在线|二轮辅导[06][三角函数+解三角形02]

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思维导图

典例剖析

例1求值(cfrac{1+cos20^{circ}}{2sin20^{circ}}-sin10^{circ}(cfrac{1}{tan5^{circ}}-tan5^{circ}))

分析:原式(=cfrac{2cos^210^{circ}}{2cdot 2sin10^{circ}cos10^{circ}}-sin10^{circ}(cfrac{cos5^{circ}}{sin5^{circ}}-cfrac{sin5^{circ}}{cos5^{circ}}))

(=cfrac{cos10^{circ}}{2sin10^{circ}}-sin10^{circ}(cfrac{cos^25^{circ}-sin^25^{circ}}{sin5^{circ}cos5^{circ}}))

(=cfrac{cos10^{circ}}{2sin10^{circ}}-sin10^{circ}cfrac{2cos10^{circ}}{2sin5^{circ}cos5^{circ}}))

(=cfrac{cos10^{circ}}{2sin10^{circ}}-2cos10^{circ})

(==cfrac{cos10^{circ}}{2sin10^{circ}}-cfrac{2cos10^{circ}cdot 2sin10^{circ}}{2sin10^{circ}})

(=cfrac{cos10^{circ}-2sin20^{circ}}{2sin10^{circ}})

(=cfrac{cos10^{circ}-2sin(30^{circ}-10^{circ})}{2sin10^{circ}})

(=cfrac{cos10^{circ}-cos10^{circ}+2cdot cfrac{sqrt{3}}{2}sin10^{circ}}{2sin10^{circ}})

(=cfrac{sqrt{3}}{2})

例2求值(cfrac{cos10^{circ}-sqrt{3}cos(-100^{circ})}{sqrt{1-sin10^{circ}}})

分析:原式(=cfrac{cos10^{circ}-sqrt{3}cos(100^{circ})}{sqrt{1-sin10^{circ}}})

(=cfrac{cos10^{circ}+sqrt{3}sin10^{circ}}{sqrt{1-sin10^{circ}}})

(=cfrac{cos10^{circ}+sqrt{3}sin10^{circ}}{sqrt{(cos5^{circ}-sin5^{circ})^2}})

(=cfrac{cos10^{circ}+sqrt{3}sin10^{circ}}{(cos5^{circ}-sin5^{circ})^2})

(=cfrac{2sin(10^{circ}+30^{circ})}{-sqrt{2}sin(5^{circ}-45^{circ})})

(=cfrac{2sin40^{circ}}{sqrt{2}sin40^{circ}}=sqrt{2})

例3化简求值(cfrac{cos40^{circ}}{cos25^{circ}cdot sqrt{1-sin40^{circ}}})

分析:原式(=cfrac{cos40^{circ}}{cos25^{circ}cdot sqrt{(sin20^{circ}-cos20^{circ})^2}})

(=cfrac{cos40^{circ}}{cos25^{circ}cdot |sin20^{circ}-cos20^{circ}|})

(=cfrac{cos^220^{circ}-sin^220^{circ}}{cos25^{circ}(cos20^{circ}-sin20^{circ})})

(=cfrac{cos20^{circ}+sin20^{circ}}{cos25^{circ}})

(=cfrac{sqrt{2}sin(20^{circ}+45^{circ})}{cos25^{circ}})

(=cfrac{sqrt{2}sin65^{circ}}{cos25^{circ}}=sqrt{2}).

例4化简求值(cfrac{sqrt{3}tan12^{circ}-3}{(4cos^212^{circ}-2)sin12^{circ}})

分析:原式=(cfrac{sqrt{3}cfrac{sin12^{circ}}{cos12^{circ}}-3cfrac{cos12^{circ}}{cos12^{circ}}}{2(2cos^212^{circ}-1)sin12^{circ}})

(=cfrac{sqrt{3}cdot cfrac{sin12^{circ}-sqrt{3}cos12^{circ}}{cos12^{circ}}}{2cos24^{circ}sin12^{circ}})

(=cfrac{sqrt{3}cdot 2sin(12^{circ}-60^{circ})}{2cos24^{circ}sin12^{circ}cos12^{circ}})

(=cfrac{2sqrt{3}sin(-48^{circ})}{sin24^{circ}cos24^{circ}}=-4sqrt{3})

练-1化简求值(cfrac{sqrt{3}-tan12^{circ}}{(2cos^212^{circ}-1)sin12^{circ}}=8)

练-2化简求值(sin50^{circ}(1+sqrt{3}tan10^{circ})=1)

例5函数(f(x)=2cos(omega x+phi)(omega eq 0))对任意(x)都有(f(cfrac{pi}{4}+x)=f(cfrac{pi}{4}-x))成立,则(f(cfrac{pi}{4}))的值为【】

$A、2或0$ $B、-2或2$ $C、0$ $D、-2或0$

分析:由任意(x)都有(f(cfrac{pi}{4}+x)=f(cfrac{pi}{4}-x))成立,可知(x=cfrac{pi}{4})为函数的一条对称轴,

而正弦型或余弦型函数在对称轴处必然会取到最值,故(f(cfrac{pi}{4})=pm 2),选B。

解后反思:此题目如果不注意函数的性质,往往会想到求(omega)(phi),这样思路就跑偏了。

例6【2018云南玉溪一模】函数(f(x)=sqrt{3}sin2x+2cos^2x)的一条对称轴为直线【】

$A、x=cfrac{pi}{12}$ $B、x=cfrac{pi}{6}$ $C、x=cfrac{pi}{3}$ $D、x=cfrac{pi}{2}$

分析:(f(x)=2sin(2x+cfrac{pi}{6})+1)

法1:比较繁琐,令(2x+cfrac{pi}{6}=kpi+cfrac{pi}{2})(kin Z),则(x=cfrac{kpi}{2}+cfrac{pi}{6})(kin Z),即对称轴有无数条,

(k=0),得到其中的一条对称轴为(x=cfrac{pi}{6}),当(k)取其他的值时,都不能得到其他的选项,故选(B)

法2:比较简单,利用函数在对称轴处的函数值能取到最值,故只需验证即可,

比如,将(x=cfrac{pi}{12})代入(sin(2x+cfrac{pi}{6})),即(sincfrac{pi}{3}),并不能使得其取到最值(pm 1),故舍去(A)

(x=cfrac{pi}{6})代入(sin(2x+cfrac{pi}{6})),即(sincfrac{pi}{2}),能使得其取到最值(+1),故(B)必然满足;用同样的方法可以验证其余的选项错误;

例7【2018江西赣州5月适应性考试】若函数(f(x)=3cos(2x+cfrac{pi}{6})-a)([0,cfrac{pi}{2}])上有两个零点(x_1)(x_2),则(x_1+x_2)=【】

$A、cfrac{pi}{3}$ $B、cfrac{2pi}{3}$ $C、cfrac{5pi}{6}$ $D、2pi$

分析:只需要考虑函数(y=cos(2x+cfrac{pi}{6}))的对称性即可,由(2x+cfrac{pi}{6}=kpi)(kin Z)

得到对称轴(x=cfrac{kpi}{2}-cfrac{pi}{12}),由题可知,对称轴必须在([0,cfrac{pi}{2}])内,令(k=1),得到对称轴为(x=cfrac{5}{12})

又两个零点(x_1)(x_2)关于对称轴(x=cfrac{5}{12})对称,故(x_1+x_2=cfrac{5}{6})

例8【2019届高三理科数学三轮模拟试题】若函数(f(x)=asinx+cosx)((a)为常数,(ain R))的图像关于直线(x=cfrac{pi}{6})对称,则函数(g(x)=sinx+acosx)的图像【】

$A、关于直线x=-cfrac{pi}{3}对称$ $B、关于直线x=cfrac{pi}{6}对称$ $C、关于点(cfrac{pi}{3},0)对称$ $D、关于点(cfrac{5pi}{6},0)对称$

分析:(y=f(x)=sqrt{a^2+1}sin(x+phi)),其中(tanphi=cfrac{1}{a})

由函数(f(x)=asinx+cosx)的图像关于直线(x=cfrac{pi}{6})对称,可知(phi=cfrac{pi}{3})

(a=cfrac{sqrt{3}}{3}),又(g(x)=sinx+cfrac{sqrt{3}}{3}cosx=cfrac{2sqrt{3}}{3}sin(x+cfrac{pi}{6}))

逐项验证,可知选(D)

例9【2018广东茂名一模】【有图情形】已知函数(f(x)=Asin(omega x+phi)(A>0,omega> 0,0<phi<pi)),其导函数的图象(f'(x))如图所示,则(f(cfrac{pi}{2}))=【】

$A、2sqrt{3}$ $B、2$ $C、2sqrt{2}$ $D、4$

技术图片

分析:由于(f'(x)=omega Acos(omega x+phi)),由(cfrac{T}{4}=cfrac{3pi}{2}-cfrac{pi}{2}=pi),故(T=4pi),故(omega=cfrac{2pi}{4pi}=cfrac{1}{2})

又由图可知,(omega A=cfrac{1}{2}A=2),故(A=4),又由图(f'(cfrac{pi}{2})=0=2cos(cfrac{1}{2} imes cfrac{pi}{2}+phi)),即(cfrac{pi}{4}+phi=kpi+cfrac{pi}{2})(kin Z),故(phi=kpi+cfrac{pi}{4}),令(k=0),即(phi=cfrac{pi}{4}in (0,pi))

故函数(f(x)=4sin(cfrac{1}{2}x+cfrac{pi}{4})),则(f(cfrac{pi}{2})=4),故选(D)

例10【2019高三理科数学二轮资料用题】【有图情形】已知函数(f(x)=Atan(omega x+phi)),其中(omega >0)(|phi|<cfrac{pi}{2})(y=f(x))的部分图象如图,则(f(cfrac{pi}{24}))=__________.

技术图片

分析:由图可知,(cfrac{T}{2}=cfrac{3pi}{8}-cfrac{pi}{8}=cfrac{pi}{4}),则(T=cfrac{pi}{2}),故(omega=cfrac{pi}{T}=2)

又当(x=cfrac{3pi}{8})时,(2 imes cfrac{3pi}{8}+phi=kpi)(kin Z),则(phi=kpi-cfrac{3pi}{4})

(k=1),则(phi=pi-cfrac{3pi}{4}=cfrac{pi}{4}in (-cfrac{pi}{2},cfrac{pi}{2})),又(x=0)时,(y=1)

(Atan(2 imes 0+cfrac{pi}{4})=1),故(A=1),即(f(x)=tan(2x+cfrac{pi}{4}))

(f(cfrac{pi}{24})=tan(2 imes cfrac{pi}{24}+cfrac{pi}{4})=sqrt{3})

例11【无图情形】【2018届湖南衡阳八中第二次月考】已知函数(y=sin(ωx+φ)) ((ω>0,0<φ<π))的最小正周期为(π),且函数图象关于点((-cfrac{3pi}{8},0))对称,则该函数的解析式为________.

分析:由于函数(y=sin(ωx+φ))的最小正周期为(π),故(omega=2),又图象关于点((-cfrac{3pi}{8},0))对称,

(2 imes (-cfrac{3pi}{8})+phi=kpi),故(phi=kpi+cfrac{3pi}{4})(kin Z) ,

(k=0)时,(phi=cfrac{3pi}{4}in (0,pi)),故解析式为(y=sin(2x+cfrac{3pi}{4})).

例12【三轮模拟考试理科用题】已知(alpha)为第二象限角,(sin(alpha+cfrac{pi}{4})=cfrac{sqrt{2}}{10}),则(tancfrac{alpha}{2})的值为多少?

法1:变形得到(cfrac{sqrt{2}}{2}(sinalpha+cosalpha)=cfrac{sqrt{2}}{10})

解得(sinalpha+cosalpha=cfrac{1}{5}),又因为(alpha)为第二象限角,

再结合勾股数可得(sinalpha=cfrac{4}{5},cosalpha=-cfrac{3}{5})

(tanalpha=-cfrac{4}{3}),又由八卦图法可知(cfrac{alpha}{2})在第一、三象限,

(tancfrac{alpha}{2}>0),再由(tanalpha=-cfrac{4}{3}=cfrac{2tancfrac{alpha}{2}}{1-(tancfrac{alpha}{2})^2})

解方程得到(tancfrac{alpha}{2}=2)

法2:同上法,得到(sinalpha=cfrac{4}{5},cosalpha=-cfrac{3}{5})

(tancfrac{alpha}{2}=cfrac{sincfrac{alpha}{2}}{coscfrac{alpha}{2}})

(=cfrac{2sincfrac{alpha}{2}coscfrac{alpha}{2}}{2coscfrac{alpha}{2}coscfrac{alpha}{2}})

(=cfrac{sinalpha}{1+cosalpha}=cfrac{cfrac{4}{5}}{1-cfrac{3}{5}}=2)

例13(三轮模拟考试理科用题)已知(f(x)=2Acos^2(omega x+phi)(A>0,omega>0,0<phi<cfrac{pi}{2})),直线(x=cfrac{pi}{3})和点((cfrac{pi}{12},0))分别是函数(f(x))图象上相邻的一条对称轴和一个对称中心,则函数(f(x))的单调增区间为( ).

$A.[kpi-cfrac{2pi}{3} ,kpi-cfrac{pi}{6}](kin Z)$
$B.[kpi-cfrac{pi}{6} ,kpi+cfrac{pi}{3}](kin Z)$
$C.[kpi-cfrac{5pi}{12} ,kpi+cfrac{pi}{12}](kin Z)$
$D.[kpi+cfrac{pi}{12} ,kpi+cfrac{7pi}{12}](kin Z)$

分析:这类题目一般需要先将(f(x))转化为正弦型或者余弦型,再利用给定的条件分别求(omega)(phi),由

(f(x)=2Acos^2(omega x+phi)=A[cos2(omega x+phi)+1]-A=Acos(2omega x+2phi))

故其周期为(T=cfrac{2pi}{2omega}=cfrac{pi}{omega})

又由题目可知(cfrac{T}{4}=cfrac{pi}{3}-cfrac{pi}{12}=cfrac{pi}{4}),则(T=pi=cfrac{pi}{omega})

(omega=1),则函数简化为(f(x)=Acos(2x+2phi)),再利用直线(x=cfrac{pi}{3})是函数(f(x))图象上的一条对称轴,

(2 imes cfrac{pi}{3}+2phi=kpi,(kin Z)),解得(phi=cfrac{kpi}{2}-cfrac{pi}{3})

(k=1),则(phi=cfrac{pi}{6}in (0,cfrac{pi}{2})),满足题意,故(f(x)=Acos(2x+2phi)=Acos(2x+cfrac{pi}{3})).

(2kpi-pileq 2x+cfrac{pi}{3}leq 2kpi(kin Z)),解得(kpi-cfrac{2pi}{3}leq x leq kpi-cfrac{pi}{6}),即单调递增区间为(A.[kpi-cfrac{2pi}{3} ,kpi-cfrac{pi}{6}](kin Z));

例14【2014(cdot)新课标全国卷Ⅰ】设(alpha,eta in (0,cfrac{pi}{2})),且(tanalpha=cfrac{1+sineta}{coseta}),则【】

$A.3alpha-eta=cfrac{pi}{2}$ $B.3alpha+eta=cfrac{pi}{2}$ $C.2alpha-eta=cfrac{pi}{2}$ $D.2alpha+eta=cfrac{pi}{2}$

分析:切化弦得到,(cfrac{sinalpha}{cosalpha}=cfrac{1+sineta}{coseta})

(sinalpha coseta-cosalpha sineta=cosalpha),即(sin(alpha-eta)=cosalpha)

又由已知可得,(-cfrac{pi}{2}<alpha-eta<-cfrac{pi}{2})

再结合(sin(alpha-eta)=cosalpha)(alpha in (0,cfrac{pi}{2}))(cosalpha>0)

故可将(-cfrac{pi}{2}<alpha-eta<-cfrac{pi}{2})压缩为(0<alpha-eta<-cfrac{pi}{2}),,

这样(sin(alpha-eta)=cosalpha);且(alpha,alpha-eta in (0,cfrac{pi}{2}))

故有((alpha-eta)+alpha=cfrac{pi}{2}),即(2alpha-eta=cfrac{pi}{2}),选C.

例15【2016(cdot)上海卷】【解三角方程】方程(3sinx=1+cos2x)在区间([0,2pi])上的解为_______________。

分析:采用升幂降角公式,得到(3sinx=1+1-2sin^2x)

整理为(2sin^2x+3sinx-2=0),即((sinx+2)(2sinx-1)=0)

解得(sinx=-2(舍去))(sinx=cfrac{1}{2})

再由(sinx=cfrac{1}{2})(xin[0,2pi])

采用图像可得,(x=cfrac{pi}{6})(x=cfrac{5pi}{6})

例16【连比形式,巧设比例因子】在(Delta ABC)中,(tanA:tanB:tanC=1:2:3),求(cfrac{AC}{AB})的值;

分析:(巧设比例因子)设(tanA=k,tanB=2k,tanC=3k,(k>0)),

则由(tanA imes tanB imes tanC=tanA+tanB+tanC)可知,(6k=6k^3),解得(k=1).则有(tanA=1,tanB=2,tanC=3),

再设比例因子,比如设(sinB=2m,cosB=m,(m>0)),由平方关系可得,(5m^2=1,m=cfrac{1}{sqrt{5}}),

(sinB=cfrac{2}{sqrt{5}},sinC=cfrac{3}{sqrt{10}}),则(cfrac{AC}{AB}=cfrac{sinB}{sinC}=cfrac{cfrac{2}{sqrt{5}}}{cfrac{3}{sqrt{10}}}=cfrac{2sqrt{2}}{3}).

例17已知(tanalpha=cfrac{1}{2}),求(sin^4alpha-cos^4alpha)的值。

【法1】:方程组法,由(left{egin{array}{l}{cfrac{sinalpha}{cosalpha}=cfrac{1}{2}}\\{sin^2alpha+cos^2alpha=1}end{array} ight.)

解得(sin^2alpha=cfrac{1}{5})(cos^2alpha=cfrac{4}{5})

代入得到(sin^4alpha-cos^4alpha=-cfrac{3}{5})

【法2】:齐次式法,(sin^4alpha-cos^4alpha=(sin^2alpha-cos^2alpha)(sin^2alpha+cos^2alpha)=sin^2alpha-cos^2alpha)

(=-cos2alpha=-cfrac{cos^2alpha-sin^2alpha}{sin^2alpha+cos^2alpha}=cfrac{1-tan^2alpha}{1+tan^2alpha}=-cfrac{3}{5})

【法3】:由(cfrac{sinalpha}{cosalpha}=cfrac{1}{2}),引入比例因子,可设(sinalpha=k)(cosalpha=2k(k eq 0))

(k^2+(2k)^2=1),可得(k^2=cfrac{1}{5}),故(k^4=cfrac{1}{25})

(sin^4alpha-cos^4alpha=k^4-(2k)^4=-15k^4=-cfrac{3}{5})

例18求值:(sin^21^{circ}+sin^22^{circ}+sin^23^{circ}+cdots+sin^288^{circ}+sin^289^{circ}=)

分析:(sin^21^{circ}+sin^289^{circ}=1)(sin^22^{circ}+sin^288^{circ}=1)(cdots)(sin^244^{circ}+sin^246^{circ}=1)(sin^245^{circ}=cfrac{1}{2})

故原式=(44+cfrac{1}{2}=44.5)

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