数学北师大版必修4课时天天提分练4单位圆与诱导公式

4单位圆与诱导公式时间：45分钟满分：80分班级________姓名________分数________一、选择题：(每小题5分，共5×6＝30分)1．sin585°的值为()A．－eq\f(\r(2),2)B.eq\f(\r(2),2)C．－eq\f(\r(3),2)D.eq\f(\r(3),2)答案：A2．如果△ABC的三角内角为A、B、C，则sineq\f(B＋C,2)＝()A．－coseq\f(A,2)B．sineq\f(A,2)C．－sineq\f(A,2)D．coseq\f(A,2)答案：D3．sin(π－2)－coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－2))化简的结果为()A．0B．－1C．2sin2D．－2sin2答案：A解析：原式＝sin2－sin2＝0，所以选A.4．sin2150°＋sin2135°＋2sin210°＋cos2225°的值为()A.eq\f(1,4)B.eq\f(3,4)C.eq\f(11,4)D.eq\f(9,4)答案：A5．已知f(sinx)＝cos3x，则f(cos10°)的值为()A．－eq\f(1,2)B.eq\f(1,2)C．－eq\f(\r(3),2)D.eq\f(\r(3),2)答案：A解析：f(cos10°)＝f(sin80°)＝cos240°＝cos(180°＋60°)＝－cos60°＝－eq\f(1,2).6．若sin(π＋α)＋cos(eq\f(π,2)＋α)＝－m，则cos(eq\f(3π,2)－α)＋2sin(2π－α)＝()A．－eq\f(2,3)mB.eq\f(2,3)mC．－eq\f(3,2)mD.eq\f(3,2)m答案：C解析：因为sin(π＋α)＋cos(eq\f(π,2)＋α)＝－sinα－sinα＝－m，所以sinα＝eq\f(m,2)，所以cos(eq\f(3π,2)－α)＋2sin(2π－α)＝－sinα－2sinα＝－3sinα＝－eq\f(3,2)m.二、填空题：(每小题5分，共5×3＝15分)7．若sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,6)－θ))＝eq\f(\r(3),3)，则sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(7π,6)－θ))＝________.答案：－eq\f(\r(3),3)8.eq\f(cos－585°,sin630°＋sin－690°)的值是________．答案：eq\r(2)9．设f(x)＝asin(πx＋α)＋bcos(πx＋β)＋2，其中a，b，α，β为非零常数．若f(2014)＝1，则f(2015)＝________.答案：3解析：f(2014)＝asin(2014π＋α)＋bcos(2014π＋β)＋2＝asinα＋bcosβ＋2＝1，∴asinα＋bcosβ＝－1.∴f(2015)＝asin(2015π＋α)＋bcos(2015π＋β)＋2＝－asinα－bcosβ＋2＝3.三、解答题：(共35分，11＋12＋12)10．求下列三角函数值．(1)cos945°；(2)sineq\f(35,6)π；(3)coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)＋\f(π,3)))；(4)sineq\b\lc\(\rc\)(\a\vs4\al\co1(－\f(100,3)π)).解析：(1)cos945°＝cos(2×360°＋225°)＝cos225°＝cos(180°＋45°)＝－cos45°＝－eq\f(\r(2),2).(2)sineq\f(35,6)π＝sineq\b\lc\(\rc\)(\a\vs4\al\co1(6π－\f(π,6)))＝－sineq\f(π,6)＝－eq\f(1,2).(3)coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)＋\f(π,3)))＝coseq\b\lc\(\rc\)(\a\vs4\al\co1(π＋\f(π,2)＋\f(π,3)))＝－coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋\f(π,3)))＝－eq\b\lc\(\rc\)(\a\vs4\al\co1(－sin\f(π,3)))＝eq\f(\r(3),2).(4)sineq\b\lc\(\rc\)(\a\vs4\al\co1(－\f(100,3)π))＝－sineq\b\lc\(\rc\)(\a\vs4\al\co1(32π＋\f(4π,3)))＝－sineq\f(4π,3)＝－sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,3)＋π))＝sineq\f(π,3)＝eq\f(\r(3),2).11．已知α是第四象限角，且f(α)＝eq\f(sinπ－αcos2π－α,cos\f(π,2)－αsin－π－αcos2π＋α).(1)若cos(α－eq\f(3π,2))＝eq\f(1,5)，求f(α)的值；(2)若α＝－1860°，求f(α)的值．解析：f(α)＝eq\f(sinπ－αcos2π－α,cos\f(π,2)－αsin－π－αcos2π＋α)＝eq\f(sinαcosα,－sinαsinπ＋αcosα)＝eq\f(1,sinα).(1)因为cos(α－eq\f(3π,2))＝eq\f(1,5)，所以cos(α－eq\f(3π,2)＋2π)＝eq\f(1,5)，所以cos(eq\f(π,2)＋α)＝eq\f(1,5)，所以sinα＝－eq\f(1,5)所以f(α)＝eq\f(1,sinα)＝－5.(2)当α＝－1860°时，f(α)＝eq\f(1,sinα)＝eq\f(1,sin－1860°)＝eq\f(1,－sin1860°)＝eq\f(1,－sin5×360°＋60°)＝eq\f(1,－sin60°)＝－eq\f(2\r(3),3).12．设f(n)＝coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(n,2)π＋\f(π,4)))(n∈N*)，求f(1)＋f(2)＋f(3)＋…＋f(2015)的值．解：∵f(1)＋f(2)＋f(3)＋f(4)＝coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋\f(π,4)))＋coseq\b\lc\(\rc\)(\a\vs4\al\co1(π＋\f(π,4)))＋coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)＋\f(π,4)))＋coseq\b\lc\(\rc\)(\a\vs4\al\co1(2π＋\f(π,4)))＝－sineq\f(π,4)－coseq\f(π,4)＋sineq\f(π,4)＋coseq\f(π,4)＝0.∴f(1)＋f(2)＋…＋f(2012)＝503[f(1)＋f(2)＋f(3)＋f(4)]＝0.∴f(1)＋f(2)＋…＋f(2015)＝f(2013)＋f(2014)＋f(2015)＝coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(2013,2)π＋\f(π,4)))＋coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(2014,2)π＋\f(π,4)))＋coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(2015,2)π＋\f(π,4)))＝coseq\b\lc\(\rc\)(\a\vs4\al\co1(1006π＋\f(π,2)＋\f(π,4)))＋coseq\b\lc\(\rc\)(\a\vs4\al\c