Question

a) Find sin( 2 ) + cos(20) using the picture below. 0 12 b) Find the exact value of cos s(3) if sin(x) = 13 and x is in quadrant III.

Answer 1

с. Solution:(0) In A ABC, . By Pythagoras theorem, 5 2 B A 12 AC²= BC² + BA ? 5° +12 2 - 25+ 144 = 13. » AC =√169 siu o = 5 Jh 13 and cas o = b h alio 12 13. - = cos 20 Now 1-2 siueo 32 sim ²0 = 1- cos 20 » 2 sin = I-coso

2 ms BF *l@) = 1 - 12 {icos = 12 1 2 3 13 = 1 3) sin 2 lo 13 1 13 = L 26 2 sin) y sin = 2) T = 0.19 -(i) Also, cos 20 = 2 costo - 1 = 2 2(1902 - 1 = 2 * 144 = 0.70 — lii) 288-169 119 -1 169 169 169 (i) and (ii), we get: buoppo hay sin + cos20 = 0.19 +0.70 – 0.89 (6) given : sin x = -12 and ac is in quadront III. 13 Now, sun'x + cos2x = 1 a cas²x = 1 - sinex 2 = 1 .121 13 1-1-4) 70 YOO F g -1 + / I - 144 144 169 25 169 Thus, cos 2 = -5 13 (ir lies in quadrout TL)

dlso, cos2x 2 cosa - 1 >> CAN +1 = 2 co 2 29 1. S 2cos2 일 2 R: 2014 - -S 연 18 13 8 13 2cos2 a 오 CS CL 8 13 음 X 고 오 0 8 26 20.SS. Hence, com a -0.SS 2