396. To find the real value of x which satisfies the equation x3-3x2x=2b3 where b is numerically greater than a. Put xy+ay (a homogeneous expression), Since b2> a2, y3 is real; .. y has one and only one real value [Art. 230] corresponding to each of the values of y3. Substituting for y to find x, we have x = $/{b3 + √(b® − ao)} + 3/{b3 — √(b¤ — a®)}. 397. Similarly to solve 398. By the last two articles, we have found one real solution of the equation 399. To find three values of x to satisfy the equation x3-3a2x=263 where b is numerically = or < a. Now since b3 is numerically not greater than a3, there are real values of 30 which satisfy this equation. Let the series of values be 3a, 360° – 3a, 360° + 3a, &c. Then x 2a cos 0 = = 2a cos a or 2a cos (120° -- a) or 2a cos (120° + a) &c. giving three and three only real values of x. 400. By the last article we have shown how three real roots of the equation x3-px + q = 0 can be found when 3 (p is alge Thus we should find, by means of Trigonometrical tables if Then, also with the help of Trigonometrical tables if necessary, we should find the cosine of of this angle. Thus, instead of finding the ratios of A from those of A by solving a cubic equation, we do the reverse; i.e. we solve a cubic equation by finding (with the help of tables if necessary) the ratios of A from those of A. 401. A geometrical mode of solving the equation x3- px + q = 0 thus requires us to trisect the angle whose cosine is— √(27q2÷4p3). An angle cannot be trisected by ruler and compass. But the following explanation indicates how a simple instrument may be used for the trisection of an angle. Let IOF be the given angle. Let a rod OR revolve round 0, so that R describes a circle which cuts IO produced in I. Let MV-rigidly attached to OR-bisect OR at right-angles at M; and suppose it to be slit along its central axis. Also let RI-pivoted at R to OR-be slit along its central axis, and always pass through a pin fixed at I'. Thus a pin P, through each of the slits, will mark the point of intersection of MV and RI'. Then, if OR rotates until P is on OF, OR shall trisect the angle IOF. '= For .. ▲ OROI', . ORI' LOIR, and exterior angle IOR twice interior angle ORI' twice angle ROP (because A's MOP, MRP are equal in all respects). = :. L ROP = 1 2 IOF. Ratios of Particular Angles. Q.E.D. 402. We may find the ratios of angles which have a simple relation to a right-angle by means of the general formulæ of this and the preceding chapter, without the special geometrical constructions of Chapter IV. Thus The complement of 30° is 60°, i.e. 2.30°; .. cos 30° = sin 60°, i.e. 2 sin 30° cos 30°. But cos 30° is not 0; .., dividing by cos 30°, The complement of 36° (i.e. 2.18°) is 54° (i.e. 3.18°); .. 2 sin 18° cos 18° = cos3 18° - 3 cos 18° sin2 18°. (the positive sign of the radical being taken to make sin 18° positive). EXAMPLES XV. 1. Draw and explain a figure giving all values of which satisfy the equations : 4. Trace the changes in sign and magnitude of sin 4 + cos and of sin 4-cos 14, as A varies from 0 to 360°. 5. Show that cos 4 - sin 4 =(-1)m √(1 − sin A) where m is the integral part of (A + 270°) ÷ 360°. 6. Given cot A=k, show that cot 4 or tan 14 may be obtained from the equation cot 14 - tan 14 = 2k ± 2,√(1 + k2). Show how the sign of the radical must be determined: and also the sign of the radical in the expression obtained by solving for cot 14. 7. If sin 44 s, show that the four values of tan A are given by {√(1+8) -1} { √/(1 − 8) + 1} ÷ 8. 8. Find when sin B is arithmetically and when it is algebraically greater or less than cos B. Hence deduce the variations in sign of cos B-sin B, and cos B + sin B. 9. Show that the three roots of the equation x(x-3)2 = 4 sin2 a are positive, and that the difference between the greatest and least of them is > 3 and <2√3. to π. 23/7=3x-1 6 sin (of 90°)-1. 12. Show that cot 10 >1+ cot for all values of 0 from 0 |