Question

(3) The following exercise is based on the half-angle formulas. Use the fact that sin(/6) = 1/2 to prove that tan(+/12) - 2-V3.

Answer 1

18 sen tan (8/2) = logo Here o = 1/ in %2 = 1/2 It losy Henee tan (19/2) - Siamo ( 5กรัก - Caso 2 Vi-simo costy) = V1 f = Ž ite assil-L lt - 2-83 - 2 2-13 0 tanth) = 2-13 (proved). 3 cose - + 4 sine - 5 - 1050+ sine , Leto take x are tome da tans Then cosse at & - cosa coso af sinasino =/ cos( x) = 1 = copo non 5* Ox= 10 + 2118 for integerk o = tan 4) +20k The solu? is a = tan (1 + 21k kis an integer 6 -10 6x210 x+h+2 (x-85 +2.) (x+372 M-8+2 xrtant A (2-13+2) 46(348313) 67 - 10 (A+B - A+ A+V38429 AB - -- The 3+2 (x-1362 T-B3+2) (x +13 + 2) = (x ++3+2) (1-3 +2)

LE) (3) 1o are operation fron) (16+) A 2 3 + compairing we get VA + 24 +13 B +23 = -10 7-3+2) A 113 +2) B =-10 by operating Ti 013+ to 113 11B 3423" B = 3 -IVE 60-106 0-10 are 40+1 (-B + 2)(x+ 3+2) - 3+1 13 3 2 this is the partial - function form. Egare of the line parallel to y = 3x-g in y = 3x-e a at ct passes through i 10=3.2th as an The e que of the required in n uired line's = 3x+4 26 Equr of the line. I to y = 3x-g 'n atata (ar m=3 m anz Ast passer wrough (2, 10), 10 = -1/2 12 ta 2232 The equin of the required line is y=32 - 343 x +13 +2 x- 13 +2