1、三角恒等变形【课内四基达标】一、选择题1.若sinsin+coscos0,那么sincos+sincos的值等于( )A.-1B.0C. D.12.(tan22.5+cot22.5)的值是( )A.7B. C.7 D.log273.函数f(x)cos2x+cos(x+)+sin(x+)+3sin2x的最小值是( )A.0B.2C.D.34.已知log2ab,则(cos15sin15)b等于( )A.a2B. C. D.a5.若方程sec2x+2tanx-30有两根、,则cot(+)( )A.-cot2B.- C.- D.6.已知sin+cos (0,则cos2的值为( )A.B.- C. D.
2、- 7.已知cos78约等于0.20,那么sin66约等于( )A.0.92B.0.85C.0.88D.0.958.若0290180,a(sin)cos,b(cos)sin,c(cos)cos则( )A.acbB.abcC.bacD.cab9.化简的结果应为( )A.-tan20B.-cot20C.tan20D.cot2010.若实数x、y满足x2+y24,则的最小值为( )A.-2B.- C.2-2D.2+2二、填空题11.已知,(0,),且3sinsin(2+),4tan1-tan2,则+的值是 .12.若sinxcosx+sin2x-cos2x,则x .13. .14.若sin+sin,
3、则cos+cos的取值范围是 .三、解答题15.已知sin+sinsin225,cos+coscos225,求cos(-)及cos(+)的值.16.已知tan-tan2tan2tan,且、均不等于 (kZ),试求的值.17.求值:-sin10(cot5-tan5)18.A、B、C是ABC的三内角,已知,求cos的值.【力量素养提高】1.若f(x)cos2x+2k(1-cosx),xR,f(x)对一切xR都有f(x)0,求实数k的取值范围.2.已知cos(-),求cos.3.已知数列an的前n项和为Sn,求的值.【综合实践创新】1.函数y的最小正周期是( )A. B. C.D.22.设1、2、3
4、都是区间(0,)内的实数,且1、2、3是公差不为零的等差数列,问tan、tan、tan能否成为等比数列.为什么?3.如图,ABCD是半圆O的内接等腰梯形,其中AB为半圆直径,AB2,设COB,梯形的周长为l,求l的最大值.参考答案【课内四基达标】一、1.B 2.C 3.A 4.C 5.C 6.B 7.A 8.A 9.D 10.C二、11. 12. k,kZ 13. 14.-,由2+2得:2+2cos(-)1cos(-)-cos(+)2cos2 -1-1-1cos(+)-1016.解:tan-tan2tan2tantan原式+cos2sin2+cos22sincos+cos2-sin2+317.
5、解:原式-sin10(-)-sin10-2cos1018.解:sin(A-B)sinC-sinBsin(A-B)sin(A+B)-sinB2cosAsinBsinBcosAA60cos【力量素养提高】1.解:f(x)2cos2x-2kcosx+2k-1 令tcosx 则f(t)2t2-2kt+2k-1 t-1,14k2-8(2k-1)0k2-4k+202-k2+总之 k2+2.解:cos(-) sin(-)coscos(-)+cos(-)cos-sin(-)sin-3.解:anSn-Sn-1n2+2n-(n-1)2-2(n-1) (n2+2n-n2+2n-1-2n+2) (2n+1)an-an
6、-12n+1-2(n-1)-1 (2n+1-2n+2-1)是首项为,公差为的等差数列原式cos2(an-)+cos2an+cos2(an+)(-cosan+ sinan)2+cos2an+(-cosan- sinan)2cos2an+sin2an+cos2an cos2an+sin2an【综合实践创新】1.C2.解:(1)当2时,2 (1+3) tan2tan(1+3) 若tan2tantan2tantan +tan tan,tan,tan既成等差数列又成等比数列 tantantan 123与已知公差不为零冲突2时,tan,tan,tan不行能成等比数列(2)若2时, + - tancot tantan1 又tantan1 tan2tantan当2时,tan,tan,tan可以成等比数列3.解:COB COD-2BC2sin CD2sin2cosl2+4sin+2cosl2cos+4sin+2, (0,)l-4sin2+4sin+4 当sin时,