高考数学复习第五章平面向量高考专题突破二高考中的三角函数与平面向量问题市赛课公开课一等奖省名师优质课.pptx

剖析题型 提炼方法,实验解读,构建知识网络 强化答题语句,探究高考 明确考向,高考中三角函数与平面向量问题,高考专题突破二,1/69,考点自测,课时作业,题型分类深度剖析,内容索引,2/69,考点自测,3/69,1,2,3,4,5,解析,答案,4/69,1,2,3,4,5,5/69,1,2,4,5,解析,3,答案,6/69,1,2,4,5,3,解析,3.,在直角三角形,ABC,中，点,D,是斜边,AB,中点，点,P,为线段,CD,中点，,则,等于,A.2 B.4 C.5 D.10,答案,7/69,1,2,4,5,3,解析,将,ABC,各边均赋予向量，,8/69,1,2,4,5,3,4,2,6,10.,9/69,解析,答案,1,2,4,5,3,10/69,5.,若函数,y,A,sin(,x,),在一个周期内图象如图所表示，,M,，,N,分别是这段图象最高点和最低点，且,0(,O,为坐标原点,),，,则,A,_.,解析,1,2,4,5,3,答案,11/69,1,2,4,5,3,12/69,题型分类深度剖析,13/69,例,1,(,山东,),设,f,(,x,),sin(,x,)sin,x,(sin,x,cos,x,),2,.,(1),求,f,(,x,),单调递增区间；,题型一三角函数图象和性质,解答,14/69,15/69,所以,f,(,x,),单调递增区间是,16/69,(2),把,y,f,(,x,),图象上全部点横坐标伸长到原来,2,倍,(,纵坐标不变,),，,再把得到图象向左平移,个单位长度，得到函数,y,g,(,x,),图象，求,值,.,解答,17/69,把,y,f,(,x,),图象上全部点横坐标伸长到原来,2,倍,(,纵坐标不变,),，,18/69,三角函数图象与性质是高考考查重点，通常先将三角函数化为,y,A,sin(,x,),k,形式，然后将,t,x,视为一个整体，结合,y,sin,t,图象求解,.,思维升华,19/69,解答,跟踪训练,1,已知函数,f,(,x,),5sin,x,cos,x,cos,2,x,(,其中,x,R,),，求：,(1),函数,f,(,x,),最小正周期；,20/69,解答,(2),函数,f,(,x,),单调区间；,21/69,所以函数,f,(,x,),单调递增区间为,所以函数,f,(,x,),单调递减区间为,22/69,解答,(3),函数,f,(,x,),图象对称轴和对称中心,.,23/69,题型二解三角形,例,2,(,全国,),ABC,内角,A,，,B,，,C,对边分别为,a,，,b,，,c,，已知,sin(,A,C,),.,(1),求,cos,B,；,解答,24/69,故,sin,B,4(1,cos,B,).,上式两边平方，整理得,17cos,2,B,32cos,B,15,0,，,25/69,解答,(2),若,a,c,6,，,ABC,面积为,2,，求,b,.,26/69,由余弦定理及,a,c,6,，,得,b,2,a,2,c,2,2,ac,cos,B,(,a,c,),2,2,ac,(1,cos,B,),所以,b,2.,27/69,依据三角形中已知条件，选择正弦定理或余弦定理求解；在处理相关角范围问题时，要注意挖掘题目中隐含条件，对结果进行正确取舍,.,思维升华,28/69,解答,(1),求,sin,C,值；,所以由正弦定理得,29/69,(2),若,a,7,，求,ABC,面积,.,解答,由余弦定理,a,2,b,2,c,2,2,bc,cos,A,，得,解得,b,8,或,b,5(,舍去,).,30/69,题型三三角函数和平面向量综合应用,例,3,已知向量,a,，,b,(cos,x,，,1).,(1),当,a,b,时，求,cos,2,x,sin 2,x,值；,解答,解,因为,a,b,，,31/69,解答,32/69,解,f,(,x,),2(,a,b,),b,33/69,34/69,(1),向量是一个处理问题工具，是一个载体，通常是用向量数量积运算或性质转化成三角函数问题,.,(2),三角形中三角函数要结合正弦定理、余弦定理进行转化，注意角范围对变形过程影响,.,思维升华,35/69,跟踪训练,3,在,ABC,中，内角,A,，,B,，,C,对边分别为,a,，,b,，,c,，已知向量,m,(cos,A,2cos,C,2,c,a,),，,n,(cos,B,，,b,),平行,.,解答,36/69,解,由已知得,b,(cos,A,2cos,C,),(2,c,a,)cos,B,，,由正弦定理，可设,则,(cos,A,2cos,C,),k,sin,B,(2,k,sin,C,k,sin,A,)cos,B,，,即,(cos,A,2cos,C,)sin,B,(2sin,C,sin,A,)cos,B,，,化简可得,sin(,A,B,),2sin(,B,C,),，,又,A,B,C,，,37/69,(2),若,b,cos,C,c,cos,B,1,，,ABC,周长为,5,，求,b,长,.,解答,解,由余弦定理可知，,由周长,a,b,c,5,，得,b,2.,38/69,课时作业,39/69,基础保,分练,1,2,3,4,5,6,解答,(1),求函数,f,(,x,),值域；,40/69,1,2,3,4,5,6,所以函数,f,(,x,),值域为,3,1.,41/69,解答,1,2,3,4,5,6,42/69,解,由题设条件及三角函数图象和性质可知，,y,f,(,x,),周期为,，,1,2,3,4,5,6,所以函数,y,f,(,x,),单调递增区间为,43/69,解答,1,2,3,4,5,6,44/69,解答,1,2,3,4,5,6,45/69,1,2,3,4,5,6,46/69,1,2,3,4,5,6,47/69,(1),求函数,y,f,(,x,),图象对称轴方程；,解答,1,2,3,4,5,6,48/69,1,2,3,4,5,6,49/69,解答,1,2,3,4,5,6,50/69,(1),求函数,f,(,x,),最小正周期；,解答,1,2,3,4,5,6,51/69,1,2,3,4,5,6,所以函数,f,(,x,),最小正周期为,2.,52/69,解答,(2),若,A,为,ABC,内角，,f,(,A,),4,，,BC,3,，求,ABC,周长最大值,.,1,2,3,4,5,6,53/69,因为,BC,3,，,1,2,3,4,5,6,54/69,1,2,3,4,5,6,55/69,技能提升练,1,2,3,4,5,6,56/69,解答,1,2,3,4,5,6,(1),求,A,；,57/69,1,2,3,4,5,6,58/69,1,2,3,4,5,6,在,ABC,中，,0,A,180,，则,30,A,300),，,则在,ABD,中，,AD,2,AB,2,BD,2,2,AB,BD,cos,B,，,1,2,3,4,5,6,61/69,解得,x,1(,负值舍去,),，,所以,a,7,，,c,5,，,1,2,3,4,5,6,62/69,解答,拓展冲刺练,1,2,3,4,5,6,63/69,1,2,3,4,5,6,64/69,1,2,3,4,5,6,函数,y,sin,z,单调递增区间是,65/69,1,2,3,4,5,6,66/69,解答,1,2,3,4,5,6,67/69,1,2,3,4,5,6,68/69,即,a,2,b,2,ab,3.,由,解得,a,1,，,b,2.,1,2,3,4,5,6,因为,m,(1,，,sin,A,),与向量,n,(2,，,sin,B,),共线，,所以,sin,B,2sin,A,，,由正弦定理得，,b,2,a,.,69/69,