回扣3 三角函数、三角恒等变换与解三角形,板块四 考前回扣,回归教材,易错提醒,内容索引,回扣训练,回归教材,1.三种三角函数的性质,2.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.,4.三角函数恒等变换“四大策略” (1)常值代换：特别是“1”的代换，1sin2cos2tan 45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.,易错提醒,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ysin x的图象变换得到ysin(x)的图象时，平移量为 ，而不是. 5.在已知两边和其中一边的对角利用正弦定理求解时，要注意检验解是否满足“大边对大角”，避免增解.,回扣训练,解析,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,答案,答案,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,解析,解析 化简函数的解析式，A中，ycos 2x是最小正周期为的偶函数.,答案,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,解析,解析 根据余弦定理得a2b2c22bccos A，,所以b2b20， 解得b1，或b2(舍去)，故选A.,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,答案,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,解析,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,解析 设BC边上的高AD交BC于点D，,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,cos()cos ()2 cos()cos 2sin()sin 2,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,解析,答案,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,11.函数f(x)Asin(x)(A，为常数，A0，0，0)的部分图象如图所示，则 的值为_.,1,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,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，,解析,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,答案,解析 因为sin2B8sin Asin C，由正弦定理可知， b28ac，,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,答案,解析,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,3,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,所以sinBACsin(BADCAD),1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,在ABC中运用正弦定理，可得,解答,15.在ABC中，角A，B，C所对的边分别为a，b，c，已知cos C(cos A sin A)cos B0. (1)求角B的大小；,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,解 由已知得,解答,(2)若a2，b ，求ABC的面积.,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,因为ABC，,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,解答,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,解答,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,向量m(1，a)与向量n(2，b)共线， b2a0，即b2a. ,即a2b2ab3. 由得a1，b2.,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,