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1、-必修四两角和与差的正弦、余弦、正切公式(一)-第 14 页两角和与差的正弦、余弦、正切公式(一)学习目标知识点一两角和与差的余弦公式C():cos()cos cos sin sin .C():cos()cos cos sin sin .思考你能根据两角差的余弦公式推导出两角和的余弦公式吗?请试着推导一下答案(),cos()cos ,sin()sin ,cos()cos()cos cos()sin sin()cos cos sin sin .即cos()cos cos sin sin .知识点二两角和与差的正弦公式S():sin()sin cos cos sin .S():sin()sin c
2、os cos sin .思考比较cos()与sin()之间有何区别和联系?利用诱导公式五(或六)可以实现正弦和余弦的互化,根据这种联系,请你试着从差角的余弦公式出发,推导出用任意角,的正弦、余弦值表示sin()及sin()的公式答案sin()coscoscoscos sinsin sin cos cos sin .即sin()sin cos cos sin .从而,sin()sin()sin cos()cos sin()sin cos cos sin .题型一化简求值例1化简求值:(1)sin(x27)cos(18x)sin(63x)sin(x18);(2)(tan 10).解(1)原式sin
3、(x27)cos(18x)cos(x27)sin(x18)sin(x27)cos(18x)cos(x27)sin(18x)sin(x27)(18x)sin 45.(2)(tan 10)(tan 10tan 60)2.跟踪训练1(1)sin 14cos 16sin 76cos 74;(2)sin(54x)cos(36x)cos(54x)sin(36x)解(1)原式sin 14cos 16sin(9014)cos(9016)sin 14cos 16cos 14sin 16sin(1416)sin 30.(2)原式sin(54x)(36x)sin 901.题型二给值求值(角)例2已知,且cos(),
4、sin ,求.解,(0,)cos(),sin().,sin ,cos .sin sin()sin()cos cos()sin .又,.跟踪训练2已知,cos(),sin(),求sin 2的值解因为,所以0,.又cos(),sin(),所以sin() ,cos() .所以sin 2sin()()sin()cos()cos()sin()()().题型三三角函数式的化简或证明例3(1)若sin()cos cos()sin 0,则sin(2)sin(2)等于()A1 B1 C0 D1答案C解析因为sin()cos cos()sin sin()sin 0.所以sin(2)sin(2)sin cos 2c
5、os sin 2sin cos 2cos sin 22sin cos 20.(2)已知sin(2)3sin ,求证:tan()2tan .证明sin(2)3sin sin()3sin()sin()cos cos()sin 3sin()cos 3cos()sin 2sin()cos 4cos()sin tan()2tan .跟踪训练3证明:2cos().证明2cos().题型四辅助角公式例4将下列各式写成Asin(x)的形式:(1)sin xcos x;(2)sin(x)cos(x)解(1)sin xcos x2(sin xcos x)2(cos sin xsin cos x)2sin(x)(2
6、)原式sin(x)cos(x)sin sin(x)cos cos(x)cos(x)cos(x)sin(x)跟踪训练4化简:(1)(cos xsin x);(2)3sin x3cos x.解(1)(cos xsin x)(cos xsin x)2(cos cos xsin sin x)2cos(x)(2)3sin x3cos x6(sin xcos x)6(sin sin xcos cos x)6cos(x)1sin 75等于()A. B.C. D.2sin 69cos 99cos 69sin 99的值为()A. B C. D3函数f(x)sin xcos x(xR)的值域是 4已知锐角、满足s
7、in ,cos ,则 .5化简:sincoscossin.一、选择题1计算sin 43cos 13cos 43sin 13的结果等于()A. B. C. D.2sin 245sin 125sin 155sin 35的值是()A B C. D.3若锐角、满足cos ,cos(),则sin 的值是()A. B. C. D.4函数f(x)sin xcos(x)的值域为()A2,2 B,C1,1 D,5若函数f(x)(1tan x)cos x,0x,则f(x)的最大值为()A1 B2 C1 D26在ABC中,三内角分别是A、B、C,若sin C2cos Asin B,则ABC一定是()A直角三角形 B
8、正三角形C等腰三角形 D等腰直角三角形二、填空题7化简sincos的结果是 8已知sin(),sin(),则的值是 9已知sin cos 1,则cos() .10已知A(3,0),B(0,3),C(cos ,sin ),若1,则sin() .三、解答题11已知sin ,sin(),均为锐角,求.12已知cos ,sin(),且、.求:(1)cos(2)的值;(2)的值13已知sin,cos,且0,求cos()当堂检测答案1答案B解析sin 75sin(3045)sin 30cos 45cos 30sin 45.2答案B解析原式sin(6999)sin(30).3答案2,2解析f(x)22sin
9、.f(x)2,24答案解析,为锐角,sin ,cos ,cos ,sin .cos()cos cos sin sin .0,.5解原式sincossincossinsinsin cos cos sin .课时精练答案一、选择题1答案A2答案B解析原式sin 65sin 55sin 25sin 35cos 25cos 35sin 25sin 35cos(3525)cos 60.3答案C解析cos ,cos(),、,sin ,sin().sin sin()sin()cos cos()sin .4答案B解析f(x)sin xcos(x)sin xcos xcos sin xsin sin xcos
10、xsin x(sin xcos x)sin(x)(xR),f(x)的值域为,5答案B解析f(x)(1tan x)cos xcos xsin x2(cos xsin x)2sin(x),0x,x.f(x)max2.6答案C解析sin Csin(AB)sin Acos Bcos Asin B2cos Asin B,sin Acos Bcos Asin B0.即sin(AB)0,AB.二、填空题7答案cos 解析原式sin cos cos sin cos cos sin sin cos .8答案解析.9答案0解析因sin cos 1且1sin 1,1cos 1,故有或所以cos sin 0,所以cos()cos cos sin sin 0.10答案解析(cos 3,sin ),(cos ,sin 3),(cos 3)cos sin (sin 3)cos23cos sin23sin 13(sin cos )13(sin cos )13sin()1,sin().三、解答题11解为锐角,sin ,cos .0,0.所以sin ,cos(),cos(2)cos()cos cos()sin sin().(2)cos cos()cos cos()sin sin(),又因为,所以.13解0,0.又sin,cos,cos,sin.cos()sinsinsincoscossin.