RP But, since the angles RPN and NOQ are equal, we have O Q 100. As particular cases of the preceding formulae, we have, by putting B equal to 45°, Similarly as in Art. 98 we may prove that 102. As further examples of the use of the formulae of the present chapter we shall find the general value of the angle which has a given sine, cosine or tangent. This has been already found in Arts. 82-84. Find the general value of all angles having a given sine. Let a be any angle having the given sine and ◊ any other angle having the same sine. We have then to find the most general value of 0 which satisfies the equation i.e. by and Ө = a + any odd multiple of π......(1), i.e. must = a + any even multiple of ......(2), (-1)" a+nπ, where n is any positive or negative integer. For when n is odd this expression agrees with (1), and when ʼn is even it agrees with (2). 103. Find the general value of all angles having the same cosine. The equation we have now to solve is i.e. by and by 0 = a + any multiple of 2π, 0 = a + any multiple of 2π. Both these sets of values are included in the solution → = 2nπ ± α, where n is any positive or negative integer. 104. Find the general value of all angles having the same tangent. The equation we have now to solve is so that the most general solution is 0 =nπ + a. CHAPTER VIII. THE TRIGONOMETRICAL RATIOS OF MULTIPLE AND SUBMULTIPLE ANGLES. 105. To find the trigonometrical ratios of an angle 2A in terms of those of the angle A. If in the formulae of Art. 88 we put B = A, we have sin 2A = sin A cos A + cos A sin A cos 2A=cos A cos A and also 2 sin A cos A, = (1 − sin2 A) — sin2 A 1 – 2 sin2 A, = = cos2 A − (1 − cos2 A) = 2 cos2 A − 1 ; Now the formulae of Art. 88 are true for all values of A and B; hence any formulae derived from them are true for all values of the angles. In particular the above formulae are true for all values of A. |