（全國通用）2019屆高考數(shù)學(xué)二輪復(fù)習(xí) 板塊四 考前回扣 回扣3 三角函數(shù)、三角恒等變換與解三角形課件.ppt

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1、回扣3三角函數(shù)、三角恒等變換與解三角形,板塊四考前回扣,,,回歸教材,易錯(cuò)提醒,內(nèi)容索引,,回扣訓(xùn)練,回歸教材,1.三種三角函數(shù)的性質(zhì),2.函數(shù)yAsin(x)(0，A0)的圖象 (1)“五點(diǎn)法”作圖,(2)由三角函數(shù)的圖象確定解析式時(shí)，一般利用五點(diǎn)中的零點(diǎn)或最值點(diǎn)作為解題突破口.,4.三角函數(shù)恒等變換“四大策略” (1)常值代換：特別是“1”的代換，1sin2cos2tan 45等. (2)降次與升次：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.,易錯(cuò)提醒,1.利用同角三角函數(shù)的平方關(guān)系式求值時(shí)，不要忽視角的范圍，要先判斷函數(shù)值的符號. 2.在求三角函數(shù)的值域(

2、或最值)時(shí)，不要忽略x的取值范圍. 3.求函數(shù)f(x)Asin(x)的單調(diào)區(qū)間時(shí)，要注意A與的符號，當(dāng)<0時(shí)，需把的符號化為正值后求解. 4.三角函數(shù)圖象變換中，注意由ysin x的圖象變換得到y(tǒng)sin(x)的圖象時(shí)，平移量為 ，而不是. 5.在已知兩邊和其中一邊的對角利用正弦定理求解時(shí)，要注意檢驗(yàn)解是否滿足“大邊對大角”，避免增解.,回扣訓(xùn)練,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,解析化簡函數(shù)的解析式，A中，ycos 2x是最小正周期為的偶函數(shù).,

3、,答案,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,解析,解析根據(jù)余弦定理得a2b2c22bccos A，,,所以b2b20， 解得b1，或b2(舍去)，故選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,,解析設(shè)BC邊上的高AD交BC于點(diǎn)D，,1,2,3

4、,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

5、,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.函數(shù)f(x)Asin(x)(A，，為常數(shù)，A0，0，0<<)的部分圖象如圖所示，則 的值為________.,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,

6、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,解析由兩個(gè)三角函數(shù)圖象的對稱中心完全相同可知，兩函數(shù)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?yàn)閟in2B8sin 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

7、,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中運(yùn)用正弦定理，可得,解答,15.在ABC中，角A，B，C所對的邊分別為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的面積.,1,2,3,4,5,6,7,8,9,10,11,12,14,13,16,15,因?yàn)锳BC，,1,2,3,4,5,6,7,8,9,10,1

8、1,12,14,13,16,15,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)與向量n(2，b)共線， 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,