150. We can now point out the use of the ambiguous sign in the formula cos 0 = ±√√1 – sin2 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, cos3 0 = 1 − sin2 0, completely determine cos 0, for we get cos = √1 − sin3 0. This is a general formula, and we shall find that it represents an important Geometrical truth. is one of the 151. Given sina, we can say that angles represented by one or the other of the positions OP1, OP, of the revolving line in Fig. I. on page 110. If we attempt to find the cosine of these angles we get although equal in magnitude, are opposite in sign. Hence, if a be the least angle whose sine is equal to a, we have cos = cos a = ± √√1 − sin3 a. 152. The same result may be obtained from the formula Ꮎ = nπ+(-1)"α. For cos {n+(−1)"a} is of different sign according as n is even or odd. (1) If be found from the equation cosa, show geometrically that there are two values of sine and of tan 0. (2) If be found from the equation metrically that there are two values of sin tan 0=a, show geoand of cos 0. (3) If A be the least angle without regard to sign such that sin A-a, show that cos A=+1 − sin2 A. (4) If A be the least positive angle such that cos A=a, prove that sin A= +√1 – cos2 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 (A – 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. Sin A+ sin B is the sum of two fractions. 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. that_sin (A+B)=sin A. cos B + cos A . sin B, 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 NH, 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 OP OP = KN. ON HP.NP KN ON HP NP + = + ON.OP NP. OP ON OP NP OP =sin ROE. cos EOF+cos HPN.sin EOF =sin A. cos B+ cos A. sin B. = ON. OP HN.NP OK ON HN NP NP.OP ON OP NP OP =cos ROE. cos EOF-sin HPN. sin EOF =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. To prove that sin (AB)=sin A. cos B-cos A. sin B, and that cos (A-B) cos A. cos B+sin A. sin B. 2004 K 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= = KN.ON PH.NP KN ON ON.OP = PH NP NP.OP ON OP NP OP =sin ROE. cos FOE-cos HPN. sin FOE + + OK. ON NH.NP OK ON NH NP = ON.OP NP.OP ΟΝ ΟΡ NP OP =cos ROE. cos FOE+ sin HPN. sin FOE cos A. cos B+ sin A. sin B. * 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 750=sin (45o +30°) =sin 45°. cos 30° + cos 45°. sin 300 (5) If sin A= and sin B=3, find a value for sin (A + B) and for cos (A-B). (6) If sin A=6 and sin B=1, find a value for sin (A-B) and for cos (A + B). 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. |