the limiting case where one side of the angle is tangent to the circle. This is the polar coördinate case and thus, in Fig. 3, OA=sinAOX. We may note also that in a circle of unit diameter the length of the arc subtended by an inscribed angle is the measure of that angle in radians. §3. Proof of the formula sin(x+y)=sinxcosy+cosxsiny. In a circle ABO (Fig. 2)* of unit diameter, let ..AD=ABcosy=sinxcosy, DC-BCcosx-sinycost. Since AC=AD+DC, sin(x+y)=sinxcosy+cosysinx. This proof applies directly to the polar graph (see Fig. 4) if OA is taken tangent to the circle (so that A=0). The present form is intended to suggest its use to those who do not care to introduce polar coördinates in a beginning course. Pa 4. Proof that cos(x+y)=cosxcosy - sinxsiny. Let OAB (Fig. 2) be the polar cosine curve, i. e., OA is the diameter (1) of the circle OAB. In the polar sine circle, Fig. 3, OA=sinx, AB=siny. On OB, lay off AC =AB and let D be the point in which BC meets the circle. :. OC=sinx-siny. Now COD is measured by the arc AD and hence / COD ABD The proofs in this section make use of Fig. 4, in which XOF。=xo; FOF, =Ax=/Fi-10F; (1=1...n); or, if one prefers to speak of the arcs, Ax=arcFo f Fi-Pi-1 is perpendicular to OF; and FQ; is perpendicular to OF-1. To prove (1), we make use of only one of the ▲x portions of the figure, for example, the third. OG, is taken equal to OF, and OG ̧=OF. Then by elementary geometry 2 2 But if we call / XOF, x, LF, F,O=x and F,F2Q3=x+^x, = 2 3 2 .(3). sin (x+ ▲x) — sinx ̧ sin Ax A x Ax ..(4). -cos(x+▲x). -1 and cost is continuous, both extremes of this double inequality approach cost as Ax approaches zero. Therefore the middle term approaches cost and we have In this, according to our figure, Ax was always positive. But if XOF, 3 had been taken as x the same figure with similar reasoning would prove (4) for that case also. Of course the theorem that for continuous functions, integration is the inverse of differentiation shows that (2) is a corollary of (1). But for some purposes of instruction it is worth while to compute (2) directly from the definition of an integral as the limit of a sum. Assuming the existence of a definite integral for cost we have *In the familiar Cartesian figure, (5) corresponds to the inner set of rectangles and (6) to the outer. and (2) will be proved if we show that S,<sinX-sinx<S′n• 0 sin x cosxsin(x+▲x) — sinx,>sin x cos(x+▲x). From the second quadrilateral F, P, FQ, we similarly get 2 sin x cos(x+^x)>sin(x+2^x)−sin(x。+^x)>sin ▲ x cos(x。+2^x), and so on. From the last quadrilateral we obtain, calling X<XOF„=x ̧+n^x, sin^x cos[x ̧+(n−1)^x]>sinX—sin[x。+(n−1)^x]>sin ^ x cos(x ̧+n^x). Adding together these inequalities we see that the sum of the first terms is S', and of the last terms is S. In the middle terms everything else cancels, leaving only sinX-sinx.. So we have as we desired. S>sinX-sinx > Sn. This result can be seen still more directly by noting that If the rays centering at O be imagined to fold together like a fan from OF。 to OF, it is evident that S, is less and S, greater than OF-OF。. $7. Second Proof of (1) and (2). In some quarters there is a tendency to reverse the old order and present the integral calculus before the differential. The definitions of the two operations of differentiation and integration are certainly independent of each other; and whatever order may be preferred for pedagogical reasons, it is not amiss to see that in either case precisely similar methods can be used in deriving the formulas for the usual functions. That such is the case depends depends on the following theorem.* *The proof of the first part of this theorem is made possible by the fact that any monotonic function is integrable, ‚—a monotonic function being such that if a<b either alwaysЛ(a)> or =ƒ(b) or alwaysƒ(a)< or=f(b). Just as we did for the special case of sinx in the last section we can let b-a=^x and add up n inequalities like (9) and thus have Sn>F(x)—Fr>S'n. To prove the second part divide by b-a and pass to the limit as b approaches a. If on an interval x。, -----, X, two functions f(x) and F(x) have the property that for every two values of x, a and b (x ≤a<b≤ X), 0 f(a)(b−a)>F(b)− F(a)>ƒ(b)(b−a). then, first, Sf(x)dx=F(X)— (9) f(x)dx==F(X)− F(x); second, if f(x) is continuous, DÎF(x)=f(x). In view of this theorem, to prove (1) and (2) we need only to prove the inequality cosx.(x-x)>sinz-sinx>cosx(x-x。), 0 ≤x。<x ≤ π/2...(10). To this end we make use of the inner part of Fig. 5 in which and TS<US=cos USS..S.S<cosx.arcSS<cosx。(x−x。), which proves (10). $8, Second Proof of (1) and (2). This, unlike Without going into details I will add the outline of a second proof of Professor Moore's for the inequality (10) and hence for (1) and (2). the others, is not a polar coördinate proof, but uses the unit circle. part of Fig. 5, In the outer |