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【基础过关系列】2022-2023学年高一数学上学期同步知识点剖析精品讲义(人教A版2019）
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必修第一册同步巩固，难度2颗星！
基础知识
二倍角的正弦余弦正切公式
(1) \(\sin 2α=2\sin α\cdot \cos α\);
(2) \(\cos 2α=\cos ^2 α-\sin ^2 α=1-2\sin ^2 α=2\cos ^2 α-1\);
(3) \(\tan 2 \alpha=\dfrac{2 \tan \alpha}{1-\tan ^2 \alpha}\);
(由 \(S_{(\alpha \pm \beta)}\)、 \(C_{(\alpha \pm \beta)}\)、 \(T_{(\alpha \pm \beta)}\)可推导出\(\sin 2α\)，\(\cos 2α\)，\(\tan 2α\)的公式)
推导
(1) \(\sin 2α=\sin (α+α) =\sin α \cos α+\cos α \sin α=2\sin α\cos α;\)
(2) \(\cos 2α=\cos (α+α)=\cos ^2 α-\sin ^2 α\);
\(\cos 2α=\cos ^2 α-\sin ^2 α=1-\sin ^2 α-\sin ^2 α=1-2\sin ^2 α\);
\(\cos 2α=\cos ^2 α-\sin ^2 α=\cos ^2 α-(1-\cos ^2 α)=2\cos ^2 α-1\);
(3) \(\tan 2 \alpha=\tan (\alpha+\alpha)=\dfrac{2 \tan \alpha}{1-\tan ^2 \alpha}\).
【例】 求\(\sin 15 ^{\circ}\cdot \cos 15 ^{\circ}\)，\(\cos ^2 75 ^{\circ} -\sin ^2 75 ^{\circ}\), \(\dfrac{\tan 15^{\circ}}{1-\tan ^2 15^{\circ}}\)的值.
解 \(\sin 15 ^{\circ}\cdot \cos 15 ^{\circ}=\dfrac{\sin 30^{\circ}}{2}=\dfrac{1}{4}\)，\(\cos ^2 75 ^{\circ} -\sin ^2 75 ^{\circ} =\cos 150 ^{\circ} =-\dfrac{\sqrt{3}}{2}\)，
\(\dfrac{2 \tan 15^{\circ}}{1-\tan ^2 15^{\circ}}=\tan 30^{\circ}=\dfrac{\sqrt{3}}{3} \Rightarrow \dfrac{\tan 15^{\circ}}{1-\tan ^2 15^{\circ}}=\dfrac{\sqrt{3}}{6}\).
降幂公式
\(\cos ^2 \alpha=\dfrac{1+\cos 2 \alpha}{2}\) \(\sin ^2 \alpha=\dfrac{1-\cos 2 \alpha}{2}\)
(由余弦倍角公式可得)
推导
\(\cos 2 \alpha=1-2 \sin ^2 \alpha \Rightarrow 2 \sin ^2 \alpha=1-\cos 2 \alpha \Rightarrow \sin ^2 \alpha=\dfrac{1-\cos 2 \alpha}{2}\)；
\(\cos 2 \alpha=2 \cos ^2 \alpha-1 \Rightarrow 2 \cos ^2 \alpha=\cos 2 \alpha-1 \Rightarrow \cos ^2 \alpha=\dfrac{1+\cos 2 a}{2}\).
【例】 求\(\cos ^2 15 ^{\circ}\)、\(\sin ^2 67.5 ^{\circ}\) 的值.
解 \(\cos ^2 15^{\circ}=\dfrac{1+\cos 30^{\circ}}{2}=\dfrac{\sqrt{3}+2}{4}\)， \(\sin ^2 67.5^{\circ}=\dfrac{1-\cos 135^{\circ}}{2}=\dfrac{2+\sqrt{2}}{4}\).
基本方法
【题型1】 给角求值
【典题1】 求下列各式的值：
(1) \(2 \cos ^2 \dfrac{25 \pi}{12}-1\)；\(\qquad \qquad\) (2) \(\dfrac{1-\tan ^2 \dfrac{\pi}{8}}{\tan \dfrac{\pi}{8}}\) ；\(\qquad \qquad\) (3) \(\dfrac{1}{\sin 10^{\circ}}-\dfrac{\sqrt{3}}{\cos 10^{\circ}}\) ；
解析 (1)原式 \(=\cos \dfrac{25 \pi}{6}=\cos \left(4 \pi+\dfrac{\pi}{6}\right)=\cos \dfrac{\pi}{6}=\dfrac{\sqrt{3}}{2}\)．
(2) \(\because \dfrac{2 \tan _{\dfrac{\pi}{8}}}{1-\tan ^2 \dfrac{\pi}{8}}=\tan \dfrac{\pi}{4}=1\)， \(\therefore \dfrac{\tan \dfrac{\pi}{8}}{1-\tan ^2 \dfrac{\pi}{8}}=\dfrac{1}{2}\)， \(\therefore \dfrac{1-\tan ^2 \dfrac{\pi}{8}}{\tan _{\dfrac{\pi}{8}}^8}=2\)．
(3)原式 \(=\dfrac{\cos 10^{\circ}-\sqrt{3} \sin 10^{\circ}}{\sin 10^{\circ} \cdot \cos 10^{\circ}}=\dfrac{2\left(\dfrac{1}{2} \cos 10^{\circ}-\dfrac{\sqrt{3}}{2} \sin 10^{\circ}\right)}{\sin 10^{\circ} \cdot \cos 10^{\circ}}\)
\(=\dfrac{4\left(\sin 30^{\circ} \cdot \cos 10^{\circ}-\cos 30^{\circ} \cdot \sin 10^{\circ}\right)}{2 \sin 10^{\circ} \cdot \cos 10^{\circ}}=\dfrac{4 \sin 20^{\circ}}{\sin 20^{\circ}}=4\)．
【巩固练习】
1.\(\left(\cos \dfrac{\pi}{12}-\sin \dfrac{\pi}{12}\right)\left(\cos \dfrac{\pi}{12}+\sin \dfrac{\pi}{12}\right)\)的值为( )
A．\(-\dfrac{\sqrt{3}}{2}\) \(\qquad \qquad \qquad \qquad\) B．\(-\dfrac{1}{2}\) \(\qquad \qquad \qquad \qquad\) C．\(\dfrac{1}{2}\)\(\qquad \qquad \qquad \qquad\) D．\(\dfrac{\sqrt{3}}{2}\)
2.求下列各式的值：
(1) \(\cos ^2 15^∘-\sin ^2 15^∘\)； \(\qquad \qquad\) (2)\(\cos \dfrac{\pi}{12} ·\cos \dfrac{5}{12} π\)．
3.求 \(\dfrac{1}{\sin 50^{\circ}}+\dfrac{\sqrt{3}}{\cos 50^{\circ}}\)的值．
参考答案
-
答案 \(D\)
解析 原式 \(=\cos ^2 \dfrac{\pi}{12}-\sin ^2 \dfrac{\pi}{12}=\cos \dfrac{\pi}{6}=\dfrac{\sqrt{3}}{2}\)． -
答案 (1)\(\dfrac{\sqrt{3}}{2}\)；(2)\(\dfrac{1}{4}\).
解析 (1)原式\(=\cos (2×15^∘ )=\cos 30^∘=\dfrac{\sqrt{3}}{2}\)．
(2)原式\(=\cos \dfrac{\pi}{12} ·\sin \dfrac{\pi}{12} =\dfrac{1}{2} \sin \dfrac{\pi}{6} =\dfrac{1}{4}\)． -
答案 \(4\)
解析 原式 \(=\dfrac{\cos 50^{\circ}+\sqrt{3} \sin 50^{\circ}}{\sin 50^{\circ} \cdot \cos 50^{\circ}}=\dfrac{2\left(\dfrac{1}{2} \cos 50^{\circ}+\dfrac{\sqrt{3}}{2} \sin 50^{\circ}\right)}{\dfrac{1}{2} \cdot 2 \sin 50^{\circ} \cdot \cos 50^{\circ}}=\dfrac{2 \sin 80^{\circ}}{\dfrac{1}{2} \sin 100^{\circ}}\)\(=\dfrac{2 \sin 80^{\circ}}{\dfrac{1}{2} \sin 80^{\circ}}=4\)．
【题型2】 给值化简求值问题
【典题1】 设\(\tan α=\dfrac{1}{2}\)，\(\cos (π+β)=-\dfrac{4}{5}(β∈(0,π))\)，则\(\tan (2α－β)\)的值为( )
A．\(-\dfrac{7}{24}\) \(\qquad \qquad \qquad \qquad\) B．\(-\dfrac{5}{24}\)\(\qquad \qquad \qquad \qquad\) C．\(\dfrac{5}{24}\) \(\qquad \qquad \qquad \qquad\) D．\(\dfrac{7}{24}\)
解析 \(∵\tan α=\dfrac{1}{2}\)， \(\therefore \tan 2 \alpha=\dfrac{2 \tan \alpha}{1-\tan ^2 \alpha}=\dfrac{1}{1-\left(\frac{1}{2}\right)^2}=\dfrac{4}{3}\)，
\(\cos (π+β)=-\cos β=-\dfrac{4}{5}\)，\(β∈(0,π)\)，
\(∴\cos β=\dfrac{4}{5}\),\(\sin β=\dfrac{3}{5}\), \(\tan \beta=\dfrac{3}{4}\)，
\(\therefore \tan (2 \alpha-\beta)=\dfrac{\tan 2 \alpha-\tan \beta}{1+\tan 2 \alpha \cdot \tan \beta}=\dfrac{7}{24}\)．
故选：\(D\)．
【典题2】 已知\(\sin \left(\dfrac{\pi}{4} -x\right)=\dfrac{5}{13}\)，\(0<x<\dfrac{\pi}{4}\)，求\(\dfrac{\cos 2 x}{\cos \left(\dfrac{\pi}{4}+x\right)}\)的值．
解析 方法1 \(∵\sin \left(\dfrac{\pi}{4} -x \right)=\dfrac{5}{13}\)，
\(∴\dfrac{\sqrt{2}}{2}(\cos x-\sin x)=\dfrac{5}{13}\)，即 \(\cos x-\sin x=\dfrac{5 \sqrt{2}}{13}\)，
两边平方得 \(1-2 \sin x \cos x=\dfrac{50}{169}\)，则 \(2 \sin x \cos x=\dfrac{119}{169}\)，
\(\therefore(\cos x+\sin x)^2=1+2 \sin x \cos x=\dfrac{288}{169}\)，
\(\therefore \cos x+\sin x=\pm \dfrac{12 \sqrt{2}}{13}\)，
\(∵0<x<\dfrac{\pi}{4}\) ，\(∴\cos x+\sin x>0\)，
\(\therefore \cos x+\sin x=\dfrac{12 \sqrt{2}}{13}\)，
\(\therefore \dfrac{\cos 2 x}{\cos \left(\dfrac{\pi}{4}+x\right)}=\dfrac{\cos ^2 x-\sin ^2 x}{\dfrac{\sqrt{2}}{2}(\cos x-\sin x)}=\sqrt{2}(\cos x+\sin x)=\dfrac{24}{13}\).
方法2 原式 \(=\dfrac{\sin \left(\dfrac{\pi}{2}+2 x\right)}{\cos \left(\dfrac{\pi}{4}+x\right)}=\dfrac{2 \sin \left(\dfrac{\pi}{4}+x\right) \cdot \cos \left(\dfrac{\pi}{4}+x\right)}{\cos \left(\dfrac{\pi}{4}+x\right)}=2 \sin \left(\dfrac{\pi}{4}+x\right)\)．
\(\because \sin \left(\dfrac{\pi}{4}-x\right)=\cos \left(\dfrac{\pi}{4}+x\right)=\dfrac{5}{13}\)，且\(0<x<\dfrac{\pi}{4}\) ，
\(\therefore \dfrac{\pi}{4} +x∈\left(\dfrac{\pi}{4} ,\dfrac{\pi}{2} \right)\)，
\(\therefore \sin \left(\dfrac{\pi}{4}+x\right)=\sqrt{1-\cos ^2\left(\dfrac{\pi}{4}+x\right)}=\dfrac{12}{13}\)，
\(\therefore\)原式 \(=2 \times \dfrac{12}{13}=\dfrac{24}{13}\)．
点拨 解题的方法很多，解题过程中注意以下几点
1.多观察各个角的关系，比如是否存在两角和、差是一定值(\(\dfrac{\pi}{3}\)、\(\dfrac{\pi}{2}\)、\(π\)之类)，两角存在倍数关系等等；
2.多注意常用公式\(A+B=π⇒\sin A=\sin B\),\(\cos A=-\cos B\) 、\(A+B=\dfrac{\pi}{2} ⇒\sin A=\cos B\)、\((\cos x±\sin x )^2=1±2 \cos x \sin x\)；
3.最重要的是多尝试，对已经条件与求证的式子多变式，懂得分析法与综合法处理问题.
【典题3】 已知 \(\dfrac{\cos 2 \alpha}{\sin \alpha+\cos \alpha}=\dfrac{\sqrt{2}}{4}\)，则 \(\cos ^2\left(\dfrac{3}{4} \pi+\alpha\right)\) 的值是\(\underline{\quad \quad}\) .
解析 \(\because \dfrac{\cos 2 \alpha}{\sin \alpha+\cos \alpha}=\dfrac{(\cos \alpha+\sin \alpha)(\cos \alpha-\sin \alpha)}{\sin \alpha+\cos \alpha}=\cos \alpha-\sin \alpha=\dfrac{\sqrt{2}}{4}\)，
\(\therefore\) 两边平方，可得 \(1-\sin 2 \alpha=\dfrac{1}{8}\)，可得 \(\sin 2 \alpha=\dfrac{7}{8}\)，
\(\therefore \cos ^2\left(\dfrac{3}{4} \pi+\alpha\right)=\dfrac{\cos \left(\dfrac{3 \pi}{2}+2 \alpha\right)-1}{2}=\dfrac{\sin 2 \alpha-1}{2}=-\dfrac{1}{16}\)．
点拨 遇到平方，可利用降幂公式起到“降幂”的作用.
【典题4】 若\(\sin \left(θ+\dfrac{\pi}{8} \right)=\dfrac{1}{3}\)，则\(\sin \left(2θ-\dfrac{\pi}{4} \right)=\) \(\underline{\quad \quad}\).
解析 \(∵2θ-\dfrac{\pi}{4} =2\left(θ+\dfrac{\pi}{8} \right)-\dfrac{\pi}{2}\) ，
\(\therefore \sin \left(2 \theta-\dfrac{\pi}{4}\right)=\sin \left[2\left(\theta+\dfrac{\pi}{8}\right)-\dfrac{\pi}{2}\right]=-\cos 2\left(\theta+\dfrac{\pi}{8}\right)\)\(=-\left[1-2 \sin ^2\left(\theta+\dfrac{\pi}{8}\right)\right]=-\dfrac{7}{9}\)．
点拨
1.因为已知角\(θ+\dfrac{\pi}{8}\) 和所求角\(2θ-\dfrac{\pi}{4}\)中\(θ\)的系数是\(2\)倍的关系，故想到\(2\left(θ+\dfrac{\pi}{8} \right)\)与\(2θ-\dfrac{\pi}{4}\)的差\(\dfrac{\pi}{2}\)是特殊角为关键，则有\(2θ-\dfrac{\pi}{4} =2\left(θ+\dfrac{\pi}{8} \right)-\dfrac{\pi}{2}\).
2.常见的角变换有：\(α=2\cdot \dfrac{\alpha }{2}\)， \(α=(α+β)-β=β-(α+β)\)，\(\dfrac{\pi}{4} +α=\dfrac{\pi}{2} -\left(\dfrac{\pi}{4} -α \right)\)，\(β=\dfrac{1}{2}\left[(α+β)-(α-β) \right]\)等.
【巩固练习】
1.若 \(\tan \alpha+\dfrac{1}{\tan \alpha}=3\)，则\(\cos 4α=\)( )
A．\(- \dfrac{7}{9}\) \(\qquad \qquad \qquad \qquad\) B．\(-\dfrac{1}{9}\) \(\qquad \qquad \qquad \qquad\) C． \(\dfrac{7}{9}\) \(\qquad \qquad \qquad \qquad\) D．\(\dfrac{1}{9}\)
2.已知\(α∈\left(0,\dfrac{\pi}{2} \right)\)，若\(\sin 2α-2\cos 2α=2\)，则\(\sin α=\)( )
A．\(\dfrac{1}{5}\) \(\qquad \qquad \qquad \qquad\) B．\(\dfrac{\sqrt{5}}{5}\) \(\qquad \qquad \qquad \qquad\) C．\(\dfrac{\sqrt{3}}{3}\) \(\qquad \qquad \qquad \qquad\) D． \(\dfrac{2\sqrt{5}}{5}\)
3.已知\(α∈\left(\dfrac{\pi}{2} ,π \right)\), \(\tan 2 \alpha=\dfrac{3}{4}\),则\(\sin 2α+\cos ^2α=\)( )
A．\(\dfrac{3}{2}\) \(\qquad \qquad \qquad \qquad\) B．\(\dfrac{1}{2}\) \(\qquad \qquad \qquad \qquad\) C．\(-\dfrac{1}{2}\) \(\qquad \qquad \qquad \qquad\) D．\(-\dfrac{3}{2}\)
4.如果 \(\dfrac{1+\tan \alpha}{1-\tan \alpha}=2013\)，那么 \(\dfrac{1}{\cos 2 \alpha}+\tan 2 \alpha=\)\(\underline{\quad \quad}\) .
5.已知\(\tan θ\)是方程\(x^2-6x+1=0\)的一根，则\(\cos ^2 \left(θ+\dfrac{\pi}{4} \right)=\) \(\underline{\quad \quad}\)．
6.若\(\cos \left(α+\dfrac{\pi}{12} \right)=\dfrac{\sqrt{2}}{3}\)，则\(\sin \left(\dfrac{\pi}{3} -2α \right)\)的值为\(\underline{\quad \quad}\)．
参考答案
-
答案 \(D\)
解析 \(\because \tan \alpha+\dfrac{1}{\tan \alpha}=\dfrac{\sin \alpha}{\cos \alpha}+\dfrac{\cos \alpha}{\sin \alpha}=\dfrac{2}{\sin 2 \alpha}=3\)， \(\therefore \sin 2 \alpha=\dfrac{2}{3}\)，
\(\therefore \cos 4 \alpha=1-2 \sin ^2 2 \alpha=\dfrac{1}{9}\)．
故选：\(D\)． -
答案 \(D\)
解析 \(∵\sin 2α-2\cos 2α=2\)，
\(\therefore \sin 2α=2(\cos 2α+1)=4 \cos ^2α\)，可得\(\sin α\cos α=2 \cos ^2α\)，
\(∵α∈(0,\dfrac{\pi}{2} )\)，可得\(\cos α≠0\)，\(\therefore \sin α=2\cos α\)，
\(∵\sin ^2α+\cos ^2α=\sin ^2α+\dfrac{1}{4} \sin ^2α=1\)，解得\(\sin ^2α=\dfrac{4}{5}\)，
可得 \(\sin \alpha=\dfrac{2 \sqrt{5}}{5}\)．
故选：\(D\)． -
答案 \(C\)
解析 \(\because \tan 2 \alpha=\dfrac{2 \tan \alpha}{1-\tan ^2 \alpha}=\dfrac{3}{4}\)，\(α∈\left(\dfrac{\pi}{2} ,π \right)\)，
\(\therefore \tan α=-3\)或\(\dfrac{1}{3}\)(舍去)，
\(\therefore \sin 2 \alpha+\cos ^2 \alpha=\dfrac{2 \sin \alpha \cos \alpha+\cos ^2 \alpha}{\sin ^2 \alpha+\cos ^2 \alpha}=\dfrac{2 \tan \alpha+1}{\tan ^2 \alpha+1}=-\dfrac{1}{2}\)．
故选：\(C\)． -
答案 \(2013\)
解析 \(\because \dfrac{1+\tan \alpha}{1-\tan \alpha}=2013\)，
\(\therefore \dfrac{1}{\cos 2 \alpha}+\tan 2 \alpha=\dfrac{\cos ^2 \alpha+\sin ^2 \alpha}{\cos ^2 \alpha-\sin ^2 \alpha}+\dfrac{2 \tan \alpha}{1-\tan ^2 \alpha}=\dfrac{1+\tan ^2 \alpha}{1-\tan ^2 \alpha}+\dfrac{2 \tan \alpha}{1-\tan ^2 \alpha}\)\(=\dfrac{(1+\tan \alpha)^2}{1-\tan ^2 \alpha}=\dfrac{1+\tan \alpha}{1-\tan \alpha}=2013\)． -
答案 \(\dfrac{1}{3}\)
解析 \(\because \tan θ\)是方程\(x^2-6x+1=0\)的一根，
\(\therefore \tan ^2θ-6\tan θ+1=0\)，则 \(\dfrac{\sin ^2 \theta}{\cos ^2 \theta}-6 \dfrac{\sin \theta}{\cos \theta}+1=0\)，
可得\(\sin ^2θ-6\sin θ\cos θ+\cos ^2θ=0\)，可得\(\sin θ\cos θ= \dfrac{1}{6}\)，
\(\therefore \sin 2θ=2\sin θ\cos θ=\dfrac{1}{3}\)，
\(\therefore \cos ^2\left(\theta+\dfrac{\pi}{4}\right)=\dfrac{1+\cos \left(2 \theta+\dfrac{\pi}{2}\right)}{2}=\dfrac{1-\sin 2 \theta}{2}=\dfrac{1-\dfrac{1}{3}}{2}=\dfrac{1}{3}\)． -
答案 \(- \dfrac{5}{9}\)
解析 \(\because \cos (α+\dfrac{\pi}{12} )=\dfrac{\sqrt{2}}{3}\)，
\(\therefore \cos \left[2\left(\alpha+\dfrac{\pi}{12}\right)\right]=2 \cos ^2\left(\alpha+\dfrac{\pi}{12}\right)-1=2 \times\left(\dfrac{\sqrt{2}}{3}\right)^2-1=-\dfrac{5}{9}\)，
即\(\cos \left(2 \alpha+\dfrac{\pi}{6}\right)=-\dfrac{5}{9}\)，
即\(\cos \left[\dfrac{\pi}{2} -\left(\dfrac{\pi}{3} -2α \right) \right]=\sin \left(\dfrac{\pi}{3} -2α \right)=- \dfrac{5}{9}\)．
【题型3】关于三角函数式的证明问题
【典题1】 证明 \(\dfrac{1+2 \sin x \cos x}{\cos 2 x}=\dfrac{1+\tan x}{1-\tan x}\).
证明 左边 \(=\dfrac{(\sin x+\cos x)^2}{\cos ^2 \alpha-\sin ^2 \alpha}=\dfrac{\sin x+\cos x}{\cos x-\sin x}\)，右边 \(=\dfrac{1+\tan x}{1-\tan x}=\dfrac{\sin x+\cos x}{\cos x-\sin x}\)，
\(\therefore \dfrac{1+2 \sin x \cos x}{\cos ^2 \alpha-\sin ^2 \alpha}=\dfrac{1+\tan x}{1-\tan x}\).
点拨 证明三角恒等式，可同时左右两边同时化简，统一角度(本题中2x化为x)和函数名(本题中用“切化弦”)便可.
【巩固练习】
1.化简\(1+\cos 2α+2\sin ^2 α=\)\(\underline{\quad \quad}\)．
2.求证： \(\dfrac{3-4 \cos 2 A+\cos 4 A}{3+4 \cos 2 A+\cos 4 A}=\tan ^4 A\)．
参考答案
-
答案 \(2\)
解析 原式\(=2\cos ^2 α+2\sin ^2 α=2(\sin ^2 α+\cos ^2 α)=2\)． -
证明 \(\because\)左边 \(=\dfrac{3-4 \cos 2 A+2 \cos ^2 2 A-1}{3+4 \cos 2 A+2 \cos ^2 2 A-1}=\left(\dfrac{1-\cos 2 A}{1+\cos 2 A}\right)^2\)\(=\left(\dfrac{2 \sin ^2 A}{2 \cos ^2 A}\right)^2=\left(\tan ^2 A\right)^2=\tan ^4 A=\)右边，
\(\therefore\dfrac{3-4 \cos 2 A+\cos 4 A}{3+4 \cos 2 A+\cos 4 A}=\tan ^4 A\)．
分层练习
【A组---基础题】
1.\(\dfrac{1}{2} \sin 15^∘ \cos 15^∘\)的值等于( )
A．\(\dfrac{1}{4}\) \(\qquad \qquad \qquad \qquad\) B．\(\dfrac{1}{8}\) \(\qquad \qquad \qquad \qquad\) C．\(\dfrac{1}{16}\) \(\qquad \qquad \qquad \qquad\) D．\(\dfrac{1}{2}\)
2.若 \(\tan \theta+\dfrac{1}{\tan \theta}=4\)，则\(\sin 2θ=\) ( )
A．\(\dfrac{1}{5}\) \(\qquad \qquad \qquad \qquad\) B．\(\dfrac{1}{4}\) \(\qquad \qquad \qquad \qquad\) C．\(\dfrac{1}{3}\) \(\qquad \qquad \qquad \qquad\) D．\(\dfrac{1}{2}\)
3.已知\(\cos \left(\dfrac{\pi}{4} -x \right)=\dfrac{3}{5}\)，则\(\sin 2x=\) ( )
A．\(\dfrac{18}{25}\) \(\qquad \qquad \qquad \qquad\) B．\(\dfrac{7}{25}\) \(\qquad \qquad \qquad \qquad\) C．\(-\dfrac{7}{25}\) \(\qquad \qquad \qquad \qquad\) D．\(-\dfrac{16}{25}\)
4.\(\dfrac{\sqrt{3}-\tan 12^{\circ}}{\left(2 \cos ^2 12^{\circ}-1\right) \sin 12^{\circ}}=\)\(\underline{\quad \quad}\)．
5.已知\(α∈\left(\dfrac{\pi}{2} ,π \right)\)，\(\sin α=\dfrac{\sqrt{5}}{5}\)，则\(\tan 2α=\)\(\underline{\quad \quad}\)．
6.已知\(\tan \left(\dfrac{\pi}{4} +α \right)=2\)，则\(\sin 2α＝\)\(\underline{\quad \quad}\)．
7.已知\(\tan α=-\dfrac{1}{3}\)，则 \(\dfrac{\sin 2 \alpha-\cos ^2 \alpha}{1+\cos 2 \alpha}=\)\(\underline{\quad \quad}\)．
8.已知\(\cos \left(α+\dfrac{\pi}{6} \right)=\dfrac{3}{5}\)，\(α∈\left(0，\dfrac{\pi}{2} \right)\),则\(\cos \left(2α+\dfrac{7\pi}{12} \right)=\) \(\underline{\quad \quad}\)．
9.已知\(\dfrac{\pi}{2} <β<α<\dfrac{3\pi}{4}\) ，且\(\cos (α-β)=\dfrac{12}{13}\)，\(\sin (α+β)=-\dfrac{3}{5}\)，求\(\cos 2α\)的值．
10.化简\(\cos ^2(α+β)+\cos ^2(α-β)-\cos 2α\cos 2β\).
11.求证 \(\dfrac{\sin 2 x}{(\sin x-\cos x+1)(\sin x+\cos x-1)}=\dfrac{1+\cos x}{\sin x}\).
参考答案
-
答案 \(B\)
解析 \(\dfrac{1}{2} \sin 15^{\circ} \cdot \cos 15^{\circ}=\dfrac{1}{4} \times 2 \sin 15^{\circ} \cdot \cos 15^{\circ}=\dfrac{1}{4} \times \sin 30^{\circ}=\dfrac{1}{8}\)． -
答案 \(D\)
解析 \(\because \tan \theta+\dfrac{1}{\tan \theta}=4\)， \(\therefore \dfrac{\sin \theta}{\cos \theta}+\dfrac{\cos \theta}{\sin \theta}=4\)．
\(\therefore \dfrac{\sin ^2 \theta+\cos ^2 \theta}{\cos \theta \cdot \sin \theta}=4\)，即 \(\dfrac{2}{\sin 2 \theta}=4\),\(\therefore \sin 2θ=\dfrac{1}{2}\)． -
答案 \(C\)
解析 法一：\(\because \cos \left(\dfrac{\pi}{4} -x \right)=\dfrac{3}{5}\)，\(\therefore \dfrac{\sqrt{2}}{2}(\cos x+\sin x)=\dfrac{3}{5}\)，
\(\therefore \dfrac{1}{2}(1+2\sin x·\cos x)= \dfrac{9}{25}\),\(\therefore \sin 2x=-\dfrac{7}{25}\)．
法二：\(\sin 2x=\cos \left(\dfrac{\pi}{2} -2x \right)=2\cos ^2 \left(\dfrac{\pi}{4} -x \right)-1=2× \dfrac{9}{25}-1=-\dfrac{7}{25}\)． -
答案 \(8\)
解析 原式 \(=\dfrac{\sqrt{3}-\dfrac{\sin 12^{\circ}}{\cos 12^{\circ}}}{\cos 24^{\circ} \sin 12^{\circ}}=\dfrac{\sqrt{3} \cos 12^{\circ}-\sin 12^{\circ}}{\cos 24^{\circ} \sin 12^{\circ} \cos 12^{\circ}}\)\(=\dfrac{2 \sin \left(60^{\circ}-12^{\circ}\right)}{\dfrac{1}{4} \sin 48^{\circ}}=\dfrac{2 \sin 48^{\circ}}{\dfrac{1}{4} \sin 48^{\circ}}=8\)． -
答案 \(-\dfrac{4}{3}\)
解析 由已知可得 \(\cos \alpha=-\dfrac{2 \sqrt{5}}{5}\)，\(\therefore \tan α=-\dfrac{1}{2}\)，\(\therefore \tan 2 \alpha=\dfrac{2 \tan \alpha}{1-\tan ^2 \alpha}=-\dfrac{4}{3}\)． -
答案 \(\dfrac{3}{5}\)
解析 由\(\tan \left(\dfrac{\pi}{4} +α \right)=2\)，即 \(\dfrac{1+\tan \alpha}{1-\tan \alpha}=2\)，解得\(\tan α=\dfrac{1}{3}\)，
所以 \(\sin 2 \alpha=\dfrac{2 \sin \alpha \cos \alpha}{\sin ^2 \alpha+\cos ^2 \alpha}=\dfrac{2 \tan \alpha}{1+\tan ^2 \alpha}=\dfrac{\dfrac{2}{3}}{1+\dfrac{1}{9}}=\dfrac{3}{5}\)． -
答案 \(- \dfrac{5}{6}\)
解析 \(\dfrac{\sin 2 \alpha-\cos ^2 \alpha}{1+\cos 2 \alpha}=\dfrac{2 \sin \alpha \cdot \cos \alpha-\cos ^2 \alpha}{1+2 \cos ^2 \alpha-1}\)\(=\dfrac{2 \sin \alpha \cdot \cos \alpha-\cos ^2 \alpha}{2 \cos ^2 \alpha}=\tan \alpha-\dfrac{1}{2}=-\dfrac{5}{6}\)． -
答案 \(-\dfrac{31 \sqrt{2}}{25}\)
解析 \(\because\cos \left(α+\dfrac{\pi}{6} \right)=\dfrac{3}{5}\)，\(α∈\left(0，\dfrac{\pi}{2} \right)\),
\(\therefore \left(α+\dfrac{\pi}{6} \right)∈\left(0,\dfrac{\pi}{2} \right)\)，\(\left(2α+\dfrac{\pi}{3} \right)∈(0,π)\)．
\(\cos \left(2α+\dfrac{\pi}{3} \right)=2\cos ^2 \left(α+\dfrac{\pi}{6} \right)-1=2×\left(\dfrac{3}{5} \right)^2-1=-\dfrac{7}{25}\)．
\(\therefore \sin \left(2 \alpha+\dfrac{\pi}{3}\right)=\sqrt{1-\cos ^2\left(2 \alpha+\dfrac{\pi}{3}\right)}=\dfrac{24}{25}\)．
\(\therefore \cos \left(2 \alpha+\dfrac{7 \pi}{12}\right)=\cos \left(2 \alpha+\dfrac{\pi}{3}+\dfrac{\pi}{4}\right)=\cos \left(2 \alpha+\dfrac{\pi}{3}\right) \cos \dfrac{\pi}{4}-\sin \left(2 \alpha+\dfrac{\pi}{3}\right) \sin \dfrac{\pi}{4}\)
\(=-\dfrac{7}{25} \times \dfrac{\sqrt{2}}{2}-\dfrac{24}{25} \times \dfrac{\sqrt{2}}{2}=-\dfrac{31 \sqrt{2}}{25}\)． -
答案 \(-\dfrac{33}{65}\)
解析 \(\because \dfrac{\pi}{2} <β<α<\dfrac{3\pi}{4}\) ，\(\therefore 0<α-β<\dfrac{\pi}{2}\) ，\(π<α+β<\dfrac{3\pi}{2}\)，
\(\because \cos (α-β)=\dfrac{12}{13}\)，\(\sin (α+β)=-\dfrac{3}{5}\)，
\(\therefore \sin (\alpha-\beta)=\sqrt{1-\left(\dfrac{12}{13}\right)^2}=\dfrac{5}{13}\)， \(\cos (\alpha+\beta)=-\sqrt{1-\left(-\dfrac{3}{5}\right)^2}=-\dfrac{4}{5}\)，
则\(\cos 2α=\cos [(α-β)+(α+β)]=\cos (α-β)\cos (α+β)-\sin (α-β)\sin (α+β)\)
\(=\dfrac{12}{13} \times\left(-\dfrac{4}{5}\right)-\left(-\dfrac{3}{5}\right) \times \dfrac{5}{13}=-\dfrac{33}{65}\)． -
答案 \(1\)
解析 \(\cos ^2(α+β)+\cos ^2(α-β)-\cos 2α \cos 2β\)
\(=\dfrac{1}{2}\left[1+\cos 2(α+β) \right]+\dfrac{1}{2}\left[1+\cos 2(α-β) \right] -\cos 2α \cos 2β\)
\(=1+\dfrac{1}{2}(\cos (2α+2β)+\cos (2α-2β)) -\cos 2α \cos 2β\)
\(=1+\dfrac{1}{2}×2\cos 2α \cos 2β-\cos 2α \cos 2β=1\). -
证明 左边 \(=\dfrac{\sin 2 x}{\sin ^2 x-(\cos x-1)^2}=\dfrac{2 \sin x \cos x}{\sin ^2 x-1-\cos ^2 x+2 \cos x}\)\(=\dfrac{2 \sin x \cos x}{2 \cos x(1-\cos x)}=\dfrac{\sin x}{1-\cos x}=\dfrac{1+\cos x}{\sin x}=\)右边.
【B组---提高题】
1.计算\(4 \cos 50^{\circ}-\tan 40^{\circ}=\)\(\underline{\quad \quad}\) .
- 在\(∆ ABC\)中，若 \(3 \cos ^2 \dfrac{A-B}{2}+5 \sin ^2 \dfrac{A+B}{2}=4\)，求\(\tan A \tan B\).
3.已知\(\sin \left(α+\dfrac{3\pi}{4} \right)=\dfrac{4}{5}\)，\(\cos \left(\dfrac{\pi}{4} -β \right)=\dfrac{3}{5}\)，且\(-\dfrac{\pi}{4} <α<\dfrac{\pi}{4}\) ,\(\dfrac{\pi}{4} <β<\dfrac{3\pi}{4}\) ，求\(\cos 2(α－β)\)的值．
参考答案
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答案 \(\sqrt{3}\)
解析 \(4 \cos 50^{\circ}-\tan 40^{\circ}=4 \cos 50^{\circ}-\dfrac{\sin 40^{\circ}}{\cos 40^{\circ}}=\dfrac{4 \cos 50^{\circ} \cos 40^{\circ}-\sin 40^{\circ}}{\cos 40^{\circ}}\)
\(=\dfrac{4 \sin 40^{\circ} \cos 40^{\circ}-\sin 40^{\circ}}{\cos 40^{\circ}}=\dfrac{2 \sin 80^{\circ}-\sin 40^{\circ}}{\cos 40^{\circ}}=\dfrac{2 \cos 10^{\circ}-\sin 40^{\circ}}{\cos 40^{\circ}}\)
\(=\dfrac{2 \cos \left(40^{\circ}-30^{\circ}\right)-\sin 40^{\circ}}{\cos 40^{\circ}}=\dfrac{\sqrt{3} \cos 40^{\circ}}{\cos 40^{\circ}}=\sqrt{3}\). -
答案 \(\dfrac{1}{4}\)
解析 在\(∆ ABC\)中，若 \(3 \cos ^2 \dfrac{A-B}{2}+5 \sin ^2 \dfrac{A+B}{2}=4\)，
\(\therefore 3 \times \dfrac{1+\cos (A-B)}{2}+5 \times \dfrac{1-\cos (A+B)}{2}=4\)，
即 \(\dfrac{3}{2} \cos (A-B)-\dfrac{5}{2} \cos (A+B)=0\)，
即\(3(\cos A \cos B+\sin A \sin B)=5(\cos A \cos B-\sin A \sin B)\)，
即\(2 \cos A \cos B=8 \sin A \sin B\) ,\(\therefore \tan A \tan B=\dfrac{1}{4}\). -
答案 \(-\dfrac{527}{625}\)
解析 由\(-\dfrac{\pi}{4} <α<\dfrac{\pi}{4}\)得，\(\dfrac{\pi}{2}<\alpha+\dfrac{3}{4} \pi<\pi\)，
所以 \(\cos \left(\alpha+\dfrac{3}{4} \pi\right)=-\sqrt{1-\sin ^2\left(\alpha+\dfrac{3}{4} \pi\right)}=-\dfrac{3}{5}\)，
由\(\dfrac{\pi}{4} <β<\dfrac{3}{4} \pi\)得，\(-\dfrac{\pi}{2} <\dfrac{\pi}{4} -β<0\)，
所以\(\sin \left(\dfrac{\pi}{4}-\beta\right)=-\sqrt{1-\cos ^2\left(\dfrac{\pi}{4}-\beta\right)}=-\dfrac{4}{5}\)，
所以\(\cos \left[\left(\alpha+\dfrac{3}{4} \pi\right)+\left(\dfrac{\pi}{4}-\beta\right)\right]\)
\(=\cos \left(\alpha+\dfrac{3}{4} \pi\right) \cos \left(\dfrac{\pi}{4}-\beta\right)-\sin \left(\alpha+\dfrac{3}{4} \pi\right) \sin \left(\dfrac{\pi}{4}-\beta\right)\)
\(=\left(-\dfrac{3}{5} \right)×\dfrac{3}{5}-\dfrac{4}{5}×\left(-\dfrac{4}{5} \right)=\dfrac{7}{25}\)
即\(－\cos (α－β)=\dfrac{7}{25}\)，
所以 \(\cos 2(\alpha-\beta)=2 \cos ^2(\alpha-\beta)-1=2 \times\left(-\dfrac{7}{25}\right)^2-1=-\dfrac{527}{625}\).
【C组---拓展题】
1.求 \(\cos \dfrac{\pi}{11} \cos \dfrac{2 \pi}{11} \cos \dfrac{3 \pi}{11} \cos \dfrac{4 \pi}{11} \cos \dfrac{5 \pi}{11}=\)\(\underline{\quad \quad}\)．
2.设\(a\),\(b\)是非零实数，\(x∈R\)，若 \(\dfrac{\sin ^4 x}{a^2}+\dfrac{\cos ^4 x}{b^2}=\dfrac{1}{a^2+b^2}\)，则 \(\dfrac{\sin ^{2008} x}{a^{2006}}+\dfrac{\cos ^{2008} x}{b^{2006}}=\)\(\underline{\quad \quad}\).
3.在\(△ABC\)中，若\(\sin B+\cos B= \sqrt{2}\)，则 \(\dfrac{\sin 2 A}{\tan B+\tan C}\)的最大值为\(\underline{\quad \quad}\).
参考答案
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答案 \(\dfrac{1}{32}\)
解析 \(\cos \dfrac{\pi}{11} \cos \dfrac{2 \pi}{11} \cos \dfrac{3 \pi}{11} \cos \dfrac{4 \pi}{11} \cos \dfrac{5 \pi}{11}=-\cos \dfrac{\pi}{11} \cos \dfrac{2 \pi}{11} \cos \dfrac{8 \pi}{11} \cos \dfrac{4 \pi}{11} \cos \dfrac{5 \pi}{11}\)
\(=\dfrac{2 \sin \dfrac{\pi}{11} \cos \dfrac{\pi}{11} \cos \dfrac{2 \pi}{11} \cos \dfrac{4 \pi}{11} \cos \dfrac{8 \pi}{11} \cos \dfrac{5 \pi}{11}}{2 \sin \dfrac{\pi}{11}}=\dfrac{\dfrac{1}{8} \cdot \sin \dfrac{16 \pi}{11} \cos \dfrac{5 \pi}{11}}{2 \sin \dfrac{\pi}{11}}\)
\(=\dfrac{-\sin \dfrac{5 \pi}{11} \cdot \cos \dfrac{5 \pi}{11}}{16 \sin \dfrac{\pi}{11}}=\dfrac{\dfrac{1}{2} \sin \dfrac{10 \pi}{11}}{16 \sin \dfrac{\pi}{11}}=\dfrac{1}{32}\). -
答案 \(\left(a^2+b^2\right) \dfrac{1}{\left(a^2+b^2\right)^{1004}}=\dfrac{1}{\left(a^2+b^2\right)^{1003}}\)
解析 由已知 \(\dfrac{\sin ^4 x}{a^2}+\dfrac{\cos ^4 x}{b^2}=\dfrac{1}{a^2+b^2}\)得 \(1=\sin ^4 x+\cos ^4 x+\dfrac{b^2}{a^2} \sin ^4 x+\dfrac{a^2}{b^2} \cos ^4 x\)
而\(1=\left(\sin ^2 x+\cos ^2 x\right)^2=\sin ^4 x+\cos ^4 x+2 \sin ^2 x \cos ^2 x\)
所以有 \(\left(\dfrac{b}{a} \sin ^2 x-\dfrac{a}{b} \cos ^2 x\right)^2=0\)
即 \(\dfrac{\sin ^4 x}{a^4}=\dfrac{\cos ^4 x}{b^4}\)，将该值记为\(C\).
则由(1)知， \(a^2 C+b^2 C=\dfrac{1}{a^2+b^2}\)，则 \(C=\dfrac{1}{\left(a^2+b^2\right)^2}\),
而 \(\dfrac{\sin ^{2008} x}{a^{2006}}+\dfrac{\cos ^{2008} x}{b^{2006}}=a^2 C^{502}+b^2 C^{502}=\left(a^2+b^2\right) \dfrac{1}{\left(a^2+b^2\right)^{1004}}=\dfrac{1}{\left(a^2+b^2\right)^{1003}}\). -
答案 \(\dfrac{\sqrt{2}-1}{2}\)
解析 \(\because \sin B+\cos B= \sqrt{2}\)，即\(\sqrt{2}\sin \left(B+\dfrac{\pi}{4} \right)= \sqrt{2}\)，
\(\therefore \sin \left(B+\dfrac{\pi}{4} \right)=1\)，
\(\because B∈(0,π)\)，\(B+\dfrac{\pi}{4} \in \left(\dfrac{\pi}{4} ,\dfrac{5\pi}{4} \right)\)，\(\therefore B+\dfrac{\pi}{4} =\dfrac{\pi}{2}\)，
\(\therefore\)可得\(B=\dfrac{\pi}{4}\)，\(C=\dfrac{3\pi}{4} -A\)，
\(\therefore \dfrac{\sin 2 A}{\tan B+\tan C}=\dfrac{\sin 2 A}{1+\tan \left(\dfrac{3 \pi}{4}-A\right)}=\dfrac{\dfrac{2 \tan A}{1+\tan { }^2 A}}{1+\dfrac{-1-\tan A}{1-\tan A}}=\dfrac{\tan A-1}{\tan ^2 A+1}=\dfrac{\dfrac{\sin A}{\cos A}-1}{\left(\dfrac{\sin A}{\cos A}\right)^2+1}\)
\(=\sin A \cos A-\cos ^2 A=\dfrac{1}{2} \sin 2 A-\dfrac{1+\cos 2 A}{2}=\dfrac{\sqrt{2}}{2} \sin \left(2 A-\dfrac{\pi}{4}\right)-\dfrac{1}{2}\)，
\(\because A∈\left(0,\dfrac{3\pi}{4} \right)\)，可得\(2A-\dfrac{\pi}{4} ∈\left(-\dfrac{\pi}{4} ,\dfrac{5\pi}{4} \right)\)，
可得\(\sin \left(2A-\dfrac{\pi}{4} \right)∈\left(-\dfrac{\sqrt{2}}{2},1 \right]\)，
\(\therefore \dfrac{\sin 2 A}{\tan B+\tan C}=\dfrac{\sqrt{2}}{2} \sin \left(2 A-\dfrac{\pi}{4}\right)-\dfrac{1}{2} \in\left(-1, \dfrac{\sqrt{2}-1}{2}\right]\)，
\(\therefore \dfrac{\sin 2 A}{\tan B+\tan C}\)的最大值为 \(\dfrac{\sqrt{2}-1}{2}\)．