高一數(shù)學(xué)同步備好課之題型全歸納(人教A版必修第一冊)專題49誘導(dǎo)公式五和公式六(原卷版+解析)

專題49誘導(dǎo)公式五和公式六1．公式五(1)角eq\f(π,2)－α與角α的終邊關(guān)于直線y＝x對稱，如圖所示．(2)公式：sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－α))＝cosα，coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－α))＝sinα.2．公式六(1)公式五與公式六中角的聯(lián)系eq\f(π,2)＋α＝π－eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－α)).(2)公式：sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))＝cosα，coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))＝－sinα.3．誘導(dǎo)公式一～六中的角可歸納為k·eq\f(π,2)±α的形式，可概括為“奇變偶不變，符號看象限”．①“變”與“不變”是針對互余關(guān)系的函數(shù)而言的．②“奇”“偶”是對誘導(dǎo)公式k·eq\f(π,2)±α中的整數(shù)k來講的．③“象限”指k·eq\f(π,2)±α中，將α看成銳角時，k·eq\f(π,2)±α所在的象限，根據(jù)“一全正，二正弦，三正切，四余弦”的符號規(guī)律確定原函數(shù)值的符號．4．利用誘導(dǎo)公式五、六，結(jié)合誘導(dǎo)公式二，還可以推出如下公式：sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)－α))＝－cosα，coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)－α))＝－sinα，sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)＋α))＝－cosα，coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)＋α))＝sinα.題型一利用誘導(dǎo)公式化簡與求值1．下列與sinθ的值相等的是()A．sin(π＋θ)B．sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－θ))C．coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－θ)) D．coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋θ))2．化簡sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)＋α))＝________.3．下列各式中，不正確的是()A．sin(180°－α)＝sinαB．coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(180°＋α,2)))＝sineq\f(α,2)C．coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)－α))＝－sinαD．tan(－α)＝－tanα4．若sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋θ))<0，且coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－θ))>0，則θ是()A．第一象限角 B．第二象限角C．第三角限角 D．第四象限角5．化簡sin(π＋α)coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)＋α))＋sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))cos(π＋α)＝________.6．化簡：eq\f(sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－α))cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α)),cosπ＋α)－eq\f(sin2π－αcos\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－α)),sinπ－α).7．化簡：eq\f(sinθ－5πcos\b\lc\(\rc\)(\a\vs4\al\co1(－\f(π,2)－θ))cos8π－θ,sin\b\lc\(\rc\)(\a\vs4\al\co1(θ－\f(3π,2)))sin－θ－4π)＝()A．－sinθB．sinθC．cosθ D．－cosθ8．化簡：eq\f(sin2π＋αcosπ－αcos\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－α))cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(7π,2)－α)),cosπ－αsin3π－αsin－π＋αsin\b\lc\(\rc\)(\a\vs4\al\co1(\f(5π,2)＋α)))＝________.9．化簡：eq\f(cosα－π,sinπ－α)·sineq\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(π,2)))coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α)).10．eq\f(sin2π－α·cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,3)＋2α))cosπ－α,tanα－3πsin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(7π,6)－2α)))等于()A．－cosαB．cosαC．sinα D．－sinα11．化簡：eq\f(sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－α)),cos（π＋α）)＋eq\f(sin（π－α）cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α)),sin（π＋α）).12．已知sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(5π,2)＋α))＝eq\f(1,5)，那么cosα＝13．已知cosθ＝－eq\f(3,5)，則sineq\b\lc\(\rc\)(\a\vs4\al\co1(θ＋\f(π,2)))＝________.14．已知coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋φ))＝eq\f(\r(3),2)，且|φ|＜eq\f(π,2)，則tanφ＝________.15．如果cos(π＋A)＝－eq\f(1,2)，那么sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋A))＝16．已知coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))＝－eq\f(3,5)，且α是第二象限角，則sineq\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(3π,2)))的結(jié)果是17．若cos(α＋π)＝－eq\f(2,3)，則sin(－α－eq\f(3π,2))＝18．已知cosα＝eq\f(1,5)，且α為第四象限角，那么coseq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(π,2)))＝________.19．若sin(3π＋α)＝－eq\f(1,2)，則coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(7π,2)－α))等于20．已知cos(π＋α)＝－eq\f(1,2)，α為第一象限角，求coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))的值．21．已知α∈eq\b\lc\(\rc\)(\a\vs4\al\co1(0，\f(3π,2)))，coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)－α))＝eq\f(\r(,3),2)，則tan(2018π－α)＝22．已知sineq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(π,4)))＝eq\f(1,3)，則coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,4)－α))的值為23．已知sineq\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(π,4)))＝eq\f(1,3)，則coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,4)＋α))等于24．已知sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,3)－α))＝eq\f(1,2)，則coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,6)＋α))的值為________．25．已知sineq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(π,6)))＝eq\f(3,5)，則coseq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(2π,3)))的值為________．26．若sineq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(π,12)))＝eq\f(1,3)，則coseq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(7π,12)))＝________.27．已知α是第四象限角，且cos(5°＋α)＝eq\f(4,5)，則cos(α－85°)＝________.28．已知sin10°＝k，則cos620°的值為()A．k B．－kC．±k D．不確定29．已知cosα＝eq\f(1,3)，則sineq\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(π,2)))·coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)＋α))tan(π－α)＝________.30．已知cos31°＝m，則sin239°tan149°的值是()A.eq\f(1－m2,m)B.eq\r(1－m2)C．－eq\f(1－m2,m) D．－eq\r(1－m2)31．若sin(180°＋α)＋cos(90°＋α)＝－a，則cos(270°－α)＋2sin(360°－α)的值是32．化簡eq\f(sin400°sin－230°,cos850°tan－50°)的結(jié)果為________．33．若f(cosx)＝cos2x，則f(sin15°)的值為34．已知f(sinx)＝cos3x，則f(cos10°)的值為35．若f(sinx)＝3－cos2x，則f(cosx)＝()A．3－cos2xB．3－sin2xC．3＋cos2x D．3＋sin2x36．計算sin21°＋sin22°＋sin23°＋…＋sin289°＝37．在△ABC中，eq\r(3)sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－A))＝3sin(π－A)，且cosA＝－eq\r(3)cos(π－B)，則C＝________.題型二利用誘導(dǎo)公式證明恒等式1．求證：eq\f(tan2π－αcos\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)－α))cos6π－α,sin\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(3π,2)))cos\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(3π,2))))＝－tanα.2．求證：eq\f(cosπ－θ,cosθ\b\lc\[\rc\](\a\vs4\al\co1(sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)－θ))－1)))＋eq\f(cos2π－θ,cosπ＋θsin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋θ))－sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)＋θ)))＝eq\f(2,sin2θ).3．求證：eq\f(sinθ＋cosθ,sinθ－cosθ)＝eq\f(2sin\b\lc\(\rc\)(\a\vs4\al\co1(θ－\f(3π,2)))cos\b\lc\(\rc\)(\a\vs4\al\co1(θ＋\f(π,2)))－1,1－2sin2π＋θ).4．求證：eq\f(cos6π＋θsin－2π－θtan2π－θ,cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)＋θ))sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)＋θ)))＝－tanθ.5．求證：eq\f(cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(5π,2)＋x)),sin\b\lc\(\rc\)(\a\vs4\al\co1(x－\f(5π,2)))tan6π－x)＝－1.6．求證：eq\f(2sin\b\lc\(\rc\)(\a\vs4\al\co1(θ－\f(3π,2)))cos\b\lc\(\rc\)(\a\vs4\al\co1(θ＋\f(π,2)))－1,1－2sin2θ)＝eq\f(tan9π＋θ＋1,tanπ＋θ－1).題型三誘導(dǎo)公式的綜合應(yīng)用1．已知銳角α終邊上一點P的坐標是(2sin2，－2cos2)，則α等于________．2．已知f(α)＝eq\f(cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)－α)),cos－π－αtanπ－α)，則feq\b\lc\(\rc\)(\a\vs4\al\co1(－\f(25,3)π))的值為________．3．已知coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,6)－α))＝eq\f(1,3)，求coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(5,6)π＋α))·sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(2π,3)－α))的值．4．已知cos(15°＋α)＝eq\f(3,5)，α為銳角，求eq\f(tan435°－α＋sinα－165°,cos195°＋α·sin105°＋α)的值．5．已知角α的終邊經(jīng)過點Peq\b\lc\(\rc\)(\a\vs4\al\co1(\f(4,5)，－\f(3,5))).(1)求sinα的值；(2)求eq\f(sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－α))tanα－π,sinα＋πcos3π－α)的值．6．已知tanθ＝2，求eq\f(sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋θ))－cosπ－θ,sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－θ))－sinπ－θ)的值．7．已知tan(3π＋α)＝2，則eq\f(sinα－3π＋cosπ－α＋sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－α))－2cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α)),－sin－α＋cosπ＋α)＝________.8．已知eq\f(sinθ＋cosθ,sinθ－cosθ)＝2，則sin(θ－5π)sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3,2)π－θ))＝________.9．已知cosα＝－eq\f(4,5)，且α為第三象限角．求f(α)＝eq\f(tanπ－α·sinπ－α·sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－α)),cosπ＋α)的值．10．已知coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))＝2sineq\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(π,2)))，則eq\f(sin（π－α）＋cos（π＋α）,5cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(5π,2)－α))＋3sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(7π,2)－α)))＝________．11．已知sin(α－3π)＝2cos(α－4π)，求eq\f(sin（π－α）＋5cos（2π－α）,2sin（\f(3π,2)－α）－sin（－α）)的值．12．已知角θ的頂點在坐標原點，始邊與x軸正半軸重合，終邊在直線3x－y＝0上，則eq\f(sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)＋θ))＋2cosπ－θ,sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－θ))－sinπ－θ)＝________.13．已知cos(75°＋α)＝eq\f(1,3)，則sin(α－15°)＋cos(105°－α)的值是14．已知α，β∈(0，eq\f(π,2))，且α，β的終邊關(guān)于直線y＝x對稱，若sinα＝eq\f(3,5)，則sinβ＝15．已知角α的終邊在第二象限，且與單位圓交于點P(a，eq\f(3,5))，求eq\f(sin（\f(π,2)＋α）＋2sin（\f(π,2)－α）,2cos（\f(3π,2)－α）)的值．16．已知f(α)＝eq\f(tanπ－αcos2π－αsin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α)),cos－α－π).(1)化簡f(α)；(2)若feq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－α))＝－eq\f(3,5)，且α是第二象限角，求tanα.17．已知f(α)＝eq\f(sinπ－αcos2π－αcos\b\lc\(\rc\)(\a\vs4\al\co1(－α＋\f(3π,2))),cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－α))sin－π－α).(1)化簡f(α)；(2)若α為第三象限角，且coseq\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(3π,2)))＝eq\f(1,5)，求f(α)的值；(3)若α＝－eq\f(31π,3)，求f(α)的值．18．已知sinα是方程5x2－7x－6＝0的根，α是第三象限角，求eq\f(sin\b\lc\(\rc\)(\a\vs4\al\co1(－α－\f(3,2)π))cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(3,2)π－α)),cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－α))sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α)))·tan2(π－α)的值．19．若sinα＝eq\f(\r(5),5)，求eq\f(cos3π－α,sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))\b\lc\[\rc\](\a\vs4\al\co1(sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(7π,2)＋α))－1)))＋eq\f(sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(5π,2)－α)),cos3π＋αsin\b\lc\(\rc\)(\a\vs4\al\co1(\f(5π,2)＋α))－sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(7π,2)＋α)))的值．20．在△ABC中，sineq\f(A＋B－C,2)＝sineq\f(A－B＋C,2)，試判斷△ABC的形狀．21．已知sineq\b\lc\(\rc\)(\a\vs4\al\co1(－\f(π,2)－α))·coseq\b\lc\(\rc\)(\a\vs4\al\co1(－\f(5π,2)－α))＝eq\f(60,169)，且eq\f(π,4)＜α＜eq\f(π,2)，求sinα與cosα的值．22．已知sin(π－α)－cos(π＋α)＝eq\f(\r(,2),3)(eq\f(π,2)<α<π)，求下列各式的值．(1)sinα－cosα；(2)cos2(eq\f(π,2)＋α)－cos2(－α)．23．已知函數(shù)f(α)＝eq\f(sin\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(π,2)))cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)＋α))tan（2π－α）,tan（α＋π）sin（α＋π）).(1)化簡f(α)；(2)若f(α)·feq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(π,2)))＝－eq\f(1,8)，且eq\f(5π,4)≤α≤eq\f(3π,2)，求f(α)＋feq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(π,2)))的值；(3)若feq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(π,2)))＝2f(α)，求f(α)·feq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(π,2)))的值．24．是否存在角α，β，α∈(－eq\f(π,2)，eq\f(π,2))，β∈(0，π)，使等式sin(3π－α)＝eq\r(2)cos(eq\f(π,2)－β)，eq\r(3)cos(－α)＝－eq\r(2)cos(π＋β)同時成立？若存在，求出α，β的值；若不存在，請說明理由．25．已知f(cosx)＝cos17x.(1)求證：f(sinx)＝sin17x；(2)對于怎樣的整數(shù)n，能由f(sinx)＝sinnx推出f(cosx)＝cosnx?專題49誘導(dǎo)公式五和公式六1．公式五(1)角eq\f(π,2)－α與角α的終邊關(guān)于直線y＝x對稱，如圖所示．(2)公式：sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－α))＝cosα，coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－α))＝sinα.2．公式六(1)公式五與公式六中角的聯(lián)系eq\f(π,2)＋α＝π－eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－α)).(2)公式：sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))＝cosα，coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))＝－sinα.3．誘導(dǎo)公式一～六中的角可歸納為k·eq\f(π,2)±α的形式，可概括為“奇變偶不變，符號看象限”．①“變”與“不變”是針對互余關(guān)系的函數(shù)而言的．②“奇”“偶”是對誘導(dǎo)公式k·eq\f(π,2)±α中的整數(shù)k來講的．③“象限”指k·eq\f(π,2)±α中，將α看成銳角時，k·eq\f(π,2)±α所在的象限，根據(jù)“一全正，二正弦，三正切，四余弦”的符號規(guī)律確定原函數(shù)值的符號．4．利用誘導(dǎo)公式五、六，結(jié)合誘導(dǎo)公式二，還可以推出如下公式：sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)－α))＝－cosα，coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)－α))＝－sinα，sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)＋α))＝－cosα，coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)＋α))＝sinα.題型一利用誘導(dǎo)公式化簡與求值1．下列與sinθ的值相等的是()A．sin(π＋θ)B．sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－θ))C．coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－θ)) D．coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋θ))[解析]sin(π＋θ)＝－sinθ；sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－θ))＝cosθ；coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－θ))＝sinθ；coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋θ))＝－sinθ.2．化簡sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)＋α))＝________.[解析]sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)＋α))＝sineq\b\lc\(\rc\)(\a\vs4\al\co1(π＋\f(π,2)＋α))＝－sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))＝－cosα.3．下列各式中，不正確的是()A．sin(180°－α)＝sinαB．coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(180°＋α,2)))＝sineq\f(α,2)C．coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)－α))＝－sinαD．tan(－α)＝－tanα[解析]由誘導(dǎo)公式知A、D正確．coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3,2)π－α))＝coseq\b\lc\(\rc\)(\a\vs4\al\co1(π＋\f(π,2)－α))＝－coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－α))＝－sinα，故C正確．coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(180°＋α,2)))＝coseq\b\lc\(\rc\)(\a\vs4\al\co1(90°＋\f(α,2)))＝－sineq\f(α,2)，故B不正確．4．若sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋θ))<0，且coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－θ))>0，則θ是()A．第一象限角 B．第二象限角C．第三角限角 D．第四象限角[解析]由于sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋θ))＝cosθ<0，coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－θ))＝sinθ>0，所以角θ的終邊落在第二象限，故選B.5．化簡sin(π＋α)coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)＋α))＋sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))cos(π＋α)＝________.[解析]原式＝(－sinα)·sinα＋cosα·(－cosα)＝－sin2α－cos2α＝－1.6．化簡：eq\f(sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－α))cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α)),cosπ＋α)－eq\f(sin2π－αcos\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－α)),sinπ－α).[解析]原式＝eq\f(cosα－sinα,－cosα)－eq\f(sin－αsinα,sinα)＝sinα－(－sinα)＝2sinα.7．化簡：eq\f(sinθ－5πcos\b\lc\(\rc\)(\a\vs4\al\co1(－\f(π,2)－θ))cos8π－θ,sin\b\lc\(\rc\)(\a\vs4\al\co1(θ－\f(3π,2)))sin－θ－4π)＝()A．－sinθB．sinθC．cosθ D．－cosθ[解析]原式＝eq\f(sinθ－πcos\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋θ))cosθ,cosθsin－θ)＝eq\f(－sinθ－sinθcosθ,cosθ－sinθ)＝－sinθ.8．化簡：eq\f(sin2π＋αcosπ－αcos\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－α))cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(7π,2)－α)),cosπ－αsin3π－αsin－π＋αsin\b\lc\(\rc\)(\a\vs4\al\co1(\f(5π,2)＋α)))＝________.[解析]原式＝eq\f(sinα·－cosα·sinα·cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)－α)),－cosα·sinα·[－sinπ－α]sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α)))＝eq\f(sinα·－sinα,－sinα·cosα)＝tanα9．化簡：eq\f(cosα－π,sinπ－α)·sineq\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(π,2)))coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α)).[解析]原式＝eq\f(cos[－π－α],sinα)·sineq\b\lc\[\rc\](\a\vs4\al\co1(－\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－α))))(－sinα)＝eq\f(cosπ－α,sinα)·eq\b\lc\[\rc\](\a\vs4\al\co1(－sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－α))))(－sinα)＝eq\f(－cosα,sinα)·(－cosα)(－sinα)＝－cos2α.10．eq\f(sin2π－α·cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,3)＋2α))cosπ－α,tanα－3πsin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(7π,6)－2α)))等于()A．－cosαB．cosαC．sinα D．－sinα[解析]原式＝eq\f(sin－α·cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,3)＋2α))·－cosα,tanα·cosα·sin\b\lc\[\rc\](\a\vs4\al\co1(\f(3,2)π－\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,3)＋2α)))))＝eq\f(sinαcosα·cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,3)＋2α)),tanαcosα\b\lc\[\rc\](\a\vs4\al\co1(－cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,3)＋2α)))))＝－cosα.故選A.11．化簡：eq\f(sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－α)),cos（π＋α）)＋eq\f(sin（π－α）cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α)),sin（π＋α）).[解析]因為sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))＝cosα，coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－α))＝sinα，cos(π＋α)＝－cosα，sin(π－α)＝sinα，coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))＝－sinα，sin(π＋α)＝－sinα，所以原式＝eq\f(cosα·sinα,－cosα)＋eq\f(sinα·（－sinα）,－sinα)＝－sinα＋sinα＝0.12．已知sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(5π,2)＋α))＝eq\f(1,5)，那么cosα＝[解析]sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(5π,2)＋α))＝sineq\b\lc\(\rc\)(\a\vs4\al\co1(2π＋\f(π,2)＋α))＝sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))＝cosα＝eq\f(1,5).13．已知cosθ＝－eq\f(3,5)，則sineq\b\lc\(\rc\)(\a\vs4\al\co1(θ＋\f(π,2)))＝________.[解析]sineq\b\lc\(\rc\)(\a\vs4\al\co1(θ＋\f(π,2)))＝cosθ＝－eq\f(3,5).14．已知coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋φ))＝eq\f(\r(3),2)，且|φ|＜eq\f(π,2)，則tanφ＝________.[解析]coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋φ))＝－sinφ＝eq\f(\r(3),2)，sinφ＝－eq\f(\r(3),2)，又∵|φ|＜eq\f(π,2)，∴cosφ＝eq\f(1,2)，故tanφ＝－eq\r(3).15．如果cos(π＋A)＝－eq\f(1,2)，那么sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋A))＝[解析]∵cos(π＋A)＝－cosA＝－eq\f(1,2)，∴cosA＝eq\f(1,2)，∴sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋A))＝cosA＝eq\f(1,2)16．已知coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))＝－eq\f(3,5)，且α是第二象限角，則sineq\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(3π,2)))的結(jié)果是[解析]∵coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))＝－sinα＝－eq\f(3,5),∴sinα＝eq\f(3,5)，且α是第二象限角,∴cosα＝－eq\r(1－sin2α)＝－eq\f(4,5).而sineq\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(3π,2)))＝－sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)－α))＝－(－cosα)＝cosα＝－eq\f(4,5)17．若cos(α＋π)＝－eq\f(2,3)，則sin(－α－eq\f(3π,2))＝[解析]因為cos(α＋π)＝－cosα＝－eq\f(2,3)，所以cosα＝eq\f(2,3).所以sineq\b\lc\(\rc\)(\a\vs4\al\co1(－α－\f(3π,2)))＝cosα＝eq\f(2,3).18．已知cosα＝eq\f(1,5)，且α為第四象限角，那么coseq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(π,2)))＝________.[解析]因為cosα＝eq\f(1,5)，且α為第四象限角，所以sinα＝－eq\r(1－cos2α)＝－eq\f(2\r(6),5)，所以coseq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(π,2)))＝－sinα＝eq\f(2\r(6),5).19．若sin(3π＋α)＝－eq\f(1,2)，則coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(7π,2)－α))等于[解析]∵sin(3π＋α)＝－sinα＝－eq\f(1,2)，∴sinα＝eq\f(1,2).∴coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(7π,2)－α))＝coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)－α))＝－coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－α))＝－sinα＝－eq\f(1,2).20．已知cos(π＋α)＝－eq\f(1,2)，α為第一象限角，求coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))的值．[解析]因為cos(π＋α)＝－cosα＝－eq\f(1,2)，所以cosα＝eq\f(1,2)，又α為第一象限角．則coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))＝－sinα＝－eq\r(1－cos2α)＝－eq\r(1－\b\lc\(\rc\)(\a\vs4\al\co1(\f(1,2)))\s\up12(2))＝－eq\f(\r(3),2).21．已知α∈eq\b\lc\(\rc\)(\a\vs4\al\co1(0，\f(3π,2)))，coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)－α))＝eq\f(\r(,3),2)，則tan(2018π－α)＝[解析]由coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)－α))＝eq\f(\r(,3),2)得sinα＝－eq\f(\r(,3),2)，又0<α<eq\f(3π,2)，所以π<α<eq\f(3π,2)，所以cosα＝－eq\r(,1－\b\lc\(\rc\)(\a\vs4\al\co1(－\f(\r(,3),2)))\s\up12(2))＝－eq\f(1,2)，tanα＝eq\r(,3).因為tan(2018π－α)＝tan(－α)＝－tanα＝－eq\r(,3)22．已知sineq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(π,4)))＝eq\f(1,3)，則coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,4)－α))的值為[解析]∵coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,4)－α))＝coseq\b\lc\[\rc\](\a\vs4\al\co1(\f(π,2)－\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,4)＋α))))＝sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,4)＋α))＝eq\f(1,3).23．已知sineq\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(π,4)))＝eq\f(1,3)，則coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,4)＋α))等于[解析]coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,4)＋α))＝coseq\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(π,4)＋\f(π,2)))＝－sineq\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(π,4)))＝－eq\f(1,3).24．已知sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,3)－α))＝eq\f(1,2)，則coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,6)＋α))的值為________．[解析]coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,6)＋α))＝coseq\b\lc\[\rc\](\a\vs4\al\co1(\f(π,2)－\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,3)－α))))＝sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,3)－α))＝eq\f(1,2).25．已知sineq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(π,6)))＝eq\f(3,5)，則coseq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(2π,3)))的值為________．[解析]coseq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(2π,3)))＝coseq\b\lc\[\rc\](\a\vs4\al\co1(\f(π,2)＋\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(π,6)))))＝－sineq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(π,6)))＝－eq\f(3,5).26．若sineq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(π,12)))＝eq\f(1,3)，則coseq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(7π,12)))＝________.[解析]coseq\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(7π,12)))＝coseq\b\lc\[\rc\](\a\vs4\al\co1(\f(π,2)＋\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,12)＋α))))＝－sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,12)＋α))＝－eq\f(1,3).27．已知α是第四象限角，且cos(5°＋α)＝eq\f(4,5)，則cos(α－85°)＝________.[解析]因為α是第四象限角，且cos(5°＋α)＝eq\f(4,5)＞0，所以5°＋α是第四象限角，所以sin(5°＋α)＝－eq\r(1－cos25°＋α)＝－eq\f(3,5)，所以cos(α－85°)＝cos(5°＋α－90°)＝sin(5°＋α)＝－eq\f(3,5).28．已知sin10°＝k，則cos620°的值為()A．k B．－kC．±k D．不確定[解析]cos620°＝cos(360°＋260°)＝cos260°＝cos(270°－10°)＝－sin10°＝－k.29．已知cosα＝eq\f(1,3)，則sineq\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(π,2)))·coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)＋α))tan(π－α)＝________.[解析]sineq\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(π,2)))coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)＋α))tan(π－α)＝－cosαsinα(－tanα)＝sin2α＝1－cos2α＝1－eq\b\lc\(\rc\)(\a\vs4\al\co1(\f(1,3)))2＝eq\f(8,9).30．已知cos31°＝m，則sin239°tan149°的值是()A.eq\f(1－m2,m)B.eq\r(1－m2)C．－eq\f(1－m2,m) D．－eq\r(1－m2)[解析]sin239°tan149°＝sin(180°＋59°)·tan(180°－31°)＝－sin59°(－tan31°)＝－sin(90°－31°)·(－tan31°)＝－cos31°·(－tan31°)＝sin31°＝eq\r(1－cos231°)＝eq\r(1－m2).31．若sin(180°＋α)＋cos(90°＋α)＝－a，則cos(270°－α)＋2sin(360°－α)的值是[解析]由sin(180°＋α)＋cos(90°＋α)＝－a，得－sinα－sinα＝－a，即sinα＝eq\f(a,2)，cos(270°－α)＋2sin(360°－α)＝－sinα－2sinα＝－3sinα＝－eq\f(3,2)a.32．化簡eq\f(sin400°sin－230°,cos850°tan－50°)的結(jié)果為________．[解析]eq\f(sin400°sin－230°,cos850°tan－50°)＝eq\f(sin360°＋40°[－sin180°＋50°],cos720°＋90°＋40°－tan50°)＝eq\f(sin40°sin50°,sin40°tan50°)＝eq\f(sin50°,\f(sin50°,cos50°))＝cos50°.33．若f(cosx)＝cos2x，則f(sin15°)的值為[解析]因為f(sin15°)＝f(cos75°)＝cos150°＝－eq\f(\r(3),2).34．已知f(sinx)＝cos3x，則f(cos10°)的值為[解析]f(cos10°)＝f(sin80°)＝cos240°＝cos(180°＋60°)＝－cos60°＝－eq\f(1,2).35．若f(sinx)＝3－cos2x，則f(cosx)＝()A．3－cos2xB．3－sin2xC．3＋cos2x D．3＋sin2x[解析]f(cosx)＝feq\b\lc\[\rc\](\a\vs4\al\co1(sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－x))))＝3－cos(π－2x)＝3＋cos2x，故選C.36．計算sin21°＋sin22°＋sin23°＋…＋sin289°＝[解析]原式＝(sin21°＋sin289°)＋(sin22°＋sin288°)＋…＋(sin244°＋sin246°)＋sin245°＝44＋eq\f(1,2)＝eq\f(89,2).37．在△ABC中，eq\r(3)sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－A))＝3sin(π－A)，且cosA＝－eq\r(3)cos(π－B)，則C＝________.[解析]由題意得eq\b\lc\{\rc\(\a\vs4\al\co1(\r(3)cosA＝3sinA，①,cosA＝\r(3)cosB，②))由①得tanA＝eq\f(\r(3),3)，故A＝eq\f(π,6).由②得cosB＝eq\f(cos\f(π,6),\r(3))＝eq\f(1,2)，故B＝eq\f(π,3).故C＝eq\f(π,2).題型二利用誘導(dǎo)公式證明恒等式1．求證：eq\f(tan2π－αcos\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)－α))cos6π－α,sin\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(3π,2)))cos\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(3π,2))))＝－tanα.[解析]左邊＝eq\f(tan2π－αcos\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)－α))cos6π－α,sin\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(3π,2)))cos\b\lc\(\rc\)(\a\vs4\al\co1(α＋\f(3π,2))))＝eq\f(tan－α－sinαcosα,－cosαsinα)＝eq\f(－tanαsinαcosα,cosαsinα)＝－tanα＝右邊，所以原等式成立．2．求證：eq\f(cosπ－θ,cosθ\b\lc\[\rc\](\a\vs4\al\co1(sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)－θ))－1)))＋eq\f(cos2π－θ,cosπ＋θsin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋θ))－sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)＋θ)))＝eq\f(2,sin2θ).[解析]左邊＝eq\f(－cosθ,cosθ－cosθ－1)＋eq\f(cosθ,－cosθcosθ＋cosθ)＝eq\f(1,1＋cosθ)＋eq\f(1,1－cosθ)＝eq\f(1－cosθ＋1＋cosθ,1＋cosθ1－cosθ)＝eq\f(2,1－cos2θ)＝eq\f(2,sin2θ)＝右邊．∴原式成立．3．求證：eq\f(sinθ＋cosθ,sinθ－cosθ)＝eq\f(2sin\b\lc\(\rc\)(\a\vs4\al\co1(θ－\f(3π,2)))cos\b\lc\(\rc\)(\a\vs4\al\co1(θ＋\f(π,2)))－1,1－2sin2π＋θ).[解析]右邊＝eq\f(－2sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)－θ))·－sinθ－1,1－2sin2θ)＝eq\f(2sin\b\lc\[\rc\](\a\vs4\al\co1(π＋\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－θ))))sinθ－1,1－2sin2θ)＝eq\f(－2sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－θ))sinθ－1,1－2sin2θ)＝eq\f(－2cosθsinθ－1,cos2θ＋sin2θ－2sin2θ)＝eq\f(sinθ＋cosθ2,sin2θ－cos2θ)＝eq\f(sinθ＋cosθ,sinθ－cosθ)＝左邊，所以原等式成立．4．求證：eq\f(cos6π＋θsin－2π－θtan2π－θ,cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)＋θ))sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)＋θ)))＝－tanθ.[解析]左邊＝eq\f(cosθsin－θtan－θ,cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋θ))sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋θ)))＝eq\f(cosθsinθtanθ,－sinθcosθ)＝－tanθ＝右邊，所以原等式成立．5．求證：eq\f(cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(5π,2)＋x)),sin\b\lc\(\rc\)(\a\vs4\al\co1(x－\f(5π,2)))tan6π－x)＝－1.[解析]因為eq\f(cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(5π,2)＋x)),sin\b\lc\(\rc\)(\a\vs4\al\co1(x－\f(5π,2)))tan6π－x)＝eq\f(cos\b\lc\(\rc\)(\a\vs4\al\co1(2π＋\f(π,2)＋x)),sin\b\lc\(\rc\)(\a\vs4\al\co1(x－\f(π,2)－2π))tan－x)＝eq\f(cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋x)),－sin\b\lc\(\rc\)(\a\vs4\al\co1(x－\f(π,2)))tanx)＝eq\f(－sinx,cosxtanx)＝－1＝右邊，所以原等式成立．6．求證：eq\f(2sin\b\lc\(\rc\)(\a\vs4\al\co1(θ－\f(3π,2)))cos\b\lc\(\rc\)(\a\vs4\al\co1(θ＋\f(π,2)))－1,1－2sin2θ)＝eq\f(tan9π＋θ＋1,tanπ＋θ－1).[解析]左邊＝eq\f(－2cosθ·sinθ－1,sin2θ＋cos2θ－2sin2θ)＝eq\f(－sinθ＋cosθ2,cosθ＋sinθcosθ－sinθ)＝eq\f(sinθ＋cosθ,sinθ－cosθ)，右邊＝eq\f(tan8π＋π＋θ＋1,tanπ＋θ－1)＝eq\f(tanπ＋θ＋1,tanπ＋θ－1)＝eq\f(tanθ＋1,tanθ－1)＝eq\f(\f(sinθ,cosθ)＋1,\f(sinθ,cosθ)－1)＝eq\f(sinθ＋cosθ,sinθ－cosθ)，所以等式成立．題型三誘導(dǎo)公式的綜合應(yīng)用1．已知銳角α終邊上一點P的坐標是(2sin2，－2cos2)，則α等于________．[解析]cosα＝eq\f(2sin2,\r(2sin22＋－2cos22))＝sin2，∴α＝2－eq\f(π,2).2．已知f(α)＝eq\f(cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)－α)),cos－π－αtanπ－α)，則feq\b\lc\(\rc\)(\a\vs4\al\co1(－\f(25,3)π))的值為________．[解析]∵f(α)＝eq\f(－sinα－cosα,－cosα－tanα)＝cosα，∴feq\b\lc\(\rc\)(\a\vs4\al\co1(－\f(25,3)π))＝coseq\b\lc\(\rc\)(\a\vs4\al\co1(－\f(25,3)π))＝coseq\f(25,3)π＝coseq\b\lc\(\rc\)(\a\vs4\al\co1(8π＋\f(π,3)))＝coseq\f(π,3)＝eq\f(1,2).3．已知coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,6)－α))＝eq\f(1,3)，求coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(5,6)π＋α))·sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(2π,3)－α))的值．[解析]coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(5,6)π＋α))·sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(2π,3)－α))＝coseq\b\lc\[\rc\](\a\vs4\al\co1(π－\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,6)－α))))·sineq\b\lc\[\rc\](\a\vs4\al\co1(π－\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,3)＋α))))＝－coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,6)－α))·sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,3)＋α))＝－coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,6)－α))·sineq\b\lc\[\rc\](\a\vs4\al\co1(\f(π,2)－\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,6)－α))))＝－coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,6)－α))·coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,6)－α))＝－eq\f(1,3)×eq\f(1,3)＝－eq\f(1,9).4．已知cos(15°＋α)＝eq\f(3,5)，α為銳角，求eq\f(tan435°－α＋sinα－165°,cos195°＋α·sin105°＋α)的值．[解析]原式＝eq\f(tan360°＋75°－α－sinα＋15°,cos180°＋15°＋α·sin[180°＋α－75°])＝eq\f(tan75°－α－sinα＋15°,－cos15°＋α·[－sinα－75°])＝－eq\f(1,cos15°＋α·sin15°＋α)＋eq\f(sinα＋15°,cos15°＋α·cos15°＋α).因為α為銳角，所以0°<α<90°，所以15°<α＋15°<105°.又cos(15°＋α)＝eq\f(3,5)，所以sin(15°＋α)＝eq\f(4,5)，故原式＝－eq\f(1,\f(3,5)×\f(4,5))＋eq\f(\f(4,5),\f(3,5)×\f(3,5))＝eq\f(5,36).5．已知角α的終邊經(jīng)過點Peq\b\lc\(\rc\)(\a\vs4\al\co1(\f(4,5)，－\f(3,5))).(1)求sinα的值；(2)求eq\f(sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－α))tanα－π,sinα＋πcos3π－α)的值．[解析](1)因為點Peq\b\lc\(\rc\)(\a\vs4\al\co1(\f(4,5)，－\f(3,5)))，所以|OP|＝1，sinα＝－eq\f(3,5).(2)eq\f(sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－α))tanα－π,sinα＋πcos3π－α)＝eq\f(cosαtanα,－sinα－cosα)＝eq\f(1,cosα)，由三角函數(shù)定義知cosα＝eq\f(4,5)，故所求式子的值為eq\f(5,4).6．已知tanθ＝2，求eq\f(sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋θ))－cosπ－θ,sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－θ))－sinπ－θ)的值．[解析]eq\f(sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋θ))－cosπ－θ,sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－θ))－sinπ－θ)＝eq\f(cosθ－－cosθ,cosθ－sinθ)＝eq\f(2cosθ,cosθ－sinθ)＝eq\f(2,1－tanθ)＝eq\f(2,1－2)＝－2.7．已知tan(3π＋α)＝2，則eq\f(sinα－3π＋cosπ－α＋sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－α))－2cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α)),－sin－α＋cosπ＋α)＝________.[解析]由tan(3π＋α)＝2，得tanα＝2，所以原式＝eq\f(－sinα＋－cosα＋cosα－2－sinα,sinα－cosα)＝eq\f(sinα,sinα－cosα)＝eq\f(tanα,tanα－1)＝eq\f(2,2－1)＝2.8．已知eq\f(sinθ＋cosθ,sinθ－cosθ)＝2，則sin(θ－5π)sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3,2)π－θ))＝________.[解析]∵eq\f(sinθ＋cosθ,sinθ－cosθ)＝2,sinθ＝3cosθ，∴tanθ＝3.sin(θ－5π)sineq\b\lc\(\rc\)(\a\vs4\al\co1(\f(3,2)π－θ))＝sinθcosθ＝eq\f(sinθcosθ,sin2θ＋cos2θ)＝eq\f(tanθ,tan2θ＋1)＝eq\f(3,10).9．已知cosα＝－eq\f(4,5)，且α為第三象限角．求f(α)＝eq\f(tanπ－α·sinπ－α·sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－α)),cosπ＋α)的值．[解析]因為cosα＝－eq\f(4,5)，且α為第三象限角，所以sinα＝－eq\r(1－cos2α)＝－eq\r(1－\b\lc\(\rc\)(\a\vs4\al\co1(－\f(4,5)))2)＝－eq\f(3,5).所以f(α)＝eq\f(－tanα·sinα·cosα,－cosα)＝tanαsinα＝eq\f(sinα,cosα)·sinα＝eq\f(－\f(3,5),－\f(4,5))×eq\b\lc\(\rc\)(\a\vs4\al\co1(－\f(3,5)))＝－eq\f(9,20).10．已知coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))＝2sineq\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(π,2)))，則eq\f(sin（π－α）＋cos（π＋α）,5cos\b\lc\(\rc\)(\a\vs4\al\co1(\f(5π,2)－α))＋3sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(7π,2)－α)))＝________．[解析]因為coseq\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)＋α))＝2sineq\b\lc\(\rc\)(\a\vs4\al\co1(α－\f(π,2)))，所以sinα＝2cosα.原式＝eq\f(sinα－cosα,5sinα－3cosα)＝eq\f(2cosα－cosα,10cosα－3cosα)＝eq\f(1,7).11．已知sin(α－3π)＝2cos(α－4π)，求eq\f(sin（π－α）＋5cos（2π－α）,2sin（\f(3π,2)－α）－sin（－α）)的值．[解析]因為sin(α－3π)＝2cos(α－4π)，所以－sin(3π－α)＝2cos(4π－α)，所以－sin(π－α)＝2cos(－α)，所以sinα＝－2cosα，且cosα≠0，所以原式＝eq\f(sinα＋5cosα,－2cosα＋sinα)＝eq\f(－2cosα＋5cosα,－2cosα－2cosα)＝eq\f(3cosα,－4cosα)＝－eq\f(3,4).12．已知角θ的頂點在坐標原點，始邊與x軸正半軸重合，終邊在直線3x－y＝0上，則eq\f(sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)＋θ))＋2cosπ－θ,sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－θ))－sinπ－θ)＝________.[解析]設(shè)θ的終邊上一點為P(x,3x)(x≠0)，則tanθ＝eq\f(y,x)＝eq\f(3x,x)＝3.因此eq\f(sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(3π,2)＋θ))＋2cosπ－θ,sin\b\lc\(\rc\)(\a\vs4\al\co1(\f(π,2)－θ))－sinπ－θ)＝eq\f(－cosθ－2cosθ,cosθ－sinθ)＝eq\f(－3cosθ,cosθ－sinθ)＝eq\f(－3,1－tanθ)＝eq\f(－3,1－3)＝eq\f(3,2).13．已知cos(75°＋α)＝eq\f(1,3)，則sin(α－15°)＋cos(10