Hence cot A cot B:: tang B: tang A; that is, the cotangents of two arcs are reciprocally proportional to their tangents. The formula cot Axtang A=R2 might be deduced immediately, by comparing the similar triangles CAT, CDS, which give AT CA :: CD: DS, or tang A: R:: R : cot A XIX. The sines and cosines of two arcs, a and b, being given, it is required to find the sine and cosine of the sum or difference of these arcs. D Let the radius AC=R, the arc D' AB=a, the arc BD=b, and consequently ABD=a + b. From the points B and D, let fall the perpendiculars BE, DF upon AC; L from the point D, draw DI perpendicular to BC; lastly, from the point I draw IK perpendicular, and IL parallel to, AC. F' C FK KE P The similar triangles BCE, ICK, give the proportions, CB: CI :: BE: IK, or R : cos b :: sina : IK_sin a cos b R CB: CI: CE: CK, or R: cos b:: cos a: CK= cos a cos b. R The triangles DIL, CBE, having their sides perpendicular, each to each, are similar, and give the proportions, CB: DI :: CE: DL, or R : sin b :: cos a: DL= L N = CB: DI:: BE: IL, or R sin b; sin a: IL: = B cos a sin b. R sin a sin b. R M But we have IK+DL=DF=sin (a+b), and CK-IL-CF-cos (a+b). Hence sin a cos b+ sin b cos a sin (a+b)= R cos (a+b): The values of sin (a-b) and of cos (a-b) might be easily deduced from these two formulas; but they may be found directly by the same figure. For, produce the sine DI till it meets the circumference at M; then we have BM=BD=b, and MI=ID=sin b. Through the point M, draw MP perpendicular, and MN parallel to, AC: since MI-DI, we have MN =IL, and IN=DL. But we have IK-IN=MP=sin (a—b), and CK+MN=CP=cos (a-b); hence sin a= sin a cos b-sin b cos a R cos a cos b+sin a sin b sin (a-b)= cos (a—b): R These are the formulas which it was required to find. The preceding demonstration may seem defective in point of generality, since, in the figure which we have followed, the arcs a and b, and even a+b, are supposed to be less than 90°. But first the demonstration is easily extended to the case in which a and b being less than 90°, their sum a+b is greater than 90°. Then the point F would fall on the prolongation of AC, and the only change required in the demonstration would be that of taking cos (a+b)-CF'; but as we should, at the same time, have CF-I'L-CK', it would still follow that cos (a+b)=CK'—I'L', or R cos (a+b)=cos a cos b—sin a sin b. And whatever be the values of the arcs a and b, it is easily shown that the formulas are true: hence we may regard them as established for all arcs. We will repeat and number the formulas for the purpose of more convenient reference. sin a cos b+sin b cos a R sin a cos b-sin b cos a R cos a cos b-sin a sin b Ꭱ cos a cos b+ sin a sin b R sin (a+b)= (1.). sin (a—b): (2.). cos (a+b)= (3.) cos (a-b): (4.) XX. If, in the formulas of the preceding Article, we make b=a, the first and the third will give cos 2a= 2 2 sin a cos a cos2 a-sin2 a 2 cos2 a-R sin 2a= = R R R formulas which enable us to find the sine and cosine of the double arc, when we know the sine and cosine of the arc itself. To express the sin a and cos a in terms of a, put ļa for a, and we have cos2 a-sin2 a 2 sina cosa R To find the sine and cosine of a in terms of a, take the equations whence = cos2 a+sin2 a=R2, and cos2 a-sin fa-R cos a, there results by adding and subtracting cos2 a=R2+R cos a, and sin2 a=R2-R cos a: sin a=√(}R2—R cos a)=√2R2—2R cos a. If we put 2a in the place of a, we shall have, sin a=√(}R2—R cos 2a)=√2R2—2R cos 2a. cos a= √(¿R2+}R cos 2a)=¦√2R2+2R cos 2a. Making, in the two last formulas, a=45°, gives cos 2a=0, and sin 45°= √}R2=R√; and also, cos 45°= √ R2=R √ }. Next, make a=22° 30', which gives cos 2a=R, and we have sin 22° 30' R √(√) and cos 22° 30' R√(} + { √ }). XXI. If we multiply together formulas (1.) and (2.) Art. XIX. and substitute for cos2 a, R2-sin2 a, and for cos2 b, R2—sin2 b; we shall obtain, after reducing and dividing by R2, sin (a+b) sin (a—b)=sin2 a-sin2b= (sina + sin b) (sin a—sin b). or, sin (ab) sin a-sin b:: sin a+sin b: sin (a+b). XXII. The formulas of Art. XIX. furnish a great number of consequences; among which it will be enough to mention those of most frequent use. By adding and subtracting we obtain the four which follow, 2 sin (a+b)+sin (a—b)=sin a cos b. 2 sin (a+b)—sin (a—b)= sin b cos a. R cos (a+b)+cos (a—b)= cos (a—b)—cos (a+b)=sin a sin b. and which serve to change a product of several sines or cosines into linear sines or cosines, that is, into sines and cosines multiplied only by constant quantities. gives a= XXIII. If in these formulas we put a+b=p, a—b=q, which p+q p = 2 b 2 cosp+cos q= cos q-cos p = 2 R 2 2 sin p+sin q=sin (p+q) cos (p—q) (1.) R 2 R 2 we shall find sin p-sin q=sin(p-q) cos (p+q) (2.) } -cos a cos b. sin R cos (p+q) cos 1 (p—q) (3.) R 2 (p+q) sin } (p—q) (4.) sin P R+cos p sin P R-cos p tang p : hence R cot p cot P R = Ꭱ tangp: formulas which are often employed in trigonometrical calculations for reducing two terms to a single one. XXIV. From the first four formulas of Art XXIII. and the first sin a tang a R R cot a COS a of Art. XX., dividing, and considering that sin p+sin q_sin (p+q) cos (p—q)_tang (p+q) R } R sin p+sin q_cos (p-q)_cot (p—q) = R = = sin p-sin q_cos (p+q)_cot (p+q) sin = sin Formulas which are the expression of so many theorems. From the first, it follows that the sum of the sines of two arcs is to the difference of these sines, as the tangent of half the sum of the arcs is to the tangent of half their difference. XXV. In order likewise to develop some formulas relative to tangents, let us consider the expression R sin (a+b) tang (a+b)=2 cos (a+b) of sin (a+b) and cos (a+b), we shall find tang (a+b) in which by substituting the values 9 R (sin a cos b+sin b cos a) cos a cos b-sin b sin a cos b tang b Now we have sin a= and sin b R R substitute these values, dividing all the terms by cos a cos b; we shall have cos a tang a 9 tang (a+b)= R2 (tang a+tang b) ; which is the value of the tangent of the sum of two arcs, expressed by the tangents of each of these arcs. For the tangent of their difference, we should in like manner find R2 (tang a-tang b) R2+tang a tang b. tang (a-b): Suppose b=a; for the duplication of the arcs, we shall have the formula 2 R2 tang a = R-tanga tang 2a: Suppose b=2a; for their triplication, we shall have the for mula tang 3 a= ; R2 (tang a+tang 2 a) R2-tang a tang 2 a in which, substituting the value of tang 2 a, we shall have 3R2 tang a-tang 3a tang 3 a= R2-3 tang 2a. XXVI. Scholium. The radius R being entirely arbitrary, is generally taken equal to 1, in which case it does not appear in the trigonometrical formulas. For example the expression for the tangent of twice an arc when R=1, becomes, If we have an analytical formula calculated to the radius of 1, and wish to apply it to another circle in which the radius is R, we must multiply each term by such a power of R as will make all the terms homogeneous: that is, so that each shall contain the same number of literal factors. |