1 as a approaches 0, since cos a approaches 1. must approach Then for very small angles sin a may be replaced by α, expressed in radians. The error thus introduced is so small that it may be neglected in many problems. Thus, to five decimal places, 115. To develop sin a and cos a in terms of a. By De Moivre's theorem, cos no + i sin ne = (cos + i sin 0)". On expanding the second member by the binomial theorem, we have cos no + i sin n0 = cos" 0 + in cos"-10 sin 0. Let a remain constant while n increases indefinitely. Then necessarily decreases indefinitely, since no = α, a sin 0 constant. By Art. 114, when approaches 0, арᎾ proaches 1, and cos approaches 1. Making these substitutions in equation (4), we have Equating the real parts of equation (1), we have By the same process as above, equation (5) becomes - .... (5) The series for sin a and cos a are convergent for all finite values of a.* They enable us to compute the sine and cosine of any angle. It is then possible to construct a table of natural functions, from which the logarithmic functions may be obtained. In using these series a must, of course, be expressed in radian measure. *See any College Algebra on the convergency of series. 1. Find the four fourth roots of unity by De Moivre's theorem. 2. Find the six sixth roots of unity by De Moivre's theorem. 3. Find the square root of 5 3 i. SOLUTION. Let √5 - 3ir (cos a + i sin α). Then and 5-3 i = r2 (cos 2 a + i sin 2 α) = (1) r2[cos (2 α- 2 nπ) + i sin (2α - 2 nπ)]. Equating the real and the imaginary parts, r2 cos (2 α - 2 nπ) = 5, 72 sin (2 α- 2 nπ) = Squaring and adding (2) and (3), we have - 3. p4 = 34, ... r2 = √ 34, and r = 34. (2) (3) (4) the quadrant being determined by (2) and (3). (5) (6) (7) (1). Squaring and adding (2) and (3), we have p6 = 13, .'. 73 = √13, and r = V13. 2 (2) (3) (4) 3 α - 2 nπ 303° 41'; (5) the quadrant being determined by (2) and (3). Substituting from (4) and (6) in (1), we have 3/2-3 i = - -.2987 +1.5041 i. Substituting from (4) and (7) in (1), we have 2-3 i=- - 1.1530 1.0107 i. Substituting from (4) and (8) in (1), we have 3/2-3 i = 1.4518 - .4933 i. We have thus found the three cube roots of 2-3 i. 10. Find the cube root of 1 + i. 11. Find the cube root of 1+i. 12. Find the cube root of 2 + 3 i. 13. Find the values of sin 3x and cos 3x in terms of sin x and cos x. 14. Find the values of sin 5x and cos 5x in terms of sin x and cos x.. 15. Prove by De Moivre's theorem that |