In the same way we find from (3) and (5), 2 cos a cos ẞ = cos (a + B) + cos (a cos (a + B) + cos (a — ß). § 90°, EXERCISES. (12) then cos (a + B) =2 cos a cos ẞ sin 2a = — sin 26. 2. Prove that if a +B=180°, then sin (a - ẞ) = 2 sin a cos B = sin 26 = sin 2a. 44. Sum of sines and cosines. If in the four equations (11) By substituting these values in (11) and (12) and reversing the members of the equation, we find Dividing the first of this group of equations by the third, we get If in the last two equations of (13) we suppose y = 0, which makes cos y = 1, they become a pair of equations which frequently come into use. If we put ac, these equations become which may also be derived directly from (7'). (16) (16') Dividing the second of (16) and (16') by the first, we find It is often required to cos 2α 1+ cos 2a 45. The problem of dimidiation. express the sine or cosine of half an angle in terms of the sine or cosine of the entire angle. To effect this let us put in equations They then become (6) and (7) 2a = y. Solving the equation cos y = 1-2 sin' y, we obtain Solving the equation cos y = 2 cos' y-1, we obtain (18) (19) (20) we put cos y = √1-sin' by and square both members, obtaining Considering sin ay as an unknown quantity, this equation can be solved as a quadratic. Transposing the last term and adding we put sin y= √1- cosy, and then proceed as before. Solving with respect to cos y, we shall find cos y = √1 + cos y (22) We have now to study the different values which these expressions for sin ty and cos y may have in consequence of the double signs of the surds and the double signs under the radical sign. 180 Ay -ty Equations (19) and (20) show that if the cosine of an angle y is given, the sine and cosine of half that angle may have either of two opposite values. The geometrical explanation of this is that to cos y may correspond either of two angles, y or 360° — y. The halves of the general measure of y are ty and y + 180° (§ 10). The halves of the general measure of 360° — y are 180° – ly and (see diagram). 180+2 360°-y Therefore the half angle may have either of four values. But because sin (180° + y)= sin (— $y), and sin (180° — y) = sin 1y, the sines and cosines of these four angles will have only two different values. This result agrees with equations (19) and (20). 180° YE M X But suppose the sine of y to be given. The angle y may then have either of two supplementary values, y and 180°-y. The halves of the general measure of these angles will bey, 90° – by, 180° + ły, and 270°y. The sines and cosines of these four angles are all different. Therefore the algebraic expression for the sines ought to be susceptible of four values instead of two, as in the first case. (5) and (6), because they show that E Now this is true of equations. 1. We have already found sin 30°; from this find the sine, cosine, tangent, and cotangent of 15° and of 75°. 2. From the values of the six trigonometric functions for 45° ($30) find those for 2210. 3. From the values for 18° find those for 36°. 46. Miscellaneous relations. The following equations are of occasional use in the applications of trigonometry, and can all be derived from the formulæ of the last two chapters. Their derivations are therefore presented as an interesting exercise. 1. sin (45° |