2019届高考数学二轮复习中档大题满分练(一)三角函数与解三角形A组.doc

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1、1中档大题满分练 1.三角函数与解三角形(A 组)中档大题集训练,练就慧眼和规范,筑牢高考满分根基!1.已知函数 f(x)= cos xcos +sin2 - .(-2) (-6)12(1)求 f(x)的单调递增区间.(2)若 x ,f(x)= ,求 cos 2x 的值.【解析】(1)f(x)= cos xcos +sin2 -12= sin xcos x+ -1-(2-3)2 12= sin 2x- cos12= sin 2x-12(122+322)= sin 2x- cos 2x= sin .14 12令 2k- 2x- 2k+ (kZ),2 6 22k- 2x2k+ ,k- xk+ (k

2、Z).3 23 6 3所以,f(x)的单调递增区间为 ,kZ.(2)因为 f(x)= sin = ,122所以 sin = ,因为 x ,所以- 2x- ,0,4 6 63所以 cos = ,所以 cos 2x=cos(2-6)+6=cos -sin 12= - = - .122.在ABC 中,a,b,c 分别为角 A,B,C 所对的边,已知 = .12(1)求角 B 的大小.(2)若 a=1,b= ,求ABC 的面积.【解析】(1)由 = 及 sin A=sin(B+C),12得 2sin Bcos C=2sin(B+C) - sin C=2sin Bcos C + 2cos Bsin C - sin C,所以 2cos Bsin C=sin C,又因为在ABC 中,sin C0,所以 cos B= ,12因为 0B,所以角 B= .3(2)在ABC 中,由余弦定理得 b2=a2+c2-2accos B,即 7=1+c2-c, 所以 c2-c-6=0,解得 c=3,所以ABC 的面积 S= acsin B= .12

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