高考数学二轮复习回扣4三角函数与平面向量课件.ppt

回扣4三角函数与平面向量,考前回扣,基础回归,易错提醒,回归训练,1.准确记忆六组诱导公式对于“，kZ”的三角函数值与角的三角函数值的关系口诀：奇变偶不变，符号看象限.,2.三角函数恒等变换“四大策略”(1)常值代换：特别是“1”的代换，1sin2cos2tan45等.(2)降次与升次：正用二倍角公式升次，逆用二倍角公式降次.(3)弦、切互化：一般是切化弦.,3.三种三角函数的性质,4.函数yAsin(x)(0，A0)的图象(1)“五点法”作图,(2)由三角函数的图象确定解析式时，一般利用五点中的零点或最值点作为解题突破口.,(3)图象变换,8.平面向量的数量积(1)若a，b为非零向量，夹角为，则ab|a|b|cos.(2)设a(x1，y1)，b(x2，y2)，则abx1x2y1y2.9.两个非零向量平行、垂直的充要条件若a(x1，y1)，b(x2，y2)，则(1)abab(b0)x1y2x2y10.(2)abab0x1x2y1y20.,12.三角形“四心”向量形式的充要条件设O为ABC所在平面上一点，角A，B，C所对的边长分别为a，b，c，则,1.利用同角三角函数的平方关系式求值时，不要忽视角的范围，要先判断函数值的符号.2.在求三角函数的值域(或最值)时，不要忽略x的取值范围.3.求函数f(x)Asin(x)的单调区间时，要注意A与的符号，当0时，需把的符号化为正值后求解.4.三角函数图象变换中，注意由ysinx的图象变换得到ysin(x)时，平移量为，而不是.,5.在已知两边和其中一边的对角时，要注意检验解是否满足“大边对大角”，避免增解.6.要特别注意零向量带来的问题：0的模是0，方向任意，并不是没有方向；0与任意非零向量平行.7.ab0是a，b为锐角的必要不充分条件；ab0是a，b为钝角的必要不充分条件.,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,答案,解析,2.下列函数中，最小正周期为的偶函数是,解析化简函数的解析式，A中，ycos2x是最小正周期为的偶函数.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解得b1，故选A.,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析设BC边上的高AD交BC于点D，,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,sin()sin()2sin()cos2cos()sin2,cos()cos()2cos()cos2sin()sin2,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析如图，,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解析由两个三角函数图象的对称中心完全相同可知，两函数的周期相同，故2，,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,13.在ABC中，内角A，B，C的对边分别是a，b，c，角B为锐角，且sin2B8sinAsinC，则的取值范围为_.,解析因为sin2B8sinAsinC，由正弦定理可知，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,设ABC的三个内角A，B，C所对的边分别为a，b，c，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,所以b2RsinB，c2RsinC，,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,代入上式整理，得cosBcosCcosAmsinC，,解答,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,(1)求角B的大小；,解答,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,(2)若a2，b，求ABC的面积.,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解答,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,解答,1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,(2)设ABC的内角A，B，C所对的边分别为a，b，c，且c，f(C)2，若向量m(1，a)与向量n(2，b)