三角函数式的化简和求值练习93

1、三角函数式的化简和求值练习一、选择题：本大题共12小题，每小题5分，共60分，在每小题给出的四个选项中，有且只有一项符合题目要求.1sin2cos3tan4的值( ) A小于0 B大于0 C等于0 D不存在2已知角的终边过点则的值为. . .0 . 或3设a=2cos28 EMBED Equation.3 1, b=(cos18 EMBED Equation.3 sin18 EMBED Equation.3 ), c=log, 则 ( ) A. abc B. bac C. bca D. cba4已知是第二象限角，则可化简为（ ）AB CD5若是第三象限角，且cos0，则是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的大小关系是( )Atan1 sin1 cos1 Btan1 cos1 sin1Ccos1 sin1 tan1 D sin1 cos1 tan17已知为第三象限角，则为值为 （ ）ABCD8若、终边关于y轴对称，则下列等式成立的是( )ABCD9若是锐角，且sin=，则cos(+)=( )A. B. C.- D.-10若，则的值为（）ABCD11已知且，则的取值范围是A. B. C. D. 12设pcoscos，qcos2 eq f(,2)，那么p、q的大小关系是( )Apq Bpq Cpq Dpq二、填空题：本大题共4小题，

3、每小题4分，共16分，把答案填在题中横线上.13已知A，B，C是ABC的三个内角，sinA，cosA是方程的两个实根.则_，_.14已知5cos2 +4cos2 =4cos，则cos2 +cos2 的取值范围是 .15已知，那么的值为_，的值为_.16计算= .三角函数式的化简和求值练习答题卡一、选择题（每小题5分，共60分）题号123456789101112答案二、填空题（每小题4分，共16分）13_ 14._ 15._ 16._三、解答题：本大题共6小题，共74分，解答应写出文字说明，证明过程或演算步骤.17（本小题满分12分）不查表求值:18（本小题满分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（本小题满分12分）且，20（本小题满分12分）一元二次方程的两个实数根为和.求的取值范围及其最小值.21（本小题满分12）不查表求sin220+cos280+cos20cos80的值.22（本小题满分14）已知函数的定义域为，对任意、都满足，当时， EMBED Equation.3 ，又对所有均成立.(1) 证明函数是增函数；(2) 求实数的取值范围.附加题：（本题满分10分）设通过三角形的重心任作一直线，则在此直线同一侧的两顶点与此直线的距离和等于第三顶点与此直线的距离。 三角函

5、数式的化简和求值练习答案一、每小题5分，共60分.1A 2A 3C 4B 5B 6A 7B 8A 9B10C 11A 12C二、每小题4分，共16分.13. 14. 15. 16. 三、解答题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=则 20 由方程有实根, 得, 所以的取值

6、范围为且; 由韦达定理, 代入和角公式, 得, 所以的取值范围为, 最小值为. 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：设x=sin220+cos280+sin20cos80y=cos220+sin280cos20sin80，则

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， 所以是奇函数. 当时，,由, 得，故，由此可知是增函数.(2) 由得，令，有，当且仅当时，等号成立，则.附加题：已知：ABC直线经过ABC的重心OAH、BF、CE垂直于，垂足分别为H、F、E，求证：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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11、(u$rZoWkThQeMbJ7G4D1z-w*t!qYmVjSgOdLaI6F3C0y)v%s#pXlUiRfNcK8H5E2A+x(u$rZnWkThPeMbJ7G4C1z-w&t!qYmVjRgOdL9I6F3B0y(v%s#oXlUiQfNbK8H5D2A+x*u$qZnWkShPeMaJ7G4C1z)w&t!pYmVjRgOcL9I6E3B0y(v%r#oXlTiQfNbK8G5D2A-x*u$qZnVkSh9I6F3B0y(v%s#oXlUiQfNcK8H5D2A+x*u$rZnWkShPeMaJ7G4C1z)w&t!pYmVjRgOcL9I6E3B0y(v%r#oXlTiQfNb

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13、%s#oXlUiQfNcK8H5D2A+x*u$rZnWkShPeMaJ7G4C1z-w&t!pYmVjRgOdL9I6E3B0y(v%s#oXlTiQfNbK8H5D2A-x*u$qZnWkShPdMaJ7F4C1z)w&s!pYmUjRgOcL9H6E3B+y(v%r#oXlTiQeNbK8G5D2A-x*t$qZnVkShPdMaI7F4C0z)w&s!pXmUjRfOcL9H6E2B+y(u%r#oWlThQeNbJ8G5D1A-w*t$qYnVkSgPdLaI7F3C0z)v&s!pXmUiRfOcK9H6E2B+x(u%rZoWlThQeMbJ8G4D1A-w*t!qYnVjSgP

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17、(u%rZoWkThQeMbJ8G4D1z-w*t!qYnVjSgPdLaI6F3C0y)v&s#pXlUiRfNcK9H5E2A+x(u$rZoWkThPeMF3C0z)v&s#pXmUiRfOcK9H5E2B+x(u%rZoWkThQeMbJ8G4D1A-w*t!qYnVjSgPdLaI6F3C0y)v&s#pXlUiRfNcK9H5E2A+x(u$rZoWkThPeMbJ7G4D1z-w&t!qYmVjSgOdL9I6F3B0y)v%s#pXlUiQfNcK8H5E2A+x*u$rZnWkThPeMaJ7G4C1z-w&t!pYmVjRgOdL9I6E3B0y(v%s#oXlTiQfNb

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