SECTION II. ON THE TRIGONOMETRIC FUNCTIONS OF THE SUM AND 64. PROP. Sin (a + B) = sin a. cos ẞ + cos a. sin B. N B At the points A, B, in the straight line AB draw two other straight lines, making angles with AB equal to A degrees and B degrees respectively. Let the two lines be produced to meet in C. From C draw CN perpendicularly to AB. Let the angle ACB = C. But since the angles at N are right angles.. by (art. 30.) .. sin (A + B) Let a and ẞ be the arcs to radius unity corresponding to the sin A. cos B + cos A. sin B. angles A and B, and therefore by (art. 28) sin (a + B) = sin a. cos B+ cos a. sin B. The figure applies to the case in which A and B are positive and acute angles. If one of the angles be negative, it must be measured on the opposite side of AB. And whatever be the values of A and B, the angles at the base of the triangle will be equal either to A and B, or to their defects from 180°, or from some multiple of 180°, and by paying attention to the algebraic signs the proof applies to angles or arcs of any magnitude whatsoever. 70. PROP. Sin 2a = 2 sin a. cos a cos 2a = (cos a)2= (sin a)2 } For by making a = ẞ in the formulæ (art. 48.) sin (a + ß) = sin a. cos B+ cos a. sin ß. cos (a + B) 74. PROP. a=√ 2 Cos a= 2 sin a= π π 2 a π 4 π 2 cos 2 a). - = sin 2 α. a) == cos 2 a)} . α, 2 and therefore α. (1 + sin 2 a) = √ § (1 − sin 2 a) {(2+√(2+√(2+ . . . + √(2+2 cos 2" a)} {(2−√(2+√(2+ . . . + √(2+2 cos 2" a)} Firstly, cos a = √ (1 + cos 2 a) .. 2 cos a = √(2 + 2 cos 2 a) Similarly, = √ {(2 + √ (2 + 2 cos 22 a)}. \ 2 cos a = √ {(2 + √ (2 + √ (2 + 2 cos 23 a)}. .. by continual substitution 2 cos a= {(2 + √ (2 + √ (2 + . . . + √ (2+2 cos 2" a)}. |