But, tan. (4+B ) : tan. (4≈B )=sin.4+sin.B : sin.A—sin.B 2 (eq. (19), trig.) A-B 2 Comparing the two latter proportions (th. 6, b. 2), we have, Given the three sides of any plane triangle, to find some relation which they must bear to the sines and cosines of the respective angles. Let ABC be the triangle, and let the perpendicular fall either upon, or without the base, as shown in the figures; and by recurring to theorem 38, book 1, we shall find Now, by proposition 3, trigonometry, we have, (1) (2) Equating these two values of CD, and reducing, we have, In this expression we observe that the part of the numerator which has the minus sign, is the side opposite to the angle; and that the denominator is twice the rectangle of the sides adjacent to the angle. From these observations we at once draw the following expressions for the cosine A, and cosine B. As these expressions are not convenient for logarithmic computation, we modify them as follows: If we put 2a=A, in equation (31), we have, cos.4+1=2 cos.2 A In the preceding expression (n), if we consider radius, unity, and add 1 to both members, we shall have, Considering (b+c) as one quantity, and observing that we have the difference of two squares, therefore (b+c)2—a2=(b+c+a)(b+c—a); but (b+c—a)=b+c+a—2a =s, and extracting square root, the final 2 In every triangle, the sum of the three angles must equal 180°; and if one of the angles is small, the other two must be comparatively large; if two of them are small, the third one must be large. The greater angle is always opposite the greater side; hence, by merely inspecting the given sides, any person can decide at once which is the greater angle; and of the three preceding equations, that one should be taken which applies to the greater angle, whether that be the particular angle required or not; because the equations bring out the cosines to the angles; and the cosines, to very small arcs vary so slowly, that it may be impossible to decide, with sufficient numerical accuracy to what particular arc the cosine belongs. For instance, the cosine 9.999999, carried to the table, applies to several arcs; and, of course, we should not know which one to take; but this difficulty does not exist when the angle is large; therefore, compute the largest angle first, and then compute the other angles by proposition 4. But we can deduce an expression for the sine of any of the angles, as well as the cosine. It is done as follows: EQUATIONS FOR THE SINES OF THE ANGLES. Resuming equation (m), and considering radius, unity, we have, Subtracting each member of this equation from 1, gives 1—cos. C=1— (a2 + b2—¿2 2ab (2) Making 2a=C, in equation (32), then a=1C, Equating the right hand members of (1) and (2), By taking equation (p), and operating in the same manner, we The preceding results are for radius unity; for any other radius, we must multiply by the number of units in such radius. For the radius of the tables, we write R; and if we put it under the radical sign, we must write R2; hence, for the sines corresponding with our logarithmic table, we must write the equations R2(s—b) (s—c thus, ( sin.4 = A large angle should not be determined by these equations, for the same reason that a small angle should not be determined from an equation expressing the cosine. In practice, the equations for cosine are more generally used, because more easily applied. In the preceding pages we have gone over the whole ground of theoretical plane trigonometry, although several particulars might have been enlarged upon, and more equations in relation to the combinations of the trigonometrical lines, might have been given; but enough has been given to solve every possible case that can arise in the practical application of the science; but to show more clearly the beauty and spirit of this science, and to redeem a promise, we give the following geometrical demonstrations of the truths expressed in some of the preceding equations. From C as the center, with CA as the radius, describe a circle. Take any arc, AB, and call it A; AD a less arc, and call it B; then BD is the difference of the two arcs, and must be designated by (A—B); AG—AB; therefore, DG=A+B; EG=sin.A; (See fig. p. 154.) En sin.B; Gn=sin.A+sin.B; Fm=mD=CH-cos.B; mn=cos.A; Therefore, Fm+mn=cos.A+cos.B=Fn; Because Therefore, mD—mn=cos.B—cos.A=nD; DG=2sin. (4+B) NF=AD; AB+NF=A+B; 180°—(A+B)=arc FB; Or, A+B 90°—(4+2)-jare FB; But the chord FB, is twice the sine of arc FB. In the triangle FBG, Fn is drawn from an angle perpendicular to the opposite side; therefore, by Proposition 5, we have, Gn: nB tan. GFn: tan. BFn sin.90°: DG sin.nDG: Gn; sin.nD G=cos.n GD A+B 2 : sin.A+sin.B Or, =sin. ( A-B ) :sin.A-sin.B 2 sin.A—sin.B=2cos. (4+) sin. ( 4–B) same as equation (16). In the triangle FBn, we have, 2 |