发布时间 : 星期一 文章2019届高考数学江苏版1轮复习练习:第3章 三角函数、解三角形 4 第4讲 分层演练直击高考 Word版含解析更新完毕开始阅读857b8590f02d2af90242a8956bec0975f565a475
π?1.函数y=sin xsin??2+x?的最小正周期是________.12π[解析] 因为y=sin xcos x=sin 2x,所以T==π.
22
[答案] π
1+cos 2α1
2. 若=,则tan 2α=________.
sin 2α2
1+cos 2α2cos2αcos α1
[解析] 因为===,sin 2α2sin αcos αsin α2
2tan α44
所以tan α=2,所以tan 2α===-.
31-tan2α1-4
4
[答案] -
3
3.化简2+cos 2-sin21的结果是________.[解析]2+cos 2-sin21=1+cos 2+1-sin21
=2cos21+cos21=3cos 1.
[答案]3cos 1
π
4.已知△ABC中,AB=2,C=,则△ABC的周长为________.
3
a243b243?2π
[解析] 设三边分别为a,b,c,则=,a=sin A,=,b=sin?3-A??,sin Asin C32πsin C3??sin?3-A?
4343?2π
△ABC的周长l=sin A+sin?3-A??+233
π
A+?+2.=23sin A+2cos A+2=4sin??6?
πA+?+2[答案] 4sin??6?5.函数y=3cos 4x+sin 4x的最小正周期为________.
[解析]y=3cos 4x+sin 4x=2?
31?cos 4x+sin 4x2?2?πππ
coscos 4x+sinsin 4x?=2cos?4x-?,=2?66??6??
2ππ
故T==.42
π[答案]
2
1
sin235°-26.=________.
sin 20°
1
sin235°-2
2sin235°-1-cos 70°-sin 20°1
[解析]====-.
sin 20°2sin 20°2sin 20°2sin 20°2
1
[答案] -
2
ππ?7.函数f(x)=sin2x+3sin xcos x在区间??4,2?上的最大值是________.
1-cos 2xπ13sin 2x311
2x-?+,当+=sin 2x-cos 2x+=sin?6?2?22222ππ?π?π5π?ππ13
,,x∈?时,2x-∈,所以当2x-=时,f(x)=1+=.max?42?6?36?6222
3
[答案]
2
[解析]f(x)=sin2x+3sin xcos x=
x2sin2-1
2π?8.若f(x)=2tan x-,则f??12?的值为________.xx
sincos22
x
1-2sin2
2π?2cos x244
[解析] 因为f(x)=2tan x+=2tan x+==,所以f?==8.?12?1sin xsin xcos xsin 2xπ
sin xsin26
[答案] 8
πsin3αcos3α
0,?,则9.设α∈?+的最小值为________.?2?cos αsin α
sin3αcos3αsin4α+cos4α
[解析]+==
cos αsin αsin αcos α
=
1
-2sin αcos α.
sin αcos α
(sin2α+cos2α)2-2sin2αcos2α
sin αcos α
1
令sin αcos α=t,则t=sin 2α.
2π10,?,所以t∈?0,?.因为α∈??2??2?11
0,?上是减函数,令g(t)=-2t,则g(t)在??2?t
1
所以当t=时,g(t)min=2-1=1.
2
[答案] 1
10.(2016·高考江苏卷)在锐角三角形ABC中,若sin A=2sin Bsin C,则tan Atan Btan
C的最小值是________.
[解析] 由sin A=sin(B+C)=2sin Bsin C得sin Bcos C+cos Bsin C=2sin Bsin C,两边同时除以cos Bcos C得tan B+tan C=2tan Btan C,令tan B+tan C=2tan Btan C=m,因为△ABC是锐角三角形,所以2tan Btan m1C>2tan B·tan C,则tan Btan C>1,m>2.又在三角形中有tan Atan Btan C=-tan(B+C)tan BtanC=-·121-m2
m24m==m-2++4≥2
m-2m-2
(m-2)·4
+4=8,当且仅当m-2=,即m=4时取得等号,m-2m-2
故tan Atan Btan C的最小值为8.
11.(1)化简
[答案] 8
4
4cos4x-2cos 2x-1
;ππ???tan??4+x?sin2?4-x?
[解] (1)原式=
(2)求值:4cos 50°-tan 40°.
(1+cos 2x)2-2cos 2x-1
π?π+xcos2?+x?tan??4??4?=
cos22xπ??π?sin??4+x?cos?4+x?=
2cos22x2cos22x
==2cos 2x.πcos 2x?sin??2+2x?sin 40°(2)原式=4sin 40°-
cos 40°
=
4sin 40°cos 40°-sin 40°
cos 40°
2sin 80°-sin 40°=
cos 40°
=
2cos(40°-30°)-sin 40°
cos 40°
2(cos 40°cos 30°+sin 40°sin 30°)-sin 40°=
cos 40°
=
3cos 40°
=3.
cos 40°
π??π-α?=+α·12.(2018·合肥模拟)已知cos?cos?6??3?
ππ?1
-,α∈??3,2?.4(1)求sin 2α的值;
1
的值.tan α
(2)求tan α-
π?π+α·cos?-α?[解] (1)因为cos??6??3?
π?π+α·sin?+α?=cos??6??6?π112α+?=-,=sin?3?2?4π12α+?=-.所以sin?3??2
ππ?π?4π?,因为α∈?,所以2α+∈π,3?,?32?3?
π32α+?=-,所以cos?3??2
ππ
2α+?-?所以sin 2α=sin??3?3???ππππ1
2α+?cos-cos?2α+?sin=.=sin?3?33?32??ππ??2π,π?,,(2)因为α∈?,所以2α∈?32??3?13
又由(1)知sin 2α=,所以cos 2α=-.
22
1sin αcos αsin2α-cos2α-2cos 2α
所以tan α-=-===-2×=23.
tan αcos αsin αsin αcos αsin 2α1
2
1
1.若tan θ+=4,则sin 2θ=________.
tan θ
-2
3
1+tan2θ1
[解析]法一:因为tan θ+==4,
tan θtan θ
=
所以4tan θ=1+tan2θ,所以sin 2θ=2sin θcos θ
2sin θcos θ2tan θ2tan θ1===.
sin2θ+cos2θ1+tan2θ4tanθ2
法二:因为tan θ+
1sin θcos θ1221=+==,所以4=,故sin 2θ=.tan θcos θsin θcos θsin θsin 2θsin 2θ2
1
[答案]
2
π4π
α+?=,则sin?2α+?的值为________.2.设α为锐角,若cos?12??6?5?
π4
α+?=,[解析] 因为α为锐角,cos??6?5π3π24
α+?=,sin 2?α+?=,所以sin??6?5?6?25
ππ7
α+?=,所以sin?2α+?cos 2?12??6?25?ππππα+?-?=sin?2?α+??cos -=sin?2???6????6?4?4
?α+π??sin π=172.cos?2??6??450
172
[答案]
50
3.(2018·研
南通调
∠
)如图,在△ABC中,AB=3,AC=2,BC=4,点D在边BC上,