RELAzIONI E FUNzIONI The formulas established by Werner are: 1 1. cosx cosy = _(cos(x + y) + cos(x y)) 2 1 2. senx seny = _(cos(x y) cos(x + y)) 2 1 _ 3. senx cosy = (sen(x + y) + sen(x y)) 2 Let us see, with an example, how a product of two large numbers was calculated, by using the first among these three formulas. To multiply 98 436 by 79 253, one had to follow these steps: 1. both numbers were divided by 100 000 till they were <1 thus obtaining 0,98436 and 0,79253 2. x and y were calculated so that: 0,98436 cosx = _ = 0,49218 and cosy = 0,79253 2 By using the so called goniometric tables (that is tables which give the sine and cosine values in correspondence of real arguments and viceversa), it was obtained: x = arccos(0,49218) 1,0562040150 y = arccos(0,79253) 0,6558497643 3. with the same table, it was obtained cos(x + y) = cos(1,0562040150 + 0,6558497643) = 0,1407881536 cos(x y) = cos(1,0562040150 0,6558497643) = 0,9209229845 4. then, the values determined were summed: cos(x + y) + cos(x y) = 0,7801348309 And by mutiplying twice by 100 000, it was obtained the value 7 801 348 309 [Example freely drawn from: Boyer, C.B.; Storia della matematica with an introduction by Lucio Lombardo Radice; 1990; Oscar Mondadori] Today, with the calculator, you can easily verify that 98 436 79 253 = 7 801 348 308. Instead, to convert an additive goniometric expression into a multiplicative, we usually refer to other formulas called prosthaphaeresis (perhaps a word which comes from the Greek, pr sth(esis) and apha resis, meaning addition and subtraction): x+y x y x y x+y senx + seny = 2sen _ cos _ cosx + cosy = 2cos _ cos _ 2 2 2 2 +y x_ y x_ y x_ x_ +y sen sen cosx cosy = 2sen senx seny = 2cos 2 2 2 2 For example, by using prosthaphaeresis, let us solve the following goniometric equation: cos5x sen4x = cos3x Let us reformulate it in this way: (cos5x cos3x) sen4x = 0 Now, let us apply the prosthaphaeresis on the difference of the two cosines so to obtain: 2sen4x senx sen4x = 0 Let us take out the common factor sen4x: sen4x (2senx + 1) = 0 Following the cancellation property for multiplication, we obtain two equations: sen4x = 0 x = 0 + k 1 senx = __ 2 11 o x = ___ + 2k 6 7_ 11 _ The solutions of the equation thus are: x1 = 0 + k , x2 = + 2k , x3 = ___ + 2k . 6 6 2senx + 1 = 0 116 7 x = __ + 2k 6