XVI. The square of the sine of an arc, together with the square of the cosine, is equal to the square of the radius; so that in general terms we have sin A+ cos2 A=R2. This property results immediately from the right-angled triangle CMP, in which MP2+CP2=CM2. It follows that when the sine of an arc is given, its cosine may be found, and reciprocally, by means of the R E formulas cos A=±√(R2-sin2A), and sin A=±√(R2—cos2A). The sign of these formulas is +, or because the same sine MP answers to the two arcs AM, AM', whose cosines CP, CP', are equal and have contrary signs; and the same cosine CP answers to the two arcs AM, AN, whose sines MP, PN, are also equal, and have contrary signs. Thus, for example, having found sin 30°R, we may deduce from it cos 30°, or sin 60° = √ (R2—¥R2) = √ aR2—†Ř√3. XVII. The sine and cosine of an arc A being given, it is required to find the tangent, secant, cotangent, and cosecant of the same arc. The triangles CPM, CAT, CDS, being similar, we have the proportions: CP : PM :: CA: AT; or cos A: sin A:: R: tang A= CP: CM:: CA: CT; or cos A: R:: R: sec A: = PM: CP:: CD: DS; or sin A: cos A: R: cot A= PM: CM:: CD: CS; or sin A: R:: R: cosec A R sin A cos A R2 COS A R cos A sin A R2 sin A which are the four formulas required. It may also be observed, that the two last formulas might be deduced from the first two, by simply putting 90°-A instead of A. From these formulas, may be deduced the values, with their proper signs, of the tangents, secants, &c. belonging to any arc whose sine and cosine are known; and since the progressive law of the sines and cosines, according to the different arcs to which they relate, has been developed already, it is unnecessary to say more of the law which regulates the tangents and secants. By means of these formulas, several results, which have already been obtained concerning the trigonometrical lines may be confirmed. If, for example, we make A=90°, we shall have sin A=R, cos A=0; and consequently tang 90° === R2 0' an expression which designates an infinite quantity; for the quotient of radius divided by a very small quantity, is very great, and increases as the divisor diminishes; hence, the quotient of the radius divided by zero is greater than any finite quantity. COS The tangent being equal to R. sin; and cotangent to R.Cos it follows that tangent and cotangent will both be positive when the sine and cosine have like algebraic signs, and both negative, when the sine and cosine have contrary algebraic signs. Hence, the tangent and cotangent have the same sign in the diagonal quadrants: that is, positive in the 1st and 3d, and negative in the 2d and 4th; results agreeing with those of Art. XII. The Algebraic signs of the secants and cosecants are readily determined. For, the secant is equal to radius square divided by the cosine, and since radius square is always positive, it follows that the algebraic sign of the secant will depend on that of the cosine: hence, it is positive in the 1st and 4th quadrants and negative in the 2nd and 3rd. Since the cosecant is equal to radius square divided by the sine, it follows that its sign will depend on the algebraic sign of the sine: hence, it will be positive in the 1st and 2nd quadrants and negative in the 3rd and 4th. XVIII. The formulas of the preceding Article, combined with each other and with the equation sin 2A+ cos 2A-R2. furnish some others worthy of attention. First we have R2 + tang2 A = R2 + R2 sin2 A cos A hence R2+tang2 A sec2 A, a cos A formula which might be immediately deduced from the rightangled triangle CAT. By these formulas, or by the right-angled triangle CDS, we have also R2+ cot2 A-cosec2 A. Lastly, by taking the product of the two formulas tang A= R sin A and cot A= formula which gives cot A R cos A sin A we have tang Ax cot A--R2, a tang B' 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. Let the radius AC=R, the arc 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; Ľ from the point D, draw DI perpendicular to BC; lastly, from the point I draw IK perpendicular, and IL parallel to, AC. D' D L The similar triangles BCE, ICK, give the proportions, CB: CI: BE: IK, or R: cos b:: sin a: IK: sin a cos b R cos a cos b. R CB: CI: CE: CK, or R: cos b:: cos a: CK= CB: DI:: BE: IL, or R sin b:: sin a: IL: cos a sin b. R sin a sin b. Ꭱ But we have 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 MIDI, we have MN =IL, and IN-DL. But we have IK-IN-MP=sin (a—b), and CK+MN=CP=cos (a-b); hence 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 sin (a+b)= (1.). R sin a cos b-sin b cos a (2.). R cos a cos b-sin a sin b cos (a+b)= (3.) R cos a cos b+sin a sin b cos (a—b) (4.) R XX. If, in the formulas of the preceding Article, we make b=a, the first and the third will give 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 and we have To find the sine and cosine of a in terms of a, take the equations cos la+sin a=R2, and cos a-sin2 ja R cos a, there results by adding and subtracting whence cosa R2+R cos a, and sina 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 2u. Making, in the two last formulas, a=45°, gives cos 2a=0, and = sin 45°✓R=RV; and also, cos 45°-✓R-R√. Next, make a=22° 30′, which gives cos 2a=R√1⁄2, 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-sin3b=(sina+sin b) (sin a—sin b). or, sin (a—b) 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, sin (a+b)+sin (a—b)=. b. 2 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. XXIII. If in these formulas we put a+b=p, a—b=q, which 2 R cos q-cosp=sin(p+q) sin (p—q) (4.) |