If then in 1 is even, writing na = and nẞ, we have cos > = cos p = 2n-1 p^=) ("−2) { sin2 λ=0 2λπ n n sin2 n Divide each of the above two equations by this last. Thus sin2 (2x+4)/n).. 1 + sinh2 0/n sin2 (2π ± Here make n infinite. The first formula may be easily remembered by giving to the values, 2π ±0, 4π±& &c. which make both sides vanish; and by putting 0 = 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 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 0 into two factors, we may write where P denotes the product of all values obtained by giving to r every positive and negative integral value, excluding zero. .. log sin (0 + h) = log (℗ + h) + ≥ 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 = log (cos h + cot @ sin h) = log (1 + h cot 0 - 1 h3...) =hcot 0-12 On the right we have log (0 + h) — log 0 = log ( 1 + +242) h h - 1⁄2 (h cot 0...)2 + . 1 h2 + 0 202 Ө log (1 +6+4)-log (1 + 2) = log (1++) Υπ Υπ Hence equating the coefficients of h in the new identity, where Σ denotes the sum of all values obtained by giving to r every positive and negative integral value including zero. 527. To express in terms of tan 0. 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, 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. Draw the lines OBT1, TT, TT, TT,... each perpendicular to the preceding to meet AT, and OA alternately. Then arc AB AT-AT + AT-AT7+... = 5 For, if A0= 1, AB = circular measure of AB=0 say. And AT1 =tan 0; AT2 = tan2 0; AT ̧= tan3 0; and so on, by Euclid VI. 8. 529. To express in terms of sin 0. We have shown in Art. 375, that for all odd integral values 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 0 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 = x. Thus the acute angle whose sine is equal to x. 531. Now we have tan (rπ+ 0) = tan @ where r is any integer. is any angle lying between г–‡π and r«+‡a, where If then Again we have cot (+π- 0) = tan 6, where r is any integer. If then is any angle lying between rπ+π and rî + 3⁄4π, where r has some assigned integral value, cot - cot3 + ⁄ cot3 & − = rπ + 1⁄2π- $. ... Hence may be expanded in powers of tan ø, when ø is in the right or left quadrant, and in powers of cot & when it is in the up or down quadrant. |