3.1.2两角和与差的正弦VIP免费

两角和与差的两角和与差的正弦、余弦公式的运用正弦、余弦公式的运用李伟峰（一）复习回顾1.两角和与差的正弦、余弦2.正余弦的二倍角公式________________________________值以及单调区间？最大值、最小值所对应中心，的周期，对称轴，对称函数xxysin.3sinsincoscossinsincoscossincoscossinsincoscossin2sin21ZkkkZkkkZkkxZkkxZkkZkkxT223,2222,22-.5,221-,221.40,.3,2.22.1单减区间：单增区间：，最小值，最大值对称中心：对称轴：周期：cossin222sincos1cos22（二）例题精析1.公式的正向运用——求值13553541312653354cos,0cos20，，解：1312sin,0sin020，，，，又-sinsincossincossin.2cos135cos53sin，求，都是锐角，，变式一、已知cos2cos分析：65164cos42sin4sin分析：10343cos3sin3cossin3sin分析：44sinsin42621tanx-1xsin2x2sin,471217534xcos2求，已知x能力提升：,04sin,2,354,471217xxx则可得解：由,544sinx,10274cos4sin4cos4sin44sinsinxxxx则,1024sin4sin4cos4cos44coscosxxxx则xxxxxxxxsincossincoscossin2cosxsinx-1sin2cossinx2tanx-1xsin2x2sin22又由7528-代入即可得（二）公式的逆向运用——辅助角公式1.练习例2、求函数的对称轴2360sin0224sin可得解：由正弦的和角公式030sinxy0030sincos30cossinxxykxxy0018090sin的对称轴为：又kxkx0000018060,1809030则Zkxk1806000，所以对称轴为：变式：xxyxxyxxyxxycossin2)4(cossin)3(cossin3)2(cos21sin231）（xxcos51sin52551sin,52cos令xsin56sinx6sin2x4sin2xxxcossincossin5可以整理为？归纳总结xbabxbaabacossin2222222222sin,cosbabbaa令xbasin22辅助角公式xxbacossincossin22tanx-1sinx2x2sin,471217534xcos2求，已知x能力提升：,04sin,2,354,471217xxx则可得解：由,544sinxxxxxxxsincossincoscossin2tanx-1xsin2x2sin2又由7528-代入即可得4cos4sin2sinxxx4cos4sin22cosxxx4cos4sin14cos22xxx.x21)cos(sinsin2值）求最小值及所对应的（）求最小正周期（已知函数xxxxf能力提升：xxxxfcossinsin2解：由xxxcossin2sin2212cos2sinxx由辅助角公式得142sin2x22T1则函数的最小正周期kx2242,212此时则函数的最小值为Zkkx,8即一、当堂检测61022答案：Zkk0,3答案：一、作业布置谢谢大家！