数竞大佬jhc的三角函数复习题

班主任让数竞大佬jhc整理的三角函数复习题，我参与编辑完成。个别题目来自参考书。度盘pdf格式下载：复习题提取码419d，答案提取码5a12

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“单纯”的运算

本文由蒋浩川原创，由(MiserWeyte)使用(LaTeX)编辑，采用CC BY-SA 4.0协议发布。

一、公式

1、三角比

(sinalpha=dfrac{y}{r})
(cosalpha=dfrac{x}{r})
( analpha=dfrac{y}{x})

2、三角函数线

单位圆中：

(sinalpha=|HY|)
(cosalpha=|OY|)
( analpha=| JL|)

3.诱导公式xN

(sin(-alpha)=-sinalpha cos(-alpha)=cosalpha an(-alpha)=- analpha)
(sin(pi-alpha)=sinalpha cos(pi-alpha)=-cosalpha an(pi-alpha)=- analpha)
(sin(pi+alpha)=-sinalpha cos(pi+alpha)=-cosalpha an(pi+alpha)= analpha)
(sin(frac{pi}{2}+alpha)=cosalpha cos(frac{pi}{2}+alpha)=-sinalpha an(frac{pi}{2}+alpha)=-frac{1}{ analpha})
(sin(frac{pi}{2}-alpha)=cosalpha cos(frac{pi}{2}-alpha)=sinalpha an(frac{pi}{2}-alpha)=frac{1}{ analpha})

4.和差倍半

(sin(alpha+eta)=sinalphacoseta+cosalphasineta)
(cos(alpha+eta)=cosalphacoseta-sinalphasineta)
( an(alpha+eta)=frac{ analpha+ aneta}{1- analpha aneta})

(sin(alpha-eta)=sinalphacoseta-cosalphasineta)
(cos(alpha-eta)=cosalphacoseta+sinalphasineta)
( an(alpha-eta)=frac{ analpha- aneta}{1+ analpha aneta})

(sin2alpha=2sinalphacosalpha sin^2frac{alpha}{2}=frac{1-cosalpha}{2})
(cos2alpha=cos^2alpha-sin^2alpha cos^2frac{alpha}{2}=frac{1+cosalpha}{2})
( an2alpha=frac{2 analpha}{1- an^2alpha} an^2frac{alpha}{2}=frac{1-cosalpha}{1+cosalpha})

5.辅助角公式

(asinalpha+bcosalpha=sqrt{a^2+b^2}sin(alpha+varphi))
(varphi=arctanfrac{b}{a})

6.其他

(1=sin^2alpha+cos^2alpha)
(2cos^2alpha=1+cos2alpha)
(2sin^2alpha=1-cos2alpha)

二、练习

1.已知(anglealpha)终边上一点(P(-3m, 4m))，(m ot=0)，求(alpha)的三个三角比

2.设(alphain(2kpi, 2kpi+frac{pi}{2})(piin Z))，证明(sinalpha<alpha< analpha)以及(sinalpha+cosalpha>1)

设：

[f(x)=egin{cases} sinpi x, x<0\ f(x-1)+1, xge0 end{cases} g(x)=egin{cases} cospi x, x<frac{1}{2}\ g(x-1)+1, xgefrac{1}{2} end{cases} ]

求(f(frac{1}{3})+f(frac{3}{4})+g(frac{1}{4})+g(frac{5}{6}))

4.求(dfrac{cos(2pi-alpha)cdot an(-alpha-pi)cdot an(3pi-alpha)}{sin(pi-alpha)cdot an(pi-alpha)} (alpha ot=kpi+frac{pi}{2}))

5.设(kin Z)，求证(cos(kpi+alpha)=(-1)^kcosalpha)，(sin(kpi+alpha)=(-1)^ksinalpha)。

6.求(alpha)：((1)sinalpha=frac{1}{2} (2) analpha=frac{sqrt{3}}{3})

7.已知( analpha=2)。求：
((1)dfrac{sinalpha+3cosalpha}{3sinalpha-4cosalpha} (2)dfrac{sin^2alpha+8sinalphacosalpha-6cos^2alpha}{3sin^2alpha-4cos^2alpha})

((3)sin^2alpha-3sinalphacosalpha+4cos^2alpha-2)

8.求(cos(alpha+frac{5}{12}pi)cos(alpha+frac{pi}{6})+cos(frac{pi}{12}-alpha)cos(frac{pi}{3}-alpha))

9.已知(sin(alpha+frac{pi}{6})=frac{3}{5})，(alphain(frac{pi}{3}, frac{5}{6}pi))，求( an(alpha+frac{5}{12}))

10.若方程(2sin x+sqrt5cos x=frac{1}{k})有解，求(k)范围。

11.求(sin10^circcdotsin50^circcdotsin70^circ)

12.(sinalpha+sineta=)(frac{sqrt2}{2})，求(cosalpha+coseta)范围。

三、拓展公式

(sinalpha=dfrac{2 anfrac{alpha}{2}}{1+ an^2frac{alpha}{2}} cosalpha=dfrac{1- an^2frac{alpha}{2}}{1+ an^2frac{alpha}{2}} analpha=dfrac{2 anfrac{alpha}{2}}{1- an^2frac{alpha}{2}})

和差化积与积化和差略。

((sinalphapmcosalpha)^2=1pmsin2alpha)
((1+ analpha)(1+ aneta)=2Longleftrightarrowalpha+eta=kpi+frac{pi}{4},kin Z)
(sin(alpha+eta)cdotsin(alpha-eta)=sin^2alpha-sin^2eta=cos^2eta-cos^2alpha)
(cos(alpha+eta)cdotcos(alpha-eta)=cos^2alpha-sin^2eta)

四、综合与提升

1.<指对幂与三角>已知(alphain(0, frac{pi}{4}))，(etain(0, 1))，试比较：
(x=(sinalpha)^{log_etasinalpha})，(y=(cosalpha)^{log_etacosalpha})，(z=(sinalpha)^{log_etacosalpha})

2.<提升>设锐角( heta)使关于(x)的方程(x^2+4xcos heta+frac{1}{ an heta}=0)有重根，求( heta)。

3.<提升>求证(dfrac{cosalpha}{1+sinalpha}-dfrac{sinalpha}{1+cosalpha}=dfrac{2(cosalpha-sinalpha)}{1+cosalpha+sinalpha})

4.<提升>已知(alphain(-frac{pi}{2},frac{pi}{2}))且(alpha ot=0)，(etain(0, pi))且(eta ot=frac{pi}{2})，(sinalpha=sqrt2coseta)，( analpha aneta=sqrt3)，求(alpha)，(eta)

5.<提升>已知(dfrac{sin^4alpha}{cos^2eta}+dfrac{cos^4alpha}{sin^2eta}=1)且(alpha,etain(0,dfrac{pi}{2}))，求证(alpha+eta=dfrac{pi}{2})

6.<提升>求(cosfrac{2}{5}pi+cosfrac{4}{5}pi)

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“单纯”的运算 参考答案

练习答案

(1)当(m>0)时，(sinalpha=frac{4}{5})，(cosalpha=-frac{3}{5})，(sinalpha=-frac{4}{3})
(2)当(m<0)时，(sinalpha=-frac{4}{5})，(cosalpha=frac{3}{5})，(sinalpha=-frac{4}{3})
2.
如“三角函数线“图，(sinalpha=|HY|,alpha=overset{frown} {HL}, analpha=|JL|)
( hereforesinalpha<alpha< analpha)
( hereforesinalpha+cosalpha=|OY|+|HY|>|OH|=1)
3.
(f(frac{1}{3})=f(-frac{2}{3})+1=-frac{sqrt3}{2}+1, f(frac{3}{4})=f(-frac{1}{4})+1=-frac{sqrt2}{2}+1)
(g(frac{1}{4})=frac{sqrt2}{2}, g(frac{5}{6})=frac{sqrt3}{2}+1)
( herefore)所求(=3)
4.
原式(=dfrac{cosalphacdot(- analpha)cdot(- analpha)}{sinalphacdot analpha}=1)
5.
蒋浩川说这题显而易见，略QwQ
6.
(1)(alpha=2kpi+frac{pi}{6})或(2kpi+frac{5}{6}pi (kin Z)) (2)(alpha=kpi+frac{pi}{6} (kin Z))
7.
(1)原式(=frac{ analpha+3}{3 analpha-4}=frac{5}{2})(2)原式(=frac{ an^2alpha+8 analpha-6}{3 an^2alpha-4}=frac{7}{4})
(3)原式(=frac{ an^2alpha-3 analpha+2}{ an^2alpha+1}-2=-frac{8}{5})
8.
原式(=cos(alpha+frac{pi}{6})sin(frac{pi}{12}-alpha)+cos(frac{pi}{12}-alpha)sin(alpha+frac{pi}{6})=sinfrac{pi}{4}=frac{sqrt2}{2})
9.
( an(alpha+frac{5}{12}pi)=dfrac{1+ an(alpha+frac{pi}{6})}{1- an(alpha+frac{pi}{6})} ecausesin(alpha+frac{pi}{6})=frac{3}{5},alpha+frac{pi}{6}in(frac{pi}{2},pi))
( herefore an(alpha+frac{pi}{6})=-frac{3}{4} herefore)所求(=frac{1}{7})
10.
左式$=3sin(alpha+ heta), heta=arctanfrac{sqrt5}{2} $$ herefore(左式)in[-3,3]( ) herefore-3leqfrac{1}{k}leq3 herefore kin[-infty,-frac{1}{3}]cup[frac{1}{3},infty]( 11. 所求)=cos80^circcos40circcos20^{circ=dfrac{frac{1}{8}sin160}circ}{sin20^circ}=dfrac{1}{8}( 12. )(sinalpha+sineta)^{2+(cosalpha+coseta)}2=2+2cos(x-y)in[0,4]( )Rightarrowcosalpha+coseta=pmsqrt{2+cos(alpha-y)-frac{1}{2}}in[-frac{sqrt{14}}{2},frac{sqrt{14}}{2}]$

综合与提升答案

(ecause f(x)=log_bx)为减函数(,bin(0,1) 0sinalpha<cosalpha<1,alphain(0,frac{pi}{4}))
(Rightarrowlog_bsinalpha>log_bcosalpha>0)
(Rightarrow x<z,z<y herefore x<z<y)
2.
依题意(Delta=16cos^2 heta-4cot heta=0)且(cot heta ot=0)
( hereforeDelta=4cot heta(2sin2 heta-1)=0)
(Rightarrowsin2 heta=frac{1}{2}Rightarrow heta=frac{pi}{12})或(frac{5}{12}pi)
3.
(dfrac{cosalpha}{1+sinalpha}=dfrac{1-sinalpha}{cosalpha}=dfrac{cosalpha+1-sinalpha}{1+sinalpha+cosalpha})

(dfrac{sinalpha}{1+cosalpha}=dfrac{1-cosalpha}{sinalpha}=dfrac{sinalpha+1-cosalpha}{1+cosalpha+sinalpha})

(Rightarrow)所求证成立
4.
两式平方并作商，(Rightarrowcos^2alpha=frac{2}{3}sin^2eta)
又(ecausesin^2alpha=2cos^2eta)
两式相加(Rightarrow2cos^2eta+frac{2}{3}sin^2eta=1Rightarrowsin^2eta=frac{3}{4})
(ecauseetain(0, pi),eta ot=frac{pi}{2} hereforesineta=frac{sqrt3}{2},eta=frac{pi}{3})或(frac{2}{3}pi)

[ hereforeegin{cases} alpha=frac{pi}{4}\ eta=frac{pi}{3} end{cases} 或 egin{cases} alpha=-frac{pi}{4}\ eta=frac{2}{3}pi end{cases} ]

令(dfrac{sin^2alpha}{coseta}=sin heta dfrac{cos^2alpha}{sineta}=cos heta hetain(0, dfrac{pi}{2}))
( hereforesin hetacoseta+sinetacos heta=sin^2alpha+cos^2alpha=1)
(Rightarrowsin( heta+eta)=1Rightarrowsin heta=coseta hereforedfrac{sin^2alpha}{coseta}=sin heta=coseta)
( hereforesinalpha=coseta Q.E.D.)
6.
令(x=cosfrac{2}{5}pi+cosfrac{4}{5}pi,y=cosfrac{2}{5}pi-cosfrac{4}{5}pi)
(xcdot y=cos^2frac{2}{5}pi-cos^2frac{4}{5}pi=frac{1}{2}(1+cosfrac{4}{5}pi)-frac{1}{2}(1+cosfrac{5}{8}pi))
(=frac{1}{2}(cosfrac{4}{5}pi-cosfrac{8}{5}pi)=-frac{1}{2}y)
(Rightarrow x=-frac{1}{2})