四川省各地2014届高三最新模拟试题分类汇编7：三角函数(3)

??,k??],k?Z 36???7?(Ⅱ)因为x?[0,],所以2x??[,],

26661?所以??sin(2x?)?1

26?1所以函数f(x)在[0,]上的取值范围是[?,1]

22所以函数f(x)的单调递增区间为[k??10、（乐山市第一中学2014届高三10月月考）某兴趣小组测量电视塔AE的高度H(单位：m），如示意图，垂直放置的标杆BC的高度h=4m，仰角∠ABE=?，∠ADE=?。

（1）该小组已经测得一组?、?的值，tan?=1.24，tan?=1.20，请据此算出H的值； （2）该小组分析若干测得的数据后，认为适当调整标杆到电视塔的距离d（单位：m），使?与?之差较大，可以提高测量精确度。若电视塔的实际高度为125m，试问d为多少时，?-?最大？

解：（1）由

HHH，同理：AB?，?tan?得AD?tan?ADtan?BD?h。 tan?∵

AD

－

AB=DB

，

故

得

HHh??tan?tan?tan?，解得：

H?htan?4?1.24??124。因此，算出的电视塔的高度H是124m。

tan??tan?1.24?1.20HHhH?h， , tan????dADDBdHH?h?tan??tan?hdhdd tan(???)???2?HH?hH(H?h)1?tan??tan?1??d?H(H?h)d?dddH(H?h)∵d?（当且仅当d?H(H?h)?125?121?555时，?2H(H?h)，d（2）由题设知d?AB，得tan??取等号），∴当d?555时，tan(???)最大。

∵0?????故所求的d是555m。

?2，则0??????2，∴当d?555时，?－?最大。

11、（泸州市2014届高三第一次教学质量诊断）

在△ABC中，角A、B、C的对边分别为a、b、c，设S为△ABC的面积，满足

4S?3(a2?b2?c2) ．

(Ⅰ)求角C的大小；

????????tanA2c(Ⅱ)若1?，且AB?BC??8，求c的值． ?tanBb1解：(Ⅰ) S?absinC，且a2?b2?c2?2abcosC. ····································································· ···· 2分

2因为4S?3(a2?b2?c2)，

1所以4?absinC?23abcosC，··········································································· ···· 3分

2所以tanC?3，···································································································· ···· 4分 因为0?C??， 所以C?（Ⅱ）由1?π； ·········································································································· ···· 6分 3tanA2c得： ?tanBbcosAsinB?sinAcosB2c， ·············································································· ···· 7分 ?cosAsinBbsinC2c即······························································································ ···· 8分 ?， ·cosAsinBb1又由正弦定理得cosA?， ·················································································· ···· 9分

2∴A?60?，

∴△ABC是等边三角形， ······················································································ ·· 10分 ????????∴AB?BC?c?c?cos120???8， ············································································ ·· 11分 所以c?4. ··············································································································· ·· 12分

12、（绵阳市高中2014届高三11月第一次诊断性考试）

已知函数

（I）求函数f（x）的定义域及最大值；

（II）求使f（x）≥0成立的x的取值集合。 解：（Ⅰ） cosx≠0知x≠kπ，k∈Z，

即函数f (x)的定义域为{x|x∈R，且x≠kπ，k∈Z}．………………………3分 又∵ f(x)?2sinxcosx(sinx?cosx)1?cos2x?2sin2x?2sinxcosx?2??sin2x

cosx2?1?(sin2x?cos2x) ?1?2sin(2x??4)，

∴ f(x)max?1?2． ……………………………………………………………8分

ππ2（II）由题意得1?2sin(2x?)≥0，即sin(2x?)≤，

2443ππ9π≤2x?≤2kπ?，k∈Z，

444π整理得kπ?≤x≤kπ?π，k∈Z．

4结合x≠kπ，k∈Z知满足f(x)≥0的x的取值集合为

解得2kπ?{x|kπ?π≤x

km）安通驾校拟围着一座山修建一条环形训练道路OASBCD，道路的平面图如图所示（单位：，已知曲线ASB为函数

的图象，且最高点为

S（1,2），折线段AOD为固定线路，其中AO＝3，OD＝4，折线段BCD为可变线路，但为保证驾驶安全，限定∠BCD＝120°。 （I）求

的值；

（II）应如何设计，才能使折线段道路BCD最长？

解：（I）由已知A=2，

且有2sin(??0??)?3，即sin??由|?|<

3， 2??得??．

32又∵ 最高点为(1，2)， ∴ 2sin(??．

63??∴ y?2sin(x?)．…………………………………………………………6分

63ππ（II）∵ B点的横坐标为3，代入函数解析式得yB?2sin(?3?)=1，

63∴ BD?12?(4?3)2?2．…………………………………………………8分 在△BCD中，设∠CBD=θ，则∠BDC=180o-120o-θ=60o-θ． 由正弦定理有

BDCDBC??，

sin120?sin?sin(60???)? 解得??)?2，?∴ CD?2626sin?，BC?sin(60???)， …………………………………9分 33∴ BC?CD?26[sin??sin(60???)] 3??2631[sin??cos??sin?] 32226?sin(??)． 33∴ 当且仅当???6时，折线段BCD最长，最长为

26千米．…………12分 3

14、（什邡中学高中2014届高三上学期第二次月考） 已知函数f(x)?3sin?xcos?x?cos2?x的周期为2?，其中??0． （Ⅰ）求?的值及函数f(x)的单调递增区间； （Ⅱ）在?ABC中，设内角A、B、C所对边的长分别为a、b、c，若a?3，c?2，3f(A)=，求b的值． 2

15、（资阳市2014届高三上学期第一次诊断性考试）

已知函数f(x)?2sin(x?)cosx?sinxcosx?3sin2x（x?R）.

3（Ⅰ）求f(x)在[0,?]内的单调递增区间；

（Ⅱ）在?ABC中，B为锐角，且f(B)?3，AC?43，D是BC边上一点，AB?AD，试求

?AD?DC的最大值．

13【解】（Ⅰ）f(x)?2(sinx?cosx)cosx?sinxcosx?3sin2x

22······························· 2分 ?2sinxcosx?3(cos2x?sin2x)?sin2x?3cos2x?2sin(2x?). ·3????5?由??2k??2x???2k?，得??k??x?························· 3分 ?k?（k?Z）. ·

2321212?5?5?取k?0，得??x?，又x?[0,?]，则x?[0,]； ·············································· 4分

12121211?17?11?取k?1，得，又x?[0,?]，则x?[············································· 5分 ?x?,?]. ·

1212125?11?∴f(x)在[0，?]上的单调递增区间是[0,]，[·············································· 6分 ,?]. ·

1212?3???2?（Ⅱ）由f(B)?3得sin(2B?)?.又0?B?，则??2B??，从而

322333. ······································································································· 8分

333由AB?AD知?ABD是正三角形，AB?AD?BD，∴AD?DC?BD?DC?BC， 2B?????，∴B??在?ABC中，由正弦定理，得43sin?3?BC，即BC?8sin?BAC.

sin?BAC32?，∴?sin?BAC?1，知43?BC?8.

23∵D是BC边上一点，∴当?BAC??3??BAC??2，C??6时，AD?CD取得最大值8. ···························································· 12分

【另】在?ACD中，由正弦定理，得

ADDC43，∴AD?8sinC， ??sinCsin(??C)sin2?3331??cosC?sinC) CD?8sin(?C)，则AD?DC?8sinC?8sin(?C)?8(sinC?223331?2????2?，∴0?C?，?C??， ?8(cosC?sinC)?8sin(C?)．∵?ADC?22333333[当C?

?3??2，即C??6时，AD?DC取得最大值8. ······················································· 12分

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