150. We can now point out the use of the ambiguous sign = in the formula cos 0 : #J1 – sino 0. If we know the numerical value of the sine of an angle 0, without knowing the magnitude of the angle, we cannot from the identity, cos 0 = 1 - sino 0, completely determine cos 0, for we get cos 0 = +/1 – sin* 0. This is a general formula, and we shall find that it represents an important Geometrical truth. 151. Given sin 0 = a, we can say that 0 is one of the angles represented by one or the other of the positions OP,, UP, of the revolving line in Fig. I. on page 110. If we attempt to find the cosine of these angles we get OM, two different values for the cosine ; for and ом, , ОР, , OP, although equal in magnitude, are opposite in sign. Hence, if a be the least angle whose sine is equal to a, we have cos 0 = = cos a = +/1 - sino a. 152. The same result may be obtained from the formula 0=ni + (-1)". For cos {n++ (-1)"a} is of different sign according as n is even or odd. EXAMPLES. XXXIII. (1) If @ be found from the equation cos 0 == 2, show geometrically that there are two values of sin 0 and of tan 0. (2) If A be found from the equation tan 6=a, show geometrically that there are two values of sin 0 and of cos 0. (3) If A be the least angle without regard to sign such that sin A=a, show that cos A=+11 - sino A. (4) If A be the least positive angle such that cos A=a, prove that sin A=+11 - cos? A. CHAPTER XI. ON THE TRIGONOMETRICAL RATIOS OF Two ANGLES. 153. We proceed to establish the following fundamental formulæ : sin (A + B)=sin A.cos B + cos A .sin B (i). Here, A and B are angles; so that (A + B) and (4 – B) are also angles. Hence, sin (A + B) is the sine of an angle, and must not be confounded with sin A + sin B. Sin (A + B) is a single fraction. 154. The proofs given in the next two pages are perfectly general, as will be explained below (cf. Art. 169); but the figures are drawn for the simplest case in each. The student should notice that the words of the two proofs are very nearly the same. To prove that sin (A + B)=sin A.cos B+cos A. sin B, and that cos (A + B)=cos A.cos B - sin A . sin B. + Let ROE be the angle A, and EOF the angle B. Then in the figure, ROF is the angle (A+B). In OF, the line which bounds the compound angle (A+B), take any point P, and from P draw PM, PN at right angles to OR and OE respectively. Draw VH, NK at right angles to MP and OR respectively. Then the angle NPH=90° – HNP=HNO=ROE=A*. MP MH+HP KN HP Now sin (A+B)=sin ROF= + OP ON.OP NP.OP ON OPTNP. OP OM OK - MK OK HN ОР OP OP NP.OP ON OPNPOP =cos A. cos B – sin A. sin B. * Or thus. A circle goes round OMNP, because the angles OMP and ONP are right angles ; therefore MPN and MON are angles in the same segment; so that MPN=MON=A, OP To prove that sin (A -B)=sin A.cos B-cos A. sin B, and that cos (A - B)%Acos B+ sin A. sin B. Let ROE be the angle A, and FOE the angle B. Then in the figure, ROF is the angle (A – B). In OF, the line which bounds the compound angle (A - B), take any point P, and from P draw PM, PN at right angles to OR and OE respectively. Draw NH, NK at right angles to MP and OR respectively. Then the angle NPH=90° - HNP=HNE=ROE=A*. MP MH - PH KN PH Now sin (A - B)=sin ROF= OP ОР OP OP ON.OP NP.OPONOP NPOP OM OR+KM OK . NH Also cos (A - B)=cos ROF= + OP ОР + + * Or thus. A circle goes round OMPN, because the angles OMP and ONP are right angles; therefore MPN and MON together make up two right angles; so that HPN=MON=A. Example. Find the value of sin 75o. =sin 45o. cos 30° + cos 45o. sin 300 1 1 + 4 13-1 (2) Show that sin 15°= 2./2 13+1 (3) Show that cos 15o= 2/2 (4) Show that tan 750=2+1/3. (5) If sin A= and sin B=, find a value for sin (A + B) and for cos (A – B). (6) If sin A=6 and sin B=ts, find a value for sin (A - B) and for cos (A + B). 1 1 (7) If sin A= and sin B= show that one value of 25 10' (A + B) is 45°. (8) Prove that sin 750=.9659... 155. It is important that the student should become thoroughly familiar with the formulæ proved on the last two pages, and that he should be able to work examples involving their use. |