三角函數(shù)式的化簡和求值練習93

1、三角函數(shù)式的化簡和求值練習一、選擇題：本大題共12小題，每小題5分，共60分，在每小題給出的四個選項中，有且只有一項符合題目要求.1sin2cos3tan4的值( ) A小于0 B大于0 C等于0 D不存在2已知角的終邊過點則的值為. . .0 . 或3設a=2cos28 EMBED Equation.3 1, b=(cos18 EMBED Equation.3 sin18 EMBED Equation.3 ), c=log, 則 ( ) A. abc B. bac C. bca D. cba4已知是第二象限角，則可化簡為（ ）AB CD5若是第三象限角，且cos0，則是A.第一象限角 B.第

2、二象限角C.第三象限角 D.第四象限角6sin1,cos1,tan1的大小關系是( )Atan1 sin1 cos1 Btan1 cos1 sin1Ccos1 sin1 tan1 D sin1 cos1 tan17已知為第三象限角，則為值為 （ ）ABCD8若、終邊關于y軸對稱，則下列等式成立的是( )ABCD9若是銳角，且sin=，則cos(+)=( )A. B. C.- D.-10若，則的值為（）ABCD11已知且，則的取值范圍是A. B. C. D. 12設pcoscos，qcos2 eq f(,2)，那么p、q的大小關系是( )Apq Bpq Cpq Dpq二、填空題：本大題共4小題，

3、每小題4分，共16分，把答案填在題中橫線上.13已知A，B，C是ABC的三個內角，sinA，cosA是方程的兩個實根.則_，_.14已知5cos2 +4cos2 =4cos，則cos2 +cos2 的取值范圍是 .15已知，那么的值為_，的值為_.16計算= .三角函數(shù)式的化簡和求值練習答題卡一、選擇題（每小題5分，共60分）題號123456789101112答案二、填空題（每小題4分，共16分）13_ 14._ 15._ 16._三、解答題：本大題共6小題，共74分，解答應寫出文字說明，證明過程或演算步驟.17（本小題滿分12分）不查表求值:18（本小題滿分12分）已知 eq f(,2) ,

4、0 eq f(,2) ,tan= eq f(3,4) ,cos()= eq f(5,13) ,求sin的值.19（本小題滿分12分）且，20（本小題滿分12分）一元二次方程的兩個實數(shù)根為和.求的取值范圍及其最小值.21（本小題滿分12）不查表求sin220+cos280+cos20cos80的值.22（本小題滿分14）已知函數(shù)的定義域為，對任意、都滿足，當時， EMBED Equation.3 ，又對所有均成立.(1) 證明函數(shù)是增函數(shù)；(2) 求實數(shù)的取值范圍.附加題：（本題滿分10分）設通過三角形的重心任作一直線，則在此直線同一側的兩頂點與此直線的距離和等于第三頂點與此直線的距離。 三角函

5、數(shù)式的化簡和求值練習答案一、每小題5分，共60分.1A 2A 3C 4B 5B 6A 7B 8A 9B10C 11A 12C二、每小題4分，共16分.13. 14. 15. 16. 三、解答題17答案：218 eq f(,2) ,tan= eq f(3,4) ,sin= eq f(3,5) ,cos= eq f(4,5) ,又 eq f(,2) ,0 eq f(,2) , -0,cos()= eq f(5,13) ,sin()= eq f(12,13) . sin=sin()=sincos()cossin()= eq f(63,65) .19解：由得=則 20 由方程有實根, 得, 所以的取值

6、范圍為且; 由韋達定理, 代入和角公式, 得, 所以的取值范圍為, 最小值為. 21解法一：sin220+cos280+sin220cos80= (1cos40)+ (1+cos160)+ sin20cos80=1cos40+cos160+sin20cos(60+20)=1cos40+ (cos120cos40sin120sin40)+sin20(cos60cos20sin60sin20)=1cos40cos40sin40+sin40sin220=1cos40(1cos40)= 解法二：設x=sin220+cos280+sin20cos80y=cos220+sin280cos20sin80，則

7、x+y=1+1sin60=，xy=cos40+cos160+sin100=2sin100sin60+sin100=0 x=y=，即x=sin220+cos280+sin20cos80=.22解：(1) 令得，令得， 所以是奇函數(shù). 當時，,由, 得，故，由此可知是增函數(shù).(2) 由得，令，有，當且僅當時，等號成立，則.附加題：已知：ABC直線經(jīng)過ABC的重心OAH、BF、CE垂直于，垂足分別為H、F、E，求證：BF=CE+AHH A N O M EB C FcL9H6E3B+y(v%r#oWlTiQeNbK8G5D1A-x*t$qZnVkSgPdMaI7F4C0z)v&s!pXmUjRfOcK

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9、s!pYmUjRgOcL9H6E3B+y(v%r#oWlTiQeNbK8G5D2A-x*t$qZnVkShPdMaI7F4C0z)w&s!pXmUjRfOcL9H6E2B+y(u%r#oWlThQeNbJ8G5D1A-w*t$qYnVkSgPdLaI7F3C0z)v&s!pXmUiRfOcK9H6E2B+x(u%rZoWlThQeMbJ8G4D1A-w*t!qYnVjSgPdLaI6F3C0y)v&s#pXlUiRfNcK9H5E2A+x(u$rZoWkThPeMbJ7G4D1z-w*t!qYmVjSgOdLaI6F3B0y)v%s#pXlUiQfNcK8H5E2A+x*u$rZnWkThPe

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