and show how the signs of the radicals must be determined. 15. Show that, for (i) cos a + cos (120° − a) + cos (120° + a) = 0, (ii) cos a cos (120° – a) + cos a cos (120° + a) + cos (120° — a) cos (120° + a) = − 2, (iii) 4 cos a cos (120° – a) cos (120° + a) = cos 3a. What relations between the roots and the coefficients of a cubic equation follow from the above? then {1 - 2 sec cos (a − 0) + sec2 0} {1 − 2 sec 0 cos (ẞ − 0) + sec2 0} provided the sum or difference of no two of the angles a, ß, y is 25. tan-1+tan-1=π. 26. tan-1+tan-1 + tan-1 + tan−1 } = — π. 27. If 4 cos 0 + 3 sin 0 = 1, find 0; given sin 53° 7′ 51′′ = ·8, sin 11° 31′ 10′′ = ·2. 28. If a cos 0 + b sin 0 = c, show that Show that if 1, 2 be two values of (whose tangents are unequal), then 2 cot (1 + 2) = tan ß – sec2 a cot ß. 15 (1 - cos 20 cos 24) = 17 sin 20 sin 24 and sin (0 − 6) + 1⁄2√{sin (0 + 4) sin (0 − 4)} = (tan 0 − 1) cos 0 cos 4, CHAPTER XVI. TRIGONOMETRICAL FACTORS. § 1. ALGEBRAICAL THEOREMS. 408. DEF. I. A (rational) integral function of any quantity is a finite sum of a finite number of terms, involving only positive integral powers of that quantity. α But +√x+cx2 +bx¬3 is not a rațional integral function of x: for the terms contain negative or fractional indices of x. Nor is ax2+ bx + c : for the value of this fraction found by division would contain an infinite number of terms (except for special values of a, b, c, p, q). 409. DEF. II. If Q and P÷Q can each be expressed as a (rational) integral function of x, then Q is called an x-factor of P, and P is called an x-multiple of Q. Thus Again x-a is an x-factor of x2 – a2, (x2-a2)÷(x-a)=x+a (an integral function of x); ax - a2 is an x-factor of x2 – a2, ··· (x2 — a2)÷(ax — a2): +1 (an integral function of x). α · But (x − a)-1 is not an x-factor of x+a; because although (x+α) ÷ (x − a) -12-a2 (an integral function of x), ‹ yet (x-a)-1 is not itself an integral function of x. 410. The ordinary terms factor and multiple may be substituted for x-factor and x-multiple respectively, if it is always remembered that an algebraic expression can be regarded as integral or not, only in reference to some specified symbol of quantity which it contains. The expression rational integral function is usually abbreviated into integral function; and the term polynomial is another equivalent. 411. The symbol f(x), read function x, denotes any expression involving x; where it is understood that ƒ (a), ƒ (b), ƒ(3), &c. represent the values obtained by writing a, b, 3, &c. respectively, instead of x, in the expression denoted by f(x). Other symbols, such as fi (x), (x), &c., denote other expressions involving x, in which the substitution of a for x is denoted by fi (a), & (a), &c. 412. Theorem I. If f(x) be any integral function of x arranged in descending powers of x and divided by x-a, the remainder which is independent of x will be ƒ (a). For the process of division consists in choosing such terms in the quotient as shall make each successive remainder of lower degree than the last. Hence in this way we may always obtain a remainder which does not contain x; and which will, therefore, be unaltered by any change in the value of x. Let then (x) be the quotient and R the remainder obtained by dividing f(x) by x-a. Thus f(x) = (x). (x − a) + R .(1). In (1) put xa, then R, which does not contain x, will be unaffected. .:. ƒ(a) = $(a). (a− a) + R Now since (a) must be finite, (a). (a-a) = 0, :. f(a) = R or R=ƒ (a). Substituting for R in (1), we have where .(2). .(3), (x) being the quotient is an integral function of x as well as f(x). COR. Ꮳ a is always a factor of ƒ (x) –ƒ (a). For, by the above, f(x) −ƒ (a) = (x). (x − a) where (x) is an integral function of x. 413. Theorem II. If f(a) = f(a) =ƒ(az) = ...... and if ɑ ɑ ɑ„,... are all different, then f(x) -ƒ(a) is divisible by the product (x-α) (x — α2) (x — αз)...... For, by Theorem I., dividing f(x) -ƒ (a) by x-α1, ƒ (x) −ƒ (α1) = Þ1 (x). (x − α1) +ƒ (α1) − ƒ (α1) where (x) is integral. Again, dividing this quotient 2 (x) by x − ɑ3, ..(2). since a-a1 and a - a2 are neither = 0. - ..2(x)=3(x). (x − αg). Substituting in (2), f(x)-ƒ (α)=3(x). (x − α) (x − α) (x — αз) ..........(3). In this way we may go on dividing each successive quotient by any number of factors, which are all different, as long as the quotient contains x at all. J. T. 20 |