8.解:(1)?tan(?4??)?12tan,??4?tan?1?tan?4?tan?1312
?1?ta?n1?ta?n?1,?ta?n22??
2sin?cos??cos? (2)原式=
?1?2cos??12tan??12??56.2?2sin?cos??cos?2cos?22
9.解:原式=
(1?cos?)?sin?(1?cos?)?sin??(1?cos?)?sin?(1?cos?)?sin?
2cos2?2?2sin?2sin?2coscos??2?2sin2cos2sin2cos2?2?2sin?2sin?2coscos??2 2)=2sin2cos2?2?22?2?22)?2(cos(sin?2?sin?cos???2=2sincos?2?22?)(sin(cos?2?cos?sin?2?2?2?2)2
??2?sincos??2=?2sin1=?sin?22cos?2=?2sin2?2cos?2
=?2sin?
1?cos2x23232121210.解:F(x)??sin2x?
=sin2x?cos2x?1
=sin(2x?(1)T?2?2??
?6)?1
(2)F(x)的最大值为2,最小值为0 (3)令2k???2≤2x??6≤2k???2
∴k???6≤x≤k???3(k?Z) ?62∴F(x)单调递增区间为[k??2k??,k??
?3](k?Z)
?2≤2x??6≤2k??5?63?∴k???3≤x≤k??(k?Z)
?3∴F(x)单调递减区间为[k??
,k??5?6](k?Z)
全章检测题答案
一.选择题
1.D 2.A 3.C 4.C 5.D 6.B 注: 2题选项A改为
2325
15(??)cos??cos?(??)sin?? sin?(??)??]? sin[?(?)? sin?152
15
1?sin??? ?cos2??1?2sin51223??1?2(?)?
525
5题改为tanA?tanB?tanAtanB?1
二.填空题 7.解:sin2???7????2?x??sin??2x??cos2x?1?2sinx?
9?4??2?8.解:原式=
3?cos20?2?cos10?2?3?2cos10??12?cos10?22?2.
9.解y?sin=cos(2x?∴T??
?6?3)
cos2x?cos?3sin2x?sin2x?32cos2x?12sin2x
10.解:△ABC中,C=90°,A+B=90°,∴sinA?cosB ∴y?cosB?2sinB?1?sin =?sin222B?2sinB
2B?2sinB?1??(sinB?1)?2
∵0??B?90? ∴0?sinB?1
∴1??(sinB?1)2?2?2 ∴1?y?2即y?(1,2) 三.解答题 11.解:(1)∵tan?2?2
2tan?22∴tan??1?tan?2?2?21?4??43
∴tan(???4tan??tan)??4=
tan??11?tan??43?143761?tan?tan?=??17
46(?43431?)?1(2)
6sin??cos?3sin??2cos??6tan??13tan??2=3(?=)?2
12.解:(1)?m?(sinA,cosA),n?(1,?2) ?m?n?sinA?2coA=0 s ?tanA? .2 (2)由(1)得f(x)?cos2x?2sinx
?1?2sinx?22sxin)?2
??2(sinx?123 2 ?x?R,?sinx???1?, 1. 即函数f(x)的值域为??2,?.
2??321232?3?13.解:(1)F(x)?cos2?x?sin2?x??a
=sin(2?x??3)?32?a
由题意:2???6??3??2∴??12
32]
(2)由(1)知f(x)?sin(x?∵x?[?∴?12?3)?7?6?a
?3,5?6] ∴x??3?[0,≤sin(x??3)≤1 5?6∴F(x)在x?[??3,]上取最小值?12?32?a
∴?12?32?a=3
∴a?3?12
?214.解:设?DAB??,则0???
则AD?dcos?,BD?dsin?,?CDB??DAB?? SABCD?S?ABD?S?BCD = = = = = =
121212141414dcos??dsin??dcos?sin??2212dsin??dsin?
2212dsin?
2d(cos?sin??sin?) d(2cos?sin??2sin?) d(sin2??1?cos2?) d[2sin(2??)?1即??3?82222?4)?1]
14d(2?1).
2∴当sin(2??
?4时,四边形ABCD的面积最大,最大面积为
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