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The Proofs of the Law of Tangents

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The Proofs of the Law of Tangents 798 SCHOOL SCIENCE AND MATHEMATICS THE PROOFS OF THE LAW OF TANGENTS. BY R. M. MATHEWS, " . Riverside, California. In the triangle ABC let a, /?, y be the angles at the respective vertices and a, b, c the sides opposite. The title "law of tangents" is used to denote the trigonometric formula ab tan (aP) _ 3^ a+b ~~ tan 0+/3) and each of those formed by cyclic^ permutation of the letters. The proofs of this theorem in American textbooks stand in decided contrast to those of the law of sines and law of cosines. The proofs of these are based .directly on the triangle and are such as to suggest the one needed extension in definition; when a is an obtuse angle. sin a,== sin(180a), cos a = cos (180a). These two definitions, be it observed, require no notion of co- ordinates or of quadrants. On the other hand, in ten of a group of twelve American texts, the only proof of the law of tangents is by algebraic manipulation from the law of sines and involv- ing the factor formulas for the sum and difference of two sines. The proofs of these two formulas, being based on the addition formulas, are invariably preceded by the definition of the func- tions for the general angle, reduction to the first quadrant, and proof of the regular list of trigonometric identities. Thus the attainment of a formula that applies to triangles only seems to depend on much formal development of general and analytic trigonometry. The law of tangents can be proved directly from a figure, however. Such proofs are contained in the other two of the dozen American books examined and in every one of four foreign texts that were at hand. Altogether there seemed to be eight different proofs but when reduced to uniform notation and draw- ings, it was discovered that there are three essentially different types according to the auxiliary lines. We may start at C as center and draw a circle with (1) a as radius, or (2) b as radius. Or we may start by bisecting y or its exterior angle. In the first two cases the object is to make {a--\-b) anc^ (a^) an^ find i(a+/?) and ^(a/?). The procedures are parallel. In the other case we make i(a+^), as complement of ^7, and find the other quantities. In each of the following proofs we take ch>b. 1. With a radius a, a semicircle is described about C as center THE LAW OF TANGENTS 799 cutting & extended at D and E (Figure 1). Draw BE, BD, and then DP parallel to BE cutting BA at F. Now the essential facts are arb = AD Z BDA == i (a+^) a+b==AE ZDBF==i (a/?) 6^ ZABE=903Ka/?) Z AEB = iy = 90^(a+^) ZEBD==rt. Z = ZBDF. These relations of the angles can be established in a variety of ways, as inspection of the figure will show. This constitutes one of the sources of variation in the proofs. We note that if BA cuts the circle again at G, the central angle GCD is a useful auxiliary. To use these relations there are two different procedures. First, in rt.A's DFB and BDE tan (ajQ) / BE DF 3^ _DF _ tan^(a-4-/3) ~~DS/J)B=~B:Ef , DF ab and - -BE^-a+b' because of similar triangles. On the other hand, apply the law of sines to the triangles ABD and ABE: ab sin [90-4 sin^(a/3), a+b _ (a]3)] ' c sin-^- (a+/3) sin ^y ab sin (a/3) COS (a+/3) "~a+b Sin^(a^-j8)^ cOS^(a^)^ whence the theorem. 2. With a radius b a semicircle is described about C as center cutting a at D and its extension at E. (Figure 2). Draw AE, F*<5.^ 800SCHOOL SCIENCE AND MATHEMATICS AD and then DF parallel to AE. Here the essential facts are: B,D ab = Z DAC === -| (<x+/3) BAE = 90+i (a/3) a+b == BE Z BAD == (a-/?) ^ p^ _ ^==90-4(a4-/3) Z DAE == rt. Z = Z ADF.^ ^ As before, there are several wa,ys to establish these facts, whence come differences in proofs. If AB cuts the semicircle at G, the central angle GCB is a useful auxiliary. The final steps in the proof may proceed either by using right triangles ADF and ADE to write tan^(a/3) and tan ^4-^ or by applying the law of sines to the triangles ABD and ABE. 3. (1). To the bisector of 7 drop perpendiculars AD and BE (Figure 3). Draw AH parallel to the bisector and cutting BE at H. Let 6 = Z ABH. Then 0+P+iy=^ - =^+^+17. ..^=^(a/3). , , AH ECDC (a&)cosiy tani (a-/3) ===-HB-===BE+AD==(a+^) sin^y .'. tani (a^) ^- cotiy^tan^ (a+/3). n^3. n<?.4. This proof is essentially the same as given in Hall and Frink, Trigonometry, p. 54, and Wilczynski, Plane Trigonometry, p. 105, Ex. 2. (2). The following proof, suggested by the foregoing, I have not found in print. Bisect the exterior angle of y by FCD on which the perpen- THE LAW OP TANGENTS 801 diculars AF and BD are dropped (Figure 4). Draw AG parallel to DF cutting BD at G. Now ZDCA==4(a+/3), /GAB=a--4(<x4-/3) ===^(af3), tan^a-^ __BG_BD-AF _(a-b) sin^^). 2V / GA DC+CF (a+&) cos -K^) tanK0^) ^^.^^tanKM-^).' (3). Draw the circumscribed circle to the triangle (Figure 5) and let the bisector of y cut the circle at D. Draw DA, DB, and DF and DL perpendicular to CB and AB, respectively. On CB take CE == CA. Then DE = DA =- DB, and A,.. .^/3) FIQ.S. FB === (ab) / FDiB === (<x(3) FC=^(a+b)^ /CDF===90iy^(a+^).^ FB CF FB ^-b tani(a=^) - / - ^nereloreTherefore tani(a+^)-^F7'^F -~CF -a4^ This ng-ure is of importance in that it is all ready to give the proofs of the formulas a+b cos (a^) a^ sin^(aj8), c sin^ -^y c sin ^y o-)-& == 4R cos ^(^P) cos ^y, a& === 4R sin (a/3) sin -|y, where R is the radius of the circumscribed^ circle. By using one of these direct proofs and the well-known direct proof for the formulas for the tangents of the half angles, the logarithmic solution of oblique triangles may be treated before passing on to the generalizations of analytic trigonometry..
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