We shall conclude this introductory chapter, by demonstrating two propositions which are of the highest importance in our subsequent investigations. The first is, In any right-angled triangle, the ratio which the side opposite tone of the acute angles bears to the hypothenuse, is the sine of that angle; the ratio which the side adjacent to one of the acu e angles bears to the hypothenuse, is the cosine of that angle; and the ratio which the side opposite to one of the acute angles bears to the side adjacent to that angle, is the tangent of that angle. Let CSP be a plane triangle right-angled at S. P PS CS Then, =sin. C, PS CP =cos. C, CP CS From C as a centre with the radius CP, describe a circle. Produce CS, to meet the circumference But the triangles TAC, PS, are similar; A S CS=CP cos C=CP sin. P PS=CS tan. C=CS cot. P The second proposition is. In any plane triangle, the ratio of any two of the sides is equal to the ratio of the sines of the angles opposite to them. Let ABC be a plane triangle; it is required to prove that CB sin. A CB sin. A CA sin. B From C let fall the perpendicular CD on AB. Then, since CDB is a plane triangle right-angled at D, by last proposition C Again, since CDA is a plane triangle right-angle at A, 'CD CA sin. A CA sin. B. (1) (2) In like manner, by dropping perpendiculars from B and A the side AC, CB we can prove, upon CB sin. A, CA sin. B, = BA sin. C, BA sin. C. In treating of plane triangles, it is convenient to designate the three angles by the capital letters A, B, C, and the sides opposite to these angles by the corresponding small letters a, b, c. According to this notion, the last proposition will be a sin. A sin. A b sin. B a sin. C' с sin, c Given the sines and cosines of two angles, to find the sine of their sum. Let ABC be a plane triangle; from C let fall CD perpendicular on AB, C Let angle CAB=8, and angle CBA=ß, Then, AB = BD+DA = BC cos. B+ AC cos. 4, because BDC and ADC are right-angled triangles, Dividing each member of the equation by AB, sin.Ccos +sin.cos. 6, by last Prop. in Chap. I. sin. C sin. cos. ẞ+ sin. ß cos. 4. But, since ABC is a plane triangle, 4++ C = 180° 180°—(A+B) sin (+6,) because 180°—(+6) is the supplement = sin. cos + sin. ß cos. Θ Hence, sin. (+0) (a) That is, the sine of the sum of two arcs or angles is equal the sine of the first multiplied by the cosine of the second, plus the sine of the second multiplied by the cosine of the first. This expression, from its great importance, is called the fundamental formula of Plane Trigonometry, and nearly the whole science may be derived from it. Given the sines and cosines of two angles, to find the sine of their difference. By formula (a). sin. (+6) = sin. cos. + sin. ß cos. 4. For substitute 180°-4, the above will become sin. 180°-(—)} = sin. (180°—0) cos. B + sin. cos.(180°) But sin. {180°-(8—~)} = sin. (4B) ·.· 180°—(66) is the supplement of (-8.) And, sin. (180°) sin. 4, = And, cos.(180°) cos. == Substitute, therefore, these values in the above expression, it becomes sin. (4-8) sin. cos. B-sin. ß cos. A .... (b) That is, the sine of the difference of two arcs or angles, is equal the sine of the first X cosine of the second, cond x cosine of the first. the sine of the se Given the sines and cosines of two angles, to find the cosine of By formula (a) their sum. sin. (+6)=sin. cos. B+ sin. ẞ cos. 8. For substitute 90°+8, the above will become sin. {90°+(+6)}= sin. (90°+8) cos. B +sin. B cos. (90°+4) But, sin. {90°+(4+ß)}=cos. (4+6) by Table II. And, cos. (90°+4) =—sin. 8. Substituting, therefore, these values in the above expression, it becomes, cos. (+6)=cos. cos. ß-sin. That is, the cosine of the sum of two arcs or angles, is equal to the cosine of the first multiplied by the cosine of the second, minus the sine of the first multiplied by the sine of the second. Given the sines and cosines of two angles, to find the cosine of their difference. By formula (a :) sin. (+8) sin. cos. B+ sin. B cos. 8, For substitute 90°-4, the above will become sin. {90°—(6—ẞ)}= sin. (90°-4) cos. B + sin. B cos. (90°--4), But, sin. {90°—(6—B)}=cos. (6—6), By Table II. sin. (90°—4) eos. (90°-8) =cos. Substituting, therefore, these values in the above expression, That is, the cosine of the difference of two arcs or angles, is equal to the cosine of the first multiplied by the cosine of the second, plus sine of the first into the sine of the second. Given the tangents of two angles, to find the tangent of their sum. By Table I.: tan. (+8) = sin. (+6) cos. (+6) sin. cos. 6+sin. ß cos. by (a) and (c) cos. cos. B-sin. 8 sin. ß That is, the tangent of the sum of two arcs or angles, is equal to the sum of the tangents of the two arcs, divided by 1 minus the product of the two tangents. Given the tangents of two angles, to find the tangent of their difference. Dividing both numerator and denominator by cos. cos. 6: Hence, the tangent of the difference of two arcs or angles, is equal to the difference of the tangents of the two arcs, divided by 1 plus the product of the two tangents. The student will have no difficulty in deducing the following: To determine the sine of twice a given angle. By formula (a): sin. (+6)=sin. cos. B+sin. ß cos. .. Let 3, then the above becomes sin. 2 0=sin. ê cos. ê+sin. 8 cos. ◊ =2 sin. cos. 8 (gl) That is, the sine of twice a given angle, is equal to twice the sine of the given angle multiplied by its cosine. θ In the last formula, for substitute ; then, 2 |