三角函数给值求值

时间:2024-03-06 14:33:50

前言

三角函数的求值,对许多学生而言,都是个噩梦,我们不妨借助这篇博文梳理思路。

考察角度

角度一:已知角是一个,未知角也是一个角

【教材习题改编】已知\(sin(\alpha-\cfrac{\pi}{3})=\cfrac{15}{17}\)\(\alpha\in(\cfrac{\pi}{2},\cfrac{5\pi}{6})\),则\(sin\alpha\)的值为【】

分析:如果已知的角为一个,如\(\alpha-\cfrac{\pi}{3}\),未知角也是一个,如\(\alpha\),此时二者之间的关系往往利用互余、互补、半角、倍角、特殊角的角度建立联系,

比如本题目\(\alpha=(\alpha-\cfrac{\pi}{3})+\cfrac{\pi}{3}\)

\(sin\alpha=sin[(\alpha-\cfrac{\pi}{3})+\cfrac{\pi}{3}]=sin(\alpha-\cfrac{\pi}{3})cos\cfrac{\pi}{3}+cos(\alpha-\cfrac{\pi}{3})sin\cfrac{\pi}{3}=\cfrac{15+8\sqrt{3}}{34}\).

  • 备注:简单题目只需要用到互余、互补、半角、倍角、特殊角中的某一个角度就可以求解;

【2017枣庄模拟】设\(\alpha\)为锐角,\(cos(\alpha+\cfrac{\pi}{6})=\cfrac{4}{5}\),求\(sin(2\alpha+\cfrac{\pi}{12})\)的值;

分析:注意到已知角为一个\(\alpha+\cfrac{\pi}{6}\),未知角也是一个\(2\alpha+\cfrac{\pi}{12}\)

故二者之间的联系可能是从余、补、半、倍、特的角度建立联系,

故将已知角二倍得到\(2(\alpha+\cfrac{\pi}{6})=2\alpha+\cfrac{\pi}{3}\),发现还是和未知角不一样,故做差就发现,

\[2\alpha+\cfrac{\pi}{12}=2(\alpha+\cfrac{\pi}{6})-\cfrac{\pi}{4} \]

\(sin(2\alpha+\cfrac{\pi}{12})=sin[2(\alpha+\cfrac{\pi}{6})-\cfrac{\pi}{4}]\)

\(=sin[2(\alpha+\cfrac{\pi}{6})]cos\cfrac{\pi}{4}-cos[2(\alpha+\cfrac{\pi}{6})]sin\cfrac{\pi}{4}\)

\(=2sin(\alpha+\cfrac{\pi}{6})cos(\alpha+\cfrac{\pi}{6})cos\cfrac{\pi}{4}-[2cos^2(\alpha+\cfrac{\pi}{6})-1]sin\cfrac{\pi}{4}\)

\(=\cdots=\cfrac{17\sqrt{2}}{50}\).

  • 备注:复杂一些的题目可能需要用到互余、互补、半角、倍角、特殊角中的某两个以上的角度才可以求解;

【2016·福建师大附中月考】若\(sin(\cfrac{\pi}{3}-\alpha)=\cfrac{1}{4}\),则\(cos(\cfrac{2\pi}{3}+2\alpha)\)=【】

$A、-\cfrac{7}{8}$ $B、-\cfrac{1}{4}$ $C、\cfrac{1}{4}$ $D、\cfrac{7}{8}$

分析:\(cos(\cfrac{2\pi}{3}+2\alpha)=cos[\pi-(\cfrac{\pi}{3}-2\alpha)]\)

\(=-cos[2(\cfrac{\pi}{3}-\alpha)]=-1+2sin^2(\cfrac{\pi}{3}-\alpha)\)

\(=-1+2\times (\cfrac{1}{4})^2=-\cfrac{7}{8}\),故选A。

【2017黑龙江哈尔滨六中模拟】已知\(sin(\alpha+\cfrac{\pi}{3})+cos(\alpha-\cfrac{\pi}{2})=-\cfrac{4\sqrt{3}}{5}\),则\(cos(\alpha+\cfrac{2\pi}{3})\)的值为_____。

分析:先将已知条件变形为\(\sqrt{3}sin(\alpha+\cfrac{\pi}{6})=-\cfrac{4\sqrt{3}}{5}\)

\(sin(\alpha+\cfrac{\pi}{6})=-\cfrac{4}{5}\),求\(cos(\alpha+\cfrac{2\pi}{3})\)

到此题目的要求明显多了,就是个给值求值题目;

故这样变形,\(cos(\alpha+\cfrac{2\pi}{3})=cos(\alpha+\cfrac{\pi}{6}+\cfrac{\pi}{2})=-sin(\alpha+\cfrac{\pi}{6})=\cfrac{4}{5}\)

【2020\(\cdot\)河南洋阳一模】设 \(0^{\circ}<\alpha<90^{\circ}\), 若 \(\sin(75^{\circ}+2\alpha)\)\(=\)\(-\cfrac{3}{5}\), 则\(\sin(15^{\circ}+\alpha)\)\(\cdot\)\(\sin(75^{\circ}-\alpha)\)=__________

解析:因为 \(0^{\circ}<\alpha<90^{\circ}\), 所以 \(75^{\circ}<75^{\circ}+2\alpha<255^{\circ}\)

又因为 \(\sin(75^{\circ}+2\alpha)=-\cfrac{3}{5}<0\), 所以 \(180^{\circ}<75^{\circ}+2\alpha<255^{\circ}\)

又由于角 \(75^{\circ}+2\alpha\) 为第三象限角,所以 \(\cos(75^{\circ}+2\alpha)=-\cfrac{4}{5}\)

又由于\(\sin(75^{\circ}-\alpha)=\cos(15^{\circ}+\alpha)\),[注意:动态角的互余关系]

所以 \(\sin(15^{\circ}+\alpha)\cdot\sin(75^{\circ}-\alpha)=\sin(15^{\circ}+\alpha)\cdot\cos(15^{\circ}+\alpha)\)

\(=\cfrac{1}{2}\sin(30^{\circ}+2\alpha)=\cfrac{1}{2}\sin [(75^{\circ}+2\alpha)-45^{\circ}]\)

\(=\cfrac{1}{2}[\sin(75^{\circ}+2\alpha)\cos45^{\circ}-\cos(75^{\circ}+2\alpha)\cdot\sin 45^{\circ}]\)

\(=\cfrac{1}{2}\times(-\cfrac{3}{5}\times \cfrac{\sqrt{2}}{2}+\cfrac{4}{5}\times\cfrac{\sqrt{2}}{2})=\cfrac{\sqrt{2}}{20}\)

  • 角度二:已知角为两个,未知角为一个角,此时常常用两个已知角的线性表示来刻画未知角

【2016聊城模拟】若\(tan\alpha=-\cfrac{1}{2}\)\(tan(\alpha-\beta)=-\cfrac{2}{5}\),求\(tan\beta\)的值。

分析:如果已知的角为两个,如\(\alpha\)\(\alpha-\beta\);未知角是一个,如\(\beta\),此时二者之间的关系往往利用两个已知角的和差就能凑出未知角,比如本题目\(\beta=\alpha-(\alpha-\beta)\)

\(tan\beta=tan[\alpha-(\alpha-\beta)]=\cfrac{tan\alpha-tan(\alpha-\beta)}{1+tan\alpha\cdot tan(\alpha-\beta)}=-\cfrac{1}{12}\)

已知\(sin\alpha+sin\beta=\sqrt{3}(cos\beta-cos\alpha)\)\(\alpha,\beta\in (0,\cfrac{\pi}{2})\),则\(sin3\alpha+sin3\beta=\underline{0}\)

分析:由题目\(sin\alpha+sin\beta=\sqrt{3}(cos\beta-cos\alpha)\),则\(sin\alpha+\sqrt{3}cos\alpha=\sqrt{3}cos\beta-sin\beta\)

\(cos(\alpha-\cfrac{\pi}{6})=cos(\beta+\cfrac{\pi}{6})\)

\(\alpha-\cfrac{\pi}{6}\in(-\cfrac{\pi}{6},\cfrac{\pi}{3})\)\(\beta+\cfrac{\pi}{6}\in(\cfrac{\pi}{6},\cfrac{2\pi}{3})\)

则有\(\alpha-\cfrac{\pi}{6}=\beta+\cfrac{\pi}{6}\),即\(\alpha=\beta+\cfrac{\pi}{3}\)

\(sin3\alpha+sin3\beta=sin3(\beta+\cfrac{\pi}{3})+sin3\beta\)

\(=sin(3\beta+\pi)+sin3\beta=-sin3\beta+sin3\beta=0\)

已知 \(\alpha\) 为锐角, \(\beta\) 为第二象限角,且 \(\cos(\alpha-\beta)=\cfrac{1}{2}, \sin(\alpha+\beta)=\cfrac{1}{2},\)\(\sin(3\alpha-\beta)\)=【\(\quad\)

$A.-\cfrac{1}{2}$ $B.\cfrac{1}{2}$ $C.\cfrac{\sqrt{3}}{2}$ $D.-\cfrac{\sqrt{3}}{2}$

法一: 因为 \(\alpha\) 为锐角, \(\beta\) 为第二象限角,

则有\(0<\alpha<\cfrac{\pi}{2}\)\(2k\pi+\cfrac{\pi}{2}<\beta<2k\pi+\pi\)\(k\in \Z\)

故有\(2k\pi-\pi<\alpha-\beta<2k\pi\)\(2k\pi+\cfrac{\pi}{2}<\alpha+\beta<2k\pi+\cfrac{3\pi}{2}\)\(k\in \Z\)

又由于\(\cos (\alpha-\beta)>0\)\(\sin (\alpha+\beta)>0\)

所以 \(\alpha-\beta\) 为第四象限角, \(\alpha+\beta\) 为第二象限角,

因此 \(\sin(\alpha-\beta)=-\cfrac{\sqrt{3}}{2}\)\(\cos(\alpha+\beta)=-\cfrac{\sqrt{3}}{2}\)

所以 \(\sin2\alpha=\sin(\alpha-\beta+\alpha+\beta)=-\cfrac{\sqrt{3}}{2}\times(-\cfrac{\sqrt{3}}{2})+\cfrac{1}{2}\times\cfrac{1}{2}=1\)

因为 \(\alpha\) 为锐角,所以\(2\alpha=\cfrac{\pi}{2}\)

所以 \(\sin(3\alpha-\beta)=\sin(2\alpha+\alpha-\beta)=\cos(\alpha-\beta)=\cfrac{1}{2}\),选 \(B\).

法二: 因为 \(\alpha\) 为锐角, \(\beta\) 为第二象限角,

则有\(0<\alpha<\cfrac{\pi}{2}\)\(2k\pi+\cfrac{\pi}{2}<\beta<2k\pi+\pi\)\(k\in \Z\)

故有\(2k\pi-\pi<\alpha-\beta<2k\pi\)\(2k\pi+\cfrac{\pi}{2}<\alpha+\beta<2k\pi+\cfrac{3\pi}{2}\)\(k\in \Z\)

又由于\(\cos (\alpha-\beta)>0\)\(\sin (\alpha+\beta)>0\)

所以 \(\alpha-\beta\) 为第四象限角, \(\alpha+\beta\) 为第二象限角,

因此 \(\sin(\alpha-\beta)=-\cfrac{\sqrt{3}}{2}\)\(\cos(\alpha+\beta)=-\cfrac{\sqrt{3}}{2}\)

所以 \(\cos2(\alpha-\beta)=2\cos^{2}(\alpha-\beta)-1=2\times(\cfrac{1}{2})^{2}-1=-\cfrac{1}{2}\)

\(\sin2(\alpha-\beta)=2\sin(\alpha-\beta)\cos(\alpha-\beta)=2\times(-\cfrac{\sqrt{3}}{2})\times\cfrac{1}{2}=-\cfrac{\sqrt{3}}{2}\)

所以 \(\sin(3\alpha-\beta)=\sin[2(\alpha-\beta)+(\alpha+\beta)]\)

\(=\sin2(\alpha-\beta)\cos(\alpha+\beta)+\cos2(\alpha-\beta)\sin(\alpha+\beta)\)

\(=(-\cfrac{\sqrt{3}}{2})\times(-\cfrac{\sqrt{3}}{2})+(-\cfrac{1}{2})\times\cfrac{1}{2}=\cfrac{1}{2}\),故选\(B\).

\(\alpha\)\(\beta\)都是锐角,且\(cos\alpha=\cfrac{\sqrt{5}}{5}\)\(sin(\alpha+\beta)=\cfrac{3}{5}\),则\(cos\beta\)等于\(\qquad\)

$A.\cfrac{2\sqrt{5}}{25}$ $B.\cfrac{2\sqrt{5}}{5}$ $C.\cfrac{2\sqrt{5}}{25}或\cfrac{2\sqrt{5}}{5}$ $D.\cfrac{2\sqrt{5}}{25}或\cfrac{\sqrt{5}}{5}$

分析:由已知可得:\(sin\alpha=\cfrac{2\sqrt{5}}{5}\)\(cos(\alpha+\beta)=\pm\cfrac{4}{5}\)

\(cos(\alpha+\beta)=\cfrac{4}{5}>\cfrac{\sqrt{5}}{5}=cos\alpha\),则有\(\alpha+\beta<\alpha\)

\(\beta<0\),这与\(\beta\)为锐角矛盾舍去,故\(cos(\alpha+\beta)=-\cfrac{4}{5}\)

所以\(cos\beta=cos[(\alpha+\beta)-\alpha]\)

\(=cos(\alpha+\beta)cos\alpha+sin(\alpha+\beta)sin\alpha\)

\(=\cfrac{2\sqrt{5}}{25}\),故选\(A\)

已知\(\alpha\)为第Ⅳ象限角,且\(sin(\alpha+\cfrac{\pi}{4})=\cfrac{3}{5}\),求\(tan(\alpha-\cfrac{\pi}{4})\)的值;

法1:由题目可知,\(\cfrac{\sqrt{2}}{2}(sin\alpha+cos\alpha)=\cfrac{3}{5}\),则\(sin\alpha+cos\alpha=\cfrac{3\sqrt{2}}{5}\)

\(sin^2\alpha+cos^2\alpha=1\)联立,得到\(2sin^2\alpha-\cfrac{6\sqrt{2}}{5}sin\alpha-\cfrac{7}{25}=0\),即\((\sqrt{2}sin\alpha+\cfrac{1}{5})(\sqrt{2}sin\alpha-\cfrac{7}{5})=0\)

解得\(sin\alpha=-\cfrac{\sqrt{2}}{10}\),或\(sin\alpha=\cfrac{7\sqrt{2}}{10}\)(不符,舍去),

\(sin\alpha=-\cfrac{\sqrt{2}}{10}\)\(cos\alpha=\cfrac{7\sqrt{2}}{10}\),从而\(tan\alpha=-\cfrac{1}{7}\)

代入\(tan(\alpha-\cfrac{\pi}{4})=\cfrac{tan\alpha-1}{1+tan\alpha}=-\cfrac{4}{3}\)

法2:由\(sin\alpha+cos\alpha=\cfrac{3\sqrt{2}}{5}\),得到\(2sin\alpha cos\alpha=-\cfrac{7}{25}\)

则有\((sin\alpha-cos\alpha)^2=\cfrac{32}{25}\),由于\(\alpha\)为第Ⅳ象限角,

得到\(sin\alpha-cos\alpha=-\cfrac{4\sqrt{2}}{5}\),又\(sin\alpha+cos\alpha=\cfrac{3\sqrt{2}}{5}\)

\(sin\alpha=-\cfrac{\sqrt{2}}{10}\)\(cos\alpha=\cfrac{7\sqrt{2}}{10}\),从而\(tan\alpha=-\cfrac{1}{7}\)

代入\(tan(\alpha-\cfrac{\pi}{4})=\cfrac{tan\alpha-1}{1+tan\alpha}=-\cfrac{4}{3}\)

法3:由\(2sin\alpha cos\alpha=-\cfrac{7}{25}\),得到\(\cfrac{2sin\alpha cos\alpha}{sin^2\alpha+cos^2\alpha}=-\cfrac{7}{25}\)

\(\cfrac{2tan\alpha}{tan^2\alpha+1}=-\cfrac{7}{25}\),解得\(tan\alpha=-7\)\(tan\alpha=-\cfrac{1}{7}\)

又由于又\(sin\alpha+cos\alpha=\cfrac{3\sqrt{2}}{5}\),则\(|cos\alpha|>|sin\alpha|\),即\(|tan\alpha|<1\)

故保留\(tan\alpha=-\cfrac{1}{7}\),代入\(tan(\alpha-\cfrac{\pi}{4})=\cfrac{tan\alpha-1}{1+tan\alpha}=-\cfrac{4}{3}\)

已知\(\sin(\theta-\cfrac{\pi}{4})=\cfrac{2\sqrt{2}}{3}\),则\(\sin2\theta\)的值为【】

$A.\cfrac{7}{9}$ $B.-\cfrac{7}{9}$ $C.\cfrac{2}{9}$ $D.-\cfrac{2}{9}$

[法1]:从数的角度分析,借助三角函数的变换求解;将已知的角看成两个角\(\theta\)\(\cfrac{\pi}{4}\)的差,

由于\(\sin(\theta-\cfrac{\pi}{4})=\cfrac{2\sqrt{2}}{3}\),即\(\sin\theta\cdot \cfrac{\sqrt{2}}{2}-\cos\theta\cdot \cfrac{\sqrt{2}}{2}=\cfrac{2\sqrt{2}}{3}\),

整理得到,\(\sin\theta-\cos\theta=\cfrac{4}{3}\),两边平方得到\(1-\sin2\theta=\cfrac{16}{9}\)

\(\sin2\theta=-\cfrac{7}{9}\),故选\(B\);

[法2]:从数的角度分析,借助三角函数的变换求解;将已知的角\(\theta-\cfrac{\pi}{4}\)看成一个整体角,

\(\sin2\theta=\cos(\cfrac{\pi}{2}-2\theta)=\cos2(\cfrac{\pi}{4}-\theta)=\cos2(\theta-\cfrac{\pi}{4})\)

\(=1-2\sin^2(\theta-\cfrac{\pi}{4})=1-2\cdot (\cfrac{2\sqrt{2}}{3})^2=-\cfrac{7}{9}\),故选\(B\);

[法3]:从形的角度分析,借助三角函数线求解;

做平面直角坐标系和单位圆,由\(\sin(\theta-\cfrac{\pi}{4})=\cfrac{2\sqrt{2}}{3}\)可知,

则角\(\theta-\cfrac{\pi}{4}\)的终边位于射线\(OA\)\(OB\)上,其关于\(y\)轴对称,

将其顺时针旋转\(\cfrac{\pi}{4}\),得到角\(\theta\)的终边位于射线\(OC\)\(OD\)上,其关于\(y=-x\)轴对称,

将角\(\theta\)乘以\(2\)倍,则得到角\(2\theta\)的终边位于射线\(OM\)\(ON\)上,其关于\(y\)轴对称,

结合图像,如果做其正弦线,可知首先排除选项\(A\)\(C\),比较选项\(B\)\(D\),可知应该排除\(D\),而选\(B\)

可转化型

【2021届高三文科数学第三轮模拟训练题】已知 \(\alpha\)\(\beta\in (0,\cfrac{\pi}{2})\),且 \(\cos\alpha=\cfrac{2\sqrt{5}}{5}\)\(\cos(\alpha+\beta)=\cfrac{5}{13}\),则 \(\cos(\alpha-\beta)\)=【\(\quad\)

$A.\cfrac{21}{65}$ $B.\cfrac{63}{65}$ $C.-\cfrac{33}{65}$ $D.\cfrac{33}{65}$

分析:当由两个已知角 \(\alpha\)\(\alpha+\beta\) 的代数和不能表示未知角时,我们应该考虑能不能得到另外的两个已知角,再来表示未知角,很明显就渗透有转化划归思想在里面。

解析:由题目可知, \(\cos2\alpha=2\cos^2\alpha-1=\cfrac{3}{5}\)

由于 \(\alpha\)\(\beta\in (0,\cfrac{\pi}{2})\),故 \(\sin2\alpha=\cfrac{4}{5}\)\(\sin(\alpha+\beta)=\cfrac{12}{13}\)

\(\cos(\alpha-\beta)=\cos[2\alpha-(\alpha+\beta)]\)

\(=\cos2\alpha\cdot\cos(\alpha+\beta)+\sin2\alpha\cdot\sin(\alpha+\beta)=\cfrac{63}{65}\)

故选 \(B\).

新题补充

【2019届高三理科数学第三轮模拟训练题】设\(5sin\alpha=2+4cos^2\alpha\),则\(cos2\alpha\)=________。

分析:由已知条件,解方程得到\(sin\alpha=-2\)(舍去),\(sin\alpha=\cfrac{3}{4}\),则\(cos2\alpha=1-2sin^2\alpha=-\cfrac{1}{8}\)

【2019届高三理科数学第三轮模拟训练题】已知\(\alpha\)为第\(Ⅳ\)象限角,且\(sin(\alpha+\cfrac{\pi}{3})=\cfrac{3}{5}\),则\(sin(\alpha+\cfrac{\pi}{12})\)=__________.

分析:由\(\alpha\)为第Ⅳ象限角,且\(sin(\alpha+\cfrac{\pi}{3})=\cfrac{3}{5}\),则\(cos(\alpha+\cfrac{\pi}{3})=\cfrac{4}{5}\)

\(sin(\alpha+\cfrac{\pi}{12})=sin[(\alpha+\cfrac{\pi}{3})-\cfrac{\pi}{4}]=\cfrac{3}{5}\times \cfrac{\sqrt{2}}{2}-\cfrac{4}{5}\times \cfrac{\sqrt{2}}{2}=-\cfrac{\sqrt{2}}{10}\)

【2019届高三理科数学第三轮模拟训练题】已知\(cos(\alpha+\cfrac{\pi}{4})=-\cfrac{5\sqrt{2}}{8}\),则\(sin2\alpha\)=【】

$A.-\cfrac{5}{8}$ $B.-\cfrac{9}{16}$ $C.\cfrac{9}{16}$ $D.\cfrac{5}{8}$

分析:\(sin2\alpha=-cos(\cfrac{\pi}{2}+2\alpha)=-[2cos^2(\alpha+\cfrac{\pi}{4})^2-1]=-\cfrac{9}{16}\),故选\(B\).

【2020届高三文科数学训练题】已知\(tan(\alpha+\beta)=2\)\(tan\beta=3\),则\(sin2\alpha\)=【】

$A.\cfrac{7}{25}$ $B.\cfrac{14}{25}$ $C.-\cfrac{7}{25}$ $D.-\cfrac{14}{25}$

法1:先求得\(tan\alpha=-\cfrac{1}{7}\),则\(\cfrac{sin\alpha}{cos\alpha}=\cfrac{-1}{7}=\cfrac{-k}{7k}\)\((k\neq 0)\)

\(sin\alpha=-k\)\(cos\alpha=7k\),则由\(k^2+49k^2=1\)得到\(k^2=\cfrac{1}{50}\)

\(sin2\alpha=2sin\alpha\cdot cos\alpha=2\times (-k)\times 7k=-14\times k^2=-14\times \cfrac{1}{50}=-\cfrac{7}{25}\),故选\(C\)

法2:先求得\(tan\alpha=-\cfrac{1}{7}\),由二次齐次式可得

\(sin2\alpha=\cfrac{2sina\cdot cos\alpha}{sin^2\alpha+cos^2\alpha}=\cfrac{2tan\alpha}{tan^2\alpha+1}=-\cfrac{7}{25}\),故选\(C\)

【2020届高三文科数学训练题】已知\(\alpha\)为第二象限角,\(\sin(\alpha+\cfrac{\pi}{4})=\cfrac{\sqrt{2}}{10}\),则\(\tan\cfrac{\alpha}{2}\)的值为【】

$A.-\cfrac{1}{2}$ $B.\cfrac{1}{3}$ $C.2$ $D.-3$

法1:由\(\sin(\alpha+\cfrac{\pi}{4})=\cfrac{\sqrt{2}}{2}(\sin\alpha+\cos\alpha)=\cfrac{\sqrt{2}}{10}\),可以得到\(\sin\alpha+\cos\alpha=\cfrac{1}{5}\)

\(\alpha\)为第二象限角,结合勾股数\(3,4,5\),可知\(\sin\alpha=\cfrac{4}{5}\)\(\cos\alpha=-\cfrac{3}{5}\)

\(\tan\cfrac{\alpha}{2}=\cfrac{\sin\cfrac{\alpha}{2}}{\cos\cfrac{\alpha}{2}}\)

\(=\cfrac{\sin\cfrac{\alpha}{2}\cdot 2\cdot\cos\cfrac{\alpha}{2}}{\cos\cfrac{\alpha}{2}\cdot 2\cdot\cos\cfrac{\alpha}{2}}=\cfrac{\sin\alpha}{1+\cos\alpha}\)

\(=\cfrac{\frac{4}{5}}{1-\frac{3}{5}}=2\), 故选\(C\).

法2:由\(\sin(\alpha+\cfrac{\pi}{4})=\cfrac{\sqrt{2}}{2}(\sin\alpha+\cos\alpha)=\cfrac{\sqrt{2}}{10}\),可以得到\(\sin\alpha+\cos\alpha=\cfrac{1}{5}\)

\(\sin\alpha+\cos\alpha=\cfrac{\sin\alpha+\cos\alpha}{\sin^2\cfrac{\alpha}{2}+\cos^2\cfrac{\alpha}{2}}\)

\(=\cfrac{2\cdot\sin\cfrac{\alpha}{2}\cdot \cos\cfrac{\alpha}{2}+\cos^2\cfrac{\alpha}{2}-\sin^2\cfrac{\alpha}{2}}{\sin^2\cfrac{\alpha}{2}+\cos^2\cfrac{\alpha}{2}}\)

\(=\cfrac{2\tan\cfrac{\alpha}{2}-\tan^2\cfrac{\alpha}{2}+1}{\tan^2\cfrac{\alpha}{2}+1}=\cfrac{1}{5}\)

解得,\(\tan\cfrac{\alpha}{2}=2\)\(\tan\cfrac{\alpha}{2}=-\cfrac{1}{3}\)

由于\(\alpha\)为第二象限角,由八卦图法可知,\(\tan\cfrac{\alpha}{2}>0\),故\(\tan\cfrac{\alpha}{2}=2\),故选\(C\).

【2021届高三文科数学资料训练题】已知\(\tan\cfrac{\alpha}{2}=3\),则\(\cos\alpha\)=【\(\quad\)

$A.\cfrac{4}{5}$ $B.-\cfrac{4}{5}$ $C.\cfrac{4}{15}$ $D.-\cfrac{3}{5}$

分析:\(\cos\alpha=\cos^2\cfrac{\alpha}{2}-\sin^2\cfrac{\alpha}{2}=\cfrac{\cos^2\cfrac{\alpha}{2}-\sin^2\cfrac{\alpha}{2}}{\cos^2\cfrac{\alpha}{2}+\sin^2\cfrac{\alpha}{2}}\)

\(=\cfrac{1-\tan^2\frac{\alpha}{2}}{1+\tan^2\frac{\alpha}{2}}=\cfrac{1-9}{1+9}=-\cfrac{4}{5}\);故选\(B\).

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