Divide each of the above two equations by this last. Thus The first formula may be easily remembered by giving to the values, 2π ± 4, 4π±& &c. which make both sides vanish; and by putting = 0 which gives 1 - cos = 2 sin2 1. 524. Many important results may be obtained from the factorisation of cos 0 and sin 0. Thus we may equate the coefficients of the several powers of 0 in the term-expansions of cos 0 and sin @ with those obtained by multiplying out the factor expressions. Or we may equate coefficients after taking the logarithms of the two expressions for cos or sin 0. Thus we have, from the two expressions for sin 0, 525. The method of the following article is of very general application, the process involved being in reality merely to differentiate both sides of an identity. 526. To find series for cot 0 and for tan 0. Resolving each factor of sin @ into two factors, we may write sin 0 = 0P (1+), where P denotes the product of all values obtained by giving to r every positive and negative integral value, excluding zero. Ө log sin ✪ = log ✪ + Σ log ( 1 + ~—-). Now for write 0 + h ; Υπ log sin (0 + h) = log (0 + h) + Σ log (1++). Subtract the identity in 0 from the corresponding identity in +h; and expand the remainder in powers of h. On the left we have log sin (0+ h) - log sin On the right we have = log (cos h + cot sin h) log (0 + h) – log 0 = log (1 + 2) h Ө h 0 h (h cot 0...)2 + 1 h2 202 log (1++)-log (1+) = log(1+0) ... Hence equating the coefficients of h in the new identity, denotes the sum of all values obtained by giving to r every positive and negative integral value including zero. In the same way from the identity it follows that Ө cos 6 = P(1+ (2r+1). 3) - tan 0 = Σ where, as above, r has every In these expansions of cot value including zero. and tan 6, the terms having equal and opposite coefficients of π should be added so as to form one term. [Otherwise the series assumes the indeterminate form We have shown in Art. 359 that for all positive integral If now n is not a positive integer the series on the left will be endless. If moreover tan is arithmetically equal to or less than 1, the series will be convergent as in the binomial theorem. Now in this case the above equation-as will be shown in Chap. XXI. still holds, provided that, on the right hand, we make represent the positive or negative acute angle whose tangent = tan 0; and give to cos" its positive value. With this understanding, then, we may make n as small a fraction as we please. Then, in the limit, Hence from the above equation, If tan 01 and 0 is acute; i.e. if lies between -1 and Let AOB be any angle not greater than half a right-angle; AB the arc opposite AOB; AT1 the tangent at A. 1 Draw the lines OBT1, TT, TT3, T3T,... each perpendicular to the preceding to meet AT1 and OA alternately. Then arc AB AT-AT + AT-AT7+... = }AT+}AT¿ For, if A0 = 1, AB = circular measure of AB = 0 say. 2 And AT1 = tan 0; AT2 = tan2 0; AT, = tan3 0; and so on, by Euclid VI. 8. 529. To express O in terms of sin 0. We have shown in Art. 375, that for all odd integral values of n, 3 15 n If now n is not an odd integer the series on the left will be endless. It is always convergent, because sin is arithmetically equal to or less than 1. Now the above equation—as will be shown in Chap. XXI.still holds provided that, on the right hand, we make ✪ represent the positive or negative acute angle whose sine = sin 0. Thus we may make n as small a fraction as we please. in the limit, if ◊ is acute, Then 530. If x is any quantity arithmetically not greater than 1, in Art. 527, we may write tan 0=x, and in Art. 529, sin 0 Thus = x. circular measure of the acute angle 1.3.5 x7 23 2.45 2.4.67 the acute angle whose sine is equal to x. 531. Now we have tan (rπ + 0) = tan @ where r is any integer. = Again we have cot (rπ+0) tan 6, where r is any integer. If then is any angle lying between гë+‡π and rπ + 3⁄4π, where r has some assigned integral value, Hence may be expanded in powers of tan o, when ø is in the right or left quadrant, and in powers of cot & when it is in the up or down quadrant. |