CHAPTER V. TRIGONOMETRIC PROBLEMS. 47. PROBLEM I. Having given two equations of the form r sin p = a, r cos p= =b, where a and b are given numbers, it is required to compute r and P. Solution. Dividing the first equation by the second, we find From this value of tan we find p itself, then the sine or cosine of p, and then r from either of the equations and The numbers in brackets show the order in which the numbers of the computation are written. In writing log r sin r cos o spaces are left for inserting log sin p and log cos o after p is found, so that either of the latter may be subtracted to obtain r. It is generally best to obtain r from both r sin p and r cos P, because then if any mistake is made in it will be shown by a difference in the results. On the other hand, a practical computer will not write down either sin or cos p, but will subtract in his head and write down log r only. sin and rcos In the preceding examples to be positive, and have we have supposed taken in the first quadrant. But either or both of these quantities may be negative. Whatever their signs, there are always two values of p, differing by 180°, corresponding to any given value of tan (§ 31). Hence the problem admits of two solutions in all cases. In the one will be positive, in the other negative. But in practice only that solution is sought which gives a positive value of r. This being the case, sin p and cos o must have the same algebraic signs as the given quantities r sin p and r cos p respectively. Now consider each case in order : I. rsin and r cos p both positive. The angle p must then be taken in the first quadrant, because only in this quadrant are sin p and cos o both positive. II. rsin p positive and rcos p negative. Sin is positive only in quadrants (1) and (2) (§ 21), and cos p is negative only in quadrants (2) and (3). Hence the requirement of signs can be fulfilled only in the second quadrant, and ዋ 90° < p < 180°. III. r sin and r cos q both negative. The only quadrant in which sine and cosine are both negative is the third. Hence in this case 180°<< 270°. IV. rsin o negative and r cos o positive. The only quadrant in which sin is negative and cos o positive is the fourth. Hence in this case φ 270° < p < 360°. EXERCISES. : Given r sin p =237.09, rcos +192.91; find r and p. p = 2. . 19 3. NOTE. In this last exercise compute the value of (+47° 50′) as if it were one quantity, and subtract the angle 47° 50′ from the result. 49. PROBLEM II. Having given two equations of the form x cos ay sin a = p, y cos α = q, it is required to find the values of x and The elimination is conducted by the method of addition and subtraction, as follows: Multiply the first equation by cos a and the second by sin a. We thus have x cos ay sin a cos a = p cos a; Next multiply the first equation by sin a and the second by We have cos α. By addition, x sin a cos ay sin2 a = Ρ sin a; -x sin a cos a + y cos3 α = q cos a. y= =p sin a It will be noticed that in these equations x and y are given in terms of p and q by equations of the same form as the original ones. EXERCISE. Find the value of p and q from the equations to find the values of r and B,-the other four quantities, a, b, ε, and O, being supposed known. Solution. Developing the sines of 6+a and ẞ+✪ (§ 40), we have r sin ẞ cos + cos ẞ sin &= a; r sin cos 0+r cos ẞ sin = b. S (a) Regarding r sin ẞ and r cos ẞ as the two unknown quantities, we see that their coefficients are cos &, sin &, cos 0 and sin 0. Multiplying the first equation by cos O and the second by cos ε, we have cos & sin →) = a cos e b cos ε. Subtracting, -- r cos ẞ (sin & cos Noting that the coefficient of r cos ẞ is sin (ε — 0), we find, by division, To find the value of r sin ẞ we multiply the first equation (a) by sin and the second by sin &, and subtract. We thus find Supposing the numerical values of a, b, ɛ, and ✪ to be given, these equations give the values of r sin ẞ and r cos ß, from which rand B can be computed by Problem I. EXERCISES. Find the values of rsing and rcos q from the following equations by the preceding method: 1. r cos (p+ε) = a, r cos (p − 0) = b. Ans. rsin @= rcos P = a cos Ꮎ sin (ε+0) b sina sin sin (ε + 0) 3. r sin (p+) = a cos y, r cos (9 — ε) = a sin y. cos (ε — 0) ; 4. Find the values of r and p from the equations when 31° 27'.4. P r cos (p+0) = 3.790 8, r cos (p − 0) = 2.060 7, 51. PROBLEM IV. To reduce an expression of the form to a monomial, a and b being given quantities. Solution. Determine the values of two auxiliary quantities k and & from the equations |