单元练习
21、（本小题满分12分） 已知函数f?x?满足下列关系式： （i）对于任意的x,y?R，恒有 2f?x?f?y??f?????x?y???2????f??x?y?； ?2?（ii）f??????1． ?2?求证：（1）f?0??0； （2）f?x?为奇函数；
（3）f?x?是以2?为周期的周期函数．
《三角恒等变换》单元测试题
3????,??412cos???sin??sin??????2?，∴5，5，又13，?是第三象限角，∴1、∵
5?3??12?4335?????????cos???????13?5??13?565 13，∴cos?????单元练习
sin??2、依题意，∵
5124?cos??cos?????????????13，∴13，又5，∴2，∴
sin??????312?4?5563sin?????????sin[???????]513?5?1365 5，∵sin??，因此有，
3?????2?x??2k???,k2???si?n?x???cx?osx??sin044?，∴coxs?sxi?n，即?4?2?3、∵，∴
???4si?n?x???4?5，又∵
??cox?s2??2??sxi?n?????24?????x??2s?ixn???4?，∴?cos4??3cox?s?2???2???5??524 251212sin?x?x?y??siny?????13得?13，又∵y是第四
4、由
cos?x?y?sinx?sin?x?y?cosx?cosy?象限角，∴
513，∵
tany1?cosy??22sinycosysiny22
2sin2y252?13??123?13 1?5因为
??f?x?1??sin?x?1??cos?x?1?22?cos?????????sin??x??cos??x??22??22??2x??sin?2x?f?x?，∴最小正周期是T?1
?5?、∵g??x???gx立，∴选C
fx?sin?x??f??x??f?x??，∴f??x?sin???x????，即得：成
g?x?2??g?x?，∴
f?x?为偶函数，又∵
f?x?2??fx??，即f?x?的周期为2，
??????w?F?s?sint?3cost?2sin?t???3?，∴w?2 6、∵功
?????b?2b?2co?s?2?s?in2?2?s?i?n4a5?26?、∵a?，，，因此，
单元练习
????a?b????cos???45??cosa,b????sin?45?????cos?90?45??????a?b??a,b????，∴
45?3?1??????y?3sin2x?cos2x?2?sin2x?cos2x??2sin2x??2sin2x??2?????2612??????，∵7、∵
????????y?2sin2?y?2sin2x?????x?????12向左平移得??12向右平移得12?选D
??5?cosx????x??x????413??422448、∵，∴，则，则式为
????5??????????sin??2x?2sin??x?cos??x?2???4??4??2sin???x?????4?????????2cos???x?cos??x?cos??x????4??4??4? xy?sin?3cos29、∵
x??x?2sin?x???x?????k???x?2k?????23?，令23223?k?Z?，
当k??1时，
5?3
?1??1?10
、
∵
cx??ocx??oxx2?x?si?n2sincos?nsxs?ix222????tanxxsxsin2x2c?os2sincos2??2，∴22222sxi?n1?txan42??x5t2an2
11、∵
11?127tan??tan?????????????1?1?31?????2?7?，∴
t??a??n???2???11?32?1?111tan???1??t2??an????0,??????37，，又∵，
单元练习
??????0,?4??，∴???2????0，∴2?????3?4
?x??5??f?x??6sin?????x??26??m?0对于66恒成立，即m?f?x?max?3 12、∵
13、∵
si?xn?y?????x?y?2?k?y?2?k?12，∴2，∴
???x，∴
si?ny?2?x?????co?s?x???2???s?i?kn?(?2y?s2?1xs?in3
???)i?n?y???2????yco?scosk??2?x??2???????????t?cos??x?y??cos??2x??22cos??x??3?4?，∴?2??4? 14、令
??2?2???2?t??5??2?1??5?2?22???????2?2???
y?15、∵
221?cosxx?tansinx2∴对称中心为?k?,0??k?Z?
5?????f?x??2sin??2x??2sin?2x?6?6??16、∵5?????2sin2x????12?，∴周期T??，①正???确；∵递减区间是2?2x?????5?3??,???62，解之为?63?，②错误；∵对称中心的横坐标
2x?5?k?5??k??x??6212，当k?1时，得③正确；应该是向右平移，④不正确．
tan17、解：由
?2?1tan?2?521?cos?1?cos?54????sin??0???sin?25，又2，，得sin?cos??∴
18、（1）∵
??41334?33?3sin?????????3525210 ?5，所以???f?x??a?bf?x???3sin?x?cos?x?cos2?x，∴
?5??1???1?11f?x??sin?2?x?cos2?x?3sin2?x??sin??2?x????626????2，22，即
??单元练习
2??∴
2?2?????1T?；
2k??（2）令
?2?2x?2????5??k??,k???2k??,k?Z?f?x??36???k?Z?62，解之在
????k??,k???63???k?Z?． 上递增；同理可求递减区间为???a?b1?co?s?1c?os?cos?1??????cos222?2cos?ab???0,??18?依题意：，又，则
???????cos???????0,???cos?2?sin??2?2?，∴12，同理?22?，因????,2??，所以2??2?????????????0,??2???1??2???2?2?，∴22，将?1、?2代入6有23，从而有
sin19
???82?6????????sin????sin????4?12??64?．
、
sin2??2cos2?1?tan?2cos2??tan??1??????2cos2??tan????1?tan?4????????cos???????2?4??1?cos2???2???2??????sin???????4???2????1?cos2??1???tan????4???2?2cos2??1????4?4tan???????4??2??2??2?????2?2sin??2???2?25??1??2??2?1?tan????1????4???2?
13?1?cosx1?cosx?f?x??sin2x???cos2x?2sinx?2?sinx20、
12cosx313?sin2x??cos2x?sin2x?cos2xsinx222 2
????sin?2x??3? ?
单元练习
0?x?（1）∵
??2，∴2?2x??3?4????x?3，即122时，f?x?为减函数，故f?x?的
??3?????,sinx2?????????k1223????2，∴x?k递减区间为；（2）∵
x??Z，或
?6?k??k??Z．
???2f2?0??f????2?21、（1）令x?y?0，???f???0?f?0??0?2?；
??????2ffy?fy?f?yf??????x??????1y?R?2?2，（2）令，，∵?2?，∴
?f?yf??y????，故
f?x?y?为奇函数；（3）令
?2，x?R，有x??2f?x??1?f???x??f??x?，即
f???x??f?x??2，y?x有
，
……①，再令
2???1?f?x??f???x??f???x?f???x??f?x?，即
f???x???f?x??f?x???f?t??f?2??t?f?x?令x???t，则x???2??t，所以，即是以2?为周期的周期
函数．