From the similar triangles BEO and OCH, or sin. A: R=R: cosec. A, OC: OH, or R2 sin. A cosec. A. 6. The radius is a mean proportional between the cosine and secant of an arc. By the similar triangles BGO and : OF, or cos. A: R = R: sec. A, OAF, or R2 GB: BO: = ОА cos. A sec. A. FA: AO = = 7. The tangent is to the radius as the sine to the cosine. For in the similar triangles OAF and OEB, BE: EO, or tan. A: R= sin. A: cos. A. 8. The tangent is to the secant as the radius to the cosecant. For in the similar triangles FAO and OCH, CO: OH, or tan. A sec. AR: cosec. A. FA:FO= 9. The sine of an arc is to the cosine as the secant to the cosecant. For (Cor. 5 and 6) sin. A cosec. A = cos. A sec. A, therefore (VI. 16) sin. A : cos. A = sec. A: cosec. A. 10. The square on the radius is equal to the squares on the sine and cosine of an arc. For OE - GB =cos. A, fore R2 sin. 2A + cos. 2A. and OB2 BE2 + EO2, there 11. The square on the sine of an arc is equal to the difference of the squares on the radius and the cosine. For (Cor. 10) sin. 2A + cos. 2A = R2, 12. The square on the sine of an arc is equal to the rectangle under the versed sine and suversed sine. or sin. 2A vers. A suvers. A; since AE is the vers. A and ED is the suvers. A. 13. The square on the secant of an arc is equal to the sum of the squares on the radius and tangent. For OF2 = OA2 + AF2, or sec. 2A R2+tan. 2A. 14. The chord of twice an arc is equal to twice the sine of that arc. For BI = 2BE, or chord 2A = 2 sin. A. 15. The square on the chord of an arc is equal to twice the rectangle under the radius and the versed sine of the arc. therefore BE2 + EA2 = or chord 2A = 2R. For (Cor. 12) BE2 = AE ED, vers. A. 16. The sine, tangent, and secant of an arc, are the same as those of that arc increased by any number of whole circumferences. For if to AB any number of whole circumferences he added, the compound arc will terminate at B. In some of the applications of trigonometry, it is necessary to attend to the signs of the trigonometrical lines. Quantities whose signs are +, are called positive; and those whose signs are —, are) called negative. If a line be measured from a given point or a given line as its origin, it is reckoned positive when it lies on one side of its origin, and negative when on the opposite side. Thus, if the sines in the first quadrant be measured from the diameter AD, and be reckoned positive, those in the second quadrant, as MN, will also be positive; and those in the third and fourth, as LN, IE, will be negative. So if the cosines, as BG, in the first quadrant, measured from CK be reckoned positive, those in the second and third, being measured from the opposite side of CK, will be negative, and those in the fourth will be positive. In a similar manner, the signs of the tangent, secant, &c., are determined; but this subject belongs properly to Analytical Trigonometry. If a number as m is to be divided by a number n, the quotient is expressed thus, m n If R = 1, and the whole circumference become respectively: 11. Sin. A = 1 cos. 2A = (1 + cos. A) (1 Cor.); so cos. A = 1 − sin. A 12. Sin. A vers. A. suvers. A. 13. Sec. 2A = 1 + tan. 2A, (1 + sin. A) (1 − sin. A)." and cosec. 2A= 1 + cot. A. 14. Chord 2A = 2 sin. A. 15. Chord 2A = 2 vers. A. 16. Sin. A sin. (mC+A), where m is any integer. If the hypotenuse of a right-angled triangle be made radius, the sides become the sines of the opposite angles, or the cosines of the adjacent angle. For let ABC be a right-angled triangle, if its hypotenuse AC be made radius, then BC is the sine of A; but the angle at C is the complement of A, therefore BC is the cosine of C. By A C E BD making C the centre, it may be similarly proved that AB is the sine of C, and therefore the cosine of A. PROPOSITION II. THEOREM. If one of the sides about the right angle of a right-angled triangle be made radius, the other side becomes the tangent of the opposite angle, and the hypotenuse the secant of the same; or the other side becomes the cotangent of the adjacent angle, and the hypotenuse the cosecant of the same. A, BC is tangent of and When AB (see fig. to Prop. 1) is radius, and AC is secant of A, AC is cosecant of C. The sine, tangent, and other trigonometrical lines of an angle for one radius, are proportional to those for another radius; and the ratios of the corresponding lines are the same as that of the respective radii. Let A be any angle, and BC and DE be arcs described with the radii AB or AC, and AD or AE; then CF and EK are its sines, and BG and DL its tangents for these radii respectively, CF: EK = AC: AE, or the sines of E angle A for the radii AC and AE, A or AB and AD are proportional to LD, and the secants AG and AL. these radii. The same may be proved for the tangents GB and CF: EK = GB: LDAG: AL. COR.-Hence, in a right-angled triangle AFC, = R : sin. A, CF AC sin. A. Hence, by equal ratios, FC AF tan. A, and AC= It is proved in the same way that AF = AC when AF is radius, Schol.-When two sides of a right-angled triangle are given, the third may be found by I. 47 and Cor. PROPOSITION IV. THEOREM. If any side of a right-angled triangle be made radius, the other two sides are proportional to the trigonometrical lines, which they represent for any other radius. Let ABC be a right-angled triangle; when AC is radius, AB and BC represent the cosine and sine of A; and (Prop. III.) these are proportional to the cosine and sine for any other radius. Therefore if sin. A, cos. A, be the sine and cosine of A for any other radius, sin. A: cos. A; or sin. A cos. A = CB: BA. CB: BA= The sides of a plane triangle are to one another as the sines of the opposite angles. A From A, any angle in the triangle ABC, let AD be drawn perpendicular to BC; and because the triangle ABD is right-angled at D, AB: AD = R : sin. B, or AD R = AB sin. B; and for the same reason, AC: AD = R : sin. C, or AD·R = AC sin. C; therefore AB sin. B=AC sin. C, hence (VI. 16), AB : AC = sin. C : sin. B. In the same manner it may be demonstrated that AB: BC = sin. C: sin. A. B D The sum of two sides of a triangle is to their difference as the tangent of half the sum of the angles at the base to the tangent of half their difference. Let ABC be any triangle, then if B and C denote the angles at B and C respectively, AB + AC: AB - AC = tan. (C + B): tan. (C — B). (Const.) From A as a centre, with the radius AC, describe the semicircle DCE; produce BA to D ; join DC and CE, and draw EF parallel to BC. (Dem.) Then twice angle AEC = AEC+ACE = DAC (I. 32) = ABC+ACB = B+ C, fore AEC (C + B); = FEC ECB, and AEC = ECB or CEF = (CB). AB-AC. = And since angle DCE is a right angle (III. 31), if EC be made radius, DC and FC are tangents of the angles AEC and FEC (Def. 4). But (VI. 2) DB: BE: DC: CF; C-B). or AB+ AC: AB PROPOSITION VII. = AC tan. (C+B): tan. THEOREM. If a perpendicular be drawn from the vertex upon the base of a triangle, the sum of the segments of the base is to the sum of the two sides as the difference between these sides to the difference between the segments of the base. For (II. c. Cor. 1) the rectangle under the sum and difference of the sides is equal to that under the sum and difference of the segments of the base, therefore (VI. 16) the above proportion subsists. |