在线|二轮辅导[06][三角函数+解三角形02]
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典例剖析
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})。
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})。
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}).
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})。
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)和(phi),这样思路就跑偏了。
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,
令(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);
将(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;
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)和(x_2)关于对称轴(x=cfrac{5}{12})对称,故(x_1+x_2=cfrac{5}{6})。
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}),
则(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)。
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)。
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}),
令(k=1),则(phi=pi-cfrac{3pi}{4}=cfrac{pi}{4}in (-cfrac{pi}{2},cfrac{pi}{2})),又(x=0)时,(y=1),
即(Atan(2 imes 0+cfrac{pi}{4})=1),故(A=1),即(f(x)=tan(2x+cfrac{pi}{4})),
故(f(cfrac{pi}{24})=tan(2 imes cfrac{pi}{24}+cfrac{pi}{4})=sqrt{3})。
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,
则(2 imes (-cfrac{3pi}{8})+phi=kpi),故(phi=kpi+cfrac{3pi}{4}),(kin Z) ,
当(k=0)时,(phi=cfrac{3pi}{4}in (0,pi)),故解析式为(y=sin(2x+cfrac{3pi}{4})).
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});
故(tanalpha=-cfrac{4}{3}),又由八卦图法可知(cfrac{alpha}{2})在第一、三象限,
故(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);
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)和(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}),
故(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,
故(2 imes cfrac{pi}{3}+2phi=kpi,(kin Z)),解得(phi=cfrac{kpi}{2}-cfrac{pi}{3}),
令(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})).
令(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));
分析:切化弦得到,(cfrac{sinalpha}{cosalpha}=cfrac{1+sineta}{coseta}),
即(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.
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))或(sinx=cfrac{1}{2}),
再由(sinx=cfrac{1}{2}),(xin[0,2pi]),
采用图像可得,(x=cfrac{pi}{6})或(x=cfrac{5pi}{6})。
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}}),
故(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}).
【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)),
由(k^2+(2k)^2=1),可得(k^2=cfrac{1}{5}),故(k^4=cfrac{1}{25}),
则(sin^4alpha-cos^4alpha=k^4-(2k)^4=-15k^4=-cfrac{3}{5});
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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