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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
半角公式
\(\qquad \qquad\) \(\sin \dfrac{\alpha}{2}=\pm \sqrt{\dfrac{1-\cos \alpha}{2}}\) , \(\cos \dfrac{\alpha}{2}=\pm \sqrt{\dfrac{1+\cos \alpha}{2}}\) , \(\tan \dfrac{\alpha}{2}=\pm \sqrt{\dfrac{1-\cos \alpha}{1+\cos \alpha}}\)
(由降幂公式可得)
证明
由降幂公式 \(\sin ^2 \alpha=\dfrac{1-\cos 2 \alpha}{2}\)得 \(\sin \alpha=\pm \sqrt{\dfrac{1-\cos 2 \alpha}{2}}\)，则 \(\sin \dfrac{\alpha}{2}=\pm \sqrt{\dfrac{1-\cos \alpha}{2}}\)；
由降幂公式 \(\cos ^2 \alpha=\dfrac{1+\cos 2 \alpha}{2}\)得 \(\cos \alpha=\pm \sqrt{\dfrac{1+\cos 2 \alpha}{2}}\)，则 \(\cos \dfrac{\alpha}{2}=\pm \sqrt{\dfrac{1+\cos \alpha}{2}}\)；
\(\tan \dfrac{\alpha}{2}=\dfrac{\sin \dfrac{\alpha}{2}}{\cos \dfrac{\alpha}{2}}=\pm \sqrt{\dfrac{1-\cos \alpha}{1+\cos \alpha}}\).
解释
半角公式，利用\(\cos α\)表示了\(\sin \dfrac{\alpha}{2}\) 、\(\cos \dfrac{\alpha}{2}\)、\(\tan \dfrac{\alpha}{2}\) .
万能公式
\(\qquad \qquad\) \(\sin \alpha=\dfrac{2 \tan \dfrac{\alpha}{2}}{1+\tan ^2 \dfrac{\alpha}{2}}\)， \(\cos \alpha=\dfrac{1-\tan ^2 \dfrac{\alpha}{2}}{1+\tan ^2 \dfrac{\alpha}{2}}\), \(\tan \alpha=\dfrac{2 \tan \dfrac{\alpha}{2}}{1-\tan ^2 \dfrac{\alpha}{2}}\)
(由倍角公式可得)
证明
\(\sin 2 \alpha=2 \sin \alpha \cos \alpha=\dfrac{2 \sin \alpha \cos \alpha}{\sin ^2 \alpha+\cos ^2 \alpha}=\dfrac{2 \tan \alpha}{1+\tan ^2 \alpha}\)，则 \(\sin \alpha=\dfrac{2 \tan \dfrac{\alpha}{2}}{1+\tan ^2 \dfrac{\alpha}{2}}\)；
\(\cos 2 \alpha=\cos ^2 \alpha-\sin ^2 \alpha=\dfrac{\cos ^2 \alpha-\sin ^2 \alpha}{\sin ^2 \alpha+\cos ^2 \alpha}=\dfrac{1-\tan ^2 \alpha}{1+\tan ^2 \alpha}\)，则 \(\cos \alpha=\dfrac{1-\tan ^2 \dfrac{\alpha}{2}}{1+\tan ^2 \dfrac{\alpha}{2}}\)；
\(\tan 2 \alpha=\dfrac{2 \tan \alpha}{1-\tan ^2 \alpha}\)，则\(\tan \alpha=\dfrac{2 \tan \dfrac{\alpha}{2}}{1-\tan ^2 \dfrac{\alpha}{2}}\).
解释
万能公式，利用\(\tan \dfrac{α}{2}\)表示了\(\sin α\)、\(\cos α\)和\(\tan α\).
和化积公式
\(\qquad \qquad\) \(\sin \alpha+\sin \beta=2 \sin \dfrac{\alpha+\beta}{2} \cos \dfrac{\alpha-\beta}{2}\) \(\sin \alpha-\sin \beta=2 \cos \dfrac{\alpha+\beta}{2} \sin \dfrac{\alpha-\beta}{2}\)
\(\qquad \qquad\) \(\cos \alpha+\cos \beta=2 \cos \dfrac{\alpha+\beta}{2} \cos \dfrac{\alpha-\beta}{2}\) \(\cos \alpha-\cos \beta=-2 \sin \dfrac{\alpha+\beta}{2} \sin \dfrac{\alpha-\beta}{2}\)
(由和差公式可得)
证明
\(\sin \alpha+\sin \beta=\sin \left[\dfrac{\alpha+\beta}{2}+\dfrac{\alpha-\beta}{2}\right]+\sin \left[\dfrac{\alpha+\beta}{2}-\dfrac{\alpha-\beta}{2}\right]\)
\(=\sin \dfrac{\alpha+\beta}{2} \cos \dfrac{\alpha-\beta}{2}+\cos \dfrac{\alpha+\beta}{2} \sin \dfrac{\alpha-\beta}{2}+\sin \dfrac{\alpha+\beta}{2} \cos \dfrac{\alpha-\beta}{2}-\cos \dfrac{\alpha+\beta}{2} \sin \dfrac{\alpha-\beta}{2}\)
\(=2 \sin \dfrac{\alpha+\beta}{2} \cos \dfrac{\alpha-\beta}{2}\).
其他类似证明.
积化和公式
\(\qquad \qquad\) \(\sin \alpha \cdot \cos \beta=\dfrac{1}{2}[\sin (\alpha+\beta)+\sin (\alpha-\beta)]\)
\(\qquad \qquad\) \(\cos \alpha \cdot \cos \beta=\dfrac{1}{2}[\cos (\alpha+\beta)+\cos (\alpha-\beta)]\)
\(\qquad \qquad\) \(\sin α\cdot \sin β=\dfrac{1}{2}[\cos (α-β)-\cos (α+β)]\)
(由和差公式可得)
证明
由和化积公式 \(\sin x+\sin y=2 \sin \dfrac{x+y}{2} \cos \dfrac{x-y}{2}\)可得 \(\sin \dfrac{x+y}{2} \cos \dfrac{x-y}{2}=\dfrac{1}{2}(\sin x+\sin y)\) ()
令 \(\alpha=\dfrac{x+y}{2}\)， \(\beta=\dfrac{x-y}{2}\)，则\(x=α+β\)\(，y=α-β\)，
则公式()变成 \(\sin \alpha \cdot \cos \beta=\dfrac{1}{2}[\sin (\alpha+\beta)+\sin (\alpha-\beta)]\).
其他类似证明.
解释
积化和公式相当于和化积公式的逆运算.
基本方法
【题型1】 公式证明
【典题1】 证明： \(\cos \alpha+\cos \beta=2 \cos \dfrac{\alpha+\beta}{2} \cos \dfrac{\alpha-\beta}{2}\).
证明 \(\cos \alpha+\cos \beta=\cos \left[\dfrac{\alpha+\beta}{2}+\dfrac{\alpha-\beta}{2}\right]+\cos \left[\dfrac{\alpha+\beta}{2}-\dfrac{\alpha-\beta}{2}\right]\)
\(=\cos \dfrac{\alpha+\beta}{2} \cos \dfrac{\alpha-\beta}{2}-\sin \dfrac{\alpha+\beta}{2} \sin \dfrac{\alpha-\beta}{2}+\cos \dfrac{\alpha+\beta}{2} \cos \dfrac{\alpha-\beta}{2}+\sin \dfrac{\alpha+\beta}{2} \sin \dfrac{\alpha-\beta}{2}\)
\(=2 \cos \dfrac{\alpha+\beta}{2} \cos \dfrac{\alpha-\beta}{2}\).
【巩固练习】
1.证明： \(\sin \alpha-\sin \beta=2 \cos \dfrac{\alpha+\beta}{2} \sin \dfrac{\alpha-\beta}{2}\).
2.证明： \(\cos \alpha \cdot \cos \beta=\dfrac{1}{2}[\cos (\alpha+\beta)+\cos (\alpha-\beta)]\).
参考答案
-
证明 \(\sin \alpha-\sin \beta=\sin \left[\dfrac{\alpha+\beta}{2}+\dfrac{\alpha-\beta}{2}\right]-\sin \left[\dfrac{\alpha+\beta}{2}-\dfrac{\alpha-\beta}{2}\right]\)
\(=\sin \dfrac{\alpha+\beta}{2} \cos \dfrac{\alpha-\beta}{2}+\cos \dfrac{\alpha+\beta}{2} \sin \dfrac{\alpha-\beta}{2}-\sin \dfrac{\alpha+\beta}{2} \cos \dfrac{\alpha-\beta}{2}+\cos \dfrac{\alpha+\beta}{2} \sin \dfrac{\alpha-\beta}{2}\)
\(=2 \cos \dfrac{\alpha+\beta}{2} \sin \dfrac{\alpha-\beta}{2}\). -
证明 由和化积公式 \(\cos x+\cos y=2 \cos \dfrac{x+y}{2} \cos \dfrac{x-y}{2}\)，
可得 \(\cos \dfrac{x+y}{2} \cos \dfrac{x-y}{2}=\dfrac{1}{2}(\cos x+\cos y) \quad(*)\)，
令\(\alpha=\dfrac{x+y}{2}\)，\(\beta=\dfrac{x-y}{2}\)，则\(x=α+β\)，\(y=α-β\)，
则公式\((*)\)变成 \(\cos \alpha \cdot \cos \beta=\dfrac{1}{2}[\cos (\alpha+\beta)+\cos (\alpha-\beta)]\).
【题型2】 半角公式的运用
【典题1】 已知 \(\sin \alpha=-\dfrac{8}{17}\)且\(π<α<\dfrac{3}{2} π\)，求\(\sin \dfrac{α}{2}\),\(\cos \dfrac{α}{2}\),\(\tan \dfrac{α}{2}\)的值．
解析 \(\because \sin \alpha=-\dfrac{8}{17}\)且\(π<α<\dfrac{3}{2} π\)， \(\therefore \cos \alpha=-\dfrac{15}{17}\)．
又 \(\dfrac{\pi}{2}<\dfrac{\alpha}{2}<\dfrac{3}{4} \pi\)， \(\therefore \sin \dfrac{\alpha}{2}=\sqrt{\dfrac{1-\cos \alpha}{2}}=\sqrt{\dfrac{1+\dfrac{15}{17}}{2}}=\dfrac{4 \sqrt{17}}{17}\)，
\(\cos \dfrac{\alpha}{2}=-\sqrt{\dfrac{1+\cos \alpha}{2}}=-\sqrt{\dfrac{1-\dfrac{15}{17}}{2}}=-\dfrac{\sqrt{17}}{17}\)， \(\tan \dfrac{\alpha}{2}=\dfrac{\sin \dfrac{\alpha}{2}}{\cos \dfrac{\alpha}{2}}=-4\)．
【巩固练习】
1.已知\(\cos θ=-\dfrac{1}{5}\)，\(\dfrac{5\pi}{2}<θ<3π\)，那么 \(\sin \dfrac{\theta}{2}=\) ( )
A． \(\dfrac{\sqrt{10}}{5}\) \(\qquad \qquad \qquad \qquad\) B． \(-\dfrac{\sqrt{10}}{5}\) \(\qquad \qquad \qquad \qquad\) C． \(\dfrac{\sqrt{15}}{5}\) \(\qquad \qquad \qquad \qquad\) D． \(-\dfrac{\sqrt{15}}{5}\)
2.已知\(α∈(\dfrac{\pi}{2},π)\)，且 \(\cos \alpha=-\dfrac{3}{5}\)，则\(\tan \dfrac{α}{2}\)等于( )
A．\(2\) \(\qquad \qquad \qquad \qquad\) B．\(-2\) \(\qquad \qquad \qquad \qquad\) C．\(\dfrac{1}{2}\) \(\qquad \qquad \qquad \qquad\) D．\(-\dfrac{1}{2}\)
3.已知 \(\cos \theta=-\dfrac{7}{25}\)， \(\theta \in(-\pi, 0)\)，则 \(\sin \dfrac{\theta}{2}+\cos \dfrac{\theta}{2}=\)\(\underline{\quad \quad}\) .
参考答案
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答案 \(D\)
解析 \(\because \dfrac{5\pi}{2}<θ<3π\)，\(\therefore \dfrac{5\pi}{4}<\dfrac{\theta}{2}<\dfrac{3\pi}{2}\)．
\(\therefore \sin \dfrac{\theta}{2}<0\)，
\(\therefore \sin \dfrac{\theta}{2}=-\sqrt{\dfrac{1-\cos \theta}{2}}=-\sqrt{\left(1+\dfrac{1}{5}\right) \times \dfrac{1}{2}}=-\dfrac{\sqrt{15}}{5}\)． -
答案 \(A\)
解析 \(\because α∈(\dfrac{\pi}{2},π)\)，\(\therefore \dfrac{α}{2}∈(\dfrac{\pi}{4},\dfrac{\pi}{2})\)，
\(\therefore \sin \dfrac{\alpha}{2}=\sqrt{\dfrac{1-\cos \alpha}{2}}=\dfrac{2 \sqrt{5}}{5}\)， \(\cos \dfrac{\alpha}{2}=\sqrt{\dfrac{1+\cos \alpha}{2}}=\dfrac{\sqrt{5}}{5}\)．
\(\therefore \tan \dfrac{\alpha}{2}=\dfrac{\sin \dfrac{\alpha}{2}}{\cos \dfrac{\alpha}{2}}=2\)． -
答案 \(-\dfrac{1}{5}\)
解析 \(\because θ∈(-π ,0)\)，\(\therefore \dfrac{\theta}{2}∈(-\dfrac{\pi}{2} ,0)\)，
\(\therefore \sin \dfrac{\theta}{2}=-\sqrt{\dfrac{1-\cos \alpha}{2}}=-\dfrac{4}{5}\)， \(\cos \dfrac{\theta}{2}=\sqrt{\dfrac{1+\cos \alpha}{2}}=\dfrac{3}{5}\)，
\(\therefore \sin \dfrac{\theta}{2}+\cos \dfrac{\theta}{2}=-\dfrac{1}{5}\)．
【题型3】三角函数恒等变换与化简
【典题1】证明 \(\dfrac{1+\sin 2 \varphi}{\cos \varphi+\sin \varphi}=\cos \varphi+\sin \varphi\).
证明 \(\dfrac{1+\sin 2 \varphi}{\cos \varphi+\sin \varphi}=\dfrac{\cos ^2 \varphi+\sin ^2 \varphi+2 \sin \varphi \cos \varphi}{\cos \varphi+\sin \varphi}=\dfrac{(\cos \varphi+\sin \varphi)^2}{\cos \varphi+\sin \varphi}=\cos \varphi+\sin \varphi\).
【典题2】证明 \(\dfrac{1+\sin 2 \theta-\cos 2 \theta}{1+\sin 2 \theta+\cos 2 \theta}=\tan \theta\).
证明 \(\dfrac{1+\sin 2 \theta-\cos 2 \theta}{1+\sin 2 \theta+\cos 2 \theta}=\dfrac{2 \sin ^2 \theta+2 \sin \theta \cos \theta}{2 \cos ^2 \theta+2 \sin \theta \cos \theta}=\dfrac{2 \sin \theta(\sin \theta+\cos \theta)}{2 \cos \theta(\sin \theta+\cos \theta)}=\dfrac{\sin \theta}{\cos \theta}=\tan \theta\).
【典题3】 已知\(π<α<\dfrac{3\pi}{2}\)，化简： \(\dfrac{1+\sin \alpha}{\sqrt{1+\cos \alpha}-\sqrt{1-\cos \alpha}}+\dfrac{1-\sin \alpha}{\sqrt{1+\cos \alpha}+\sqrt{1-\cos \alpha}}\)．
解析 原式 \(=\dfrac{\left(\sin \dfrac{\alpha}{2}+\cos _2^\alpha\right)^2}{\sqrt{2}\left|\cos \dfrac{\alpha}{2}\right|-\sqrt{2}\left|\sin \dfrac{\alpha}{2}\right|}+\dfrac{\left(\sin \dfrac{\alpha}{2}-\cos \dfrac{\alpha}{2}\right)^2}{\sqrt{2}\left|\cos \dfrac{\alpha}{2}\right|+\sqrt{2}\left|\sin \dfrac{\alpha}{2}\right|}\)，
\(\because π<α<\dfrac{3\pi}{2}\)，\(\therefore \dfrac{\pi}{2}<\dfrac{α}{2}<\dfrac{3\pi}{4}\)，
\(\therefore \cos \dfrac{\alpha}{2} <0\)，\(\sin \dfrac{α}{2}>0\)．
\(\therefore\) 原式 \(=\dfrac{\left(\sin \dfrac{\alpha}{2}+\cos \dfrac{\alpha}{2}\right)^2}{-\sqrt{2}\left(\sin \dfrac{\alpha}{2}+\cos \dfrac{\alpha}{2}\right)}+\dfrac{\left(\sin \dfrac{\alpha}{2}-\cos \dfrac{\alpha}{2}\right)^2}{\sqrt{2}\left(\sin \dfrac{\alpha}{2}-\cos \dfrac{\alpha}{2}\right)}\)\(=-\dfrac{\sin \dfrac{\alpha}{2}+\cos \dfrac{\alpha}{2}}{\sqrt{2}}+\dfrac{\sin \dfrac{\alpha}{2}-\cos \dfrac{\alpha}{2}}{\sqrt{2}}=-\sqrt{2} \cos \dfrac{\alpha}{2}\)．
【巩固练习】
1.化简 \(\dfrac{\sin 4 \alpha}{4 \sin ^2\left(\dfrac{\pi}{4}+\alpha\right) \tan \left(\dfrac{\pi}{4}-\alpha\right)}\)得( )
A．\(\sin 2α\)\(\qquad \qquad \qquad \qquad\) B．\(\cos 2α\) \(\qquad \qquad \qquad \qquad\) C．\(\sin α\) \(\qquad \qquad \qquad \qquad\) D．\(\cos α\)
2.证明 \(\dfrac{1-\cos 2 \theta}{1+\cos 2 \theta}=\tan ^2 \theta\).
3.证明 \(3+\cos 4α-4\cos 2α=8\sin ^4 α\).
4.已知\(α,β\)都是锐角，且 \(\tan \beta=\dfrac{\sin \alpha-\cos \alpha}{\sin \alpha+\cos \alpha}\)，求 \(\dfrac{\sin \beta}{\sin \alpha-\cos \alpha}\)的值.
参考答案
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答案 \(A\)
解析 \(4 \sin ^2\left(\dfrac{\pi}{4}+\alpha\right) \tan \left(\dfrac{\pi}{4}-\alpha\right)=4 \cos ^2\left(\dfrac{\pi}{4}-\alpha\right) \tan \left(\dfrac{\pi}{4}-\alpha\right)\)\(=4 \cos \left(\dfrac{\pi}{4}-\alpha\right) \sin \left(\dfrac{\pi}{4}-\alpha\right)\)
\(=2\sin (\dfrac{\pi}{2}-2α)=2\cos 2α\)，
原式 \(=\dfrac{\sin 4 \alpha}{4 \sin ^2\left(\dfrac{\pi}{4}+\alpha\right) \cdot \tan \left(\dfrac{\pi}{4}-\alpha\right)}=\dfrac{\sin 4 \alpha}{2 \cos 2 \alpha}=\dfrac{2 \sin 2 \alpha \cdot \cos 2 \alpha}{2 \cos 2 \alpha}=\sin 2 a\)． -
证明 左边 \(=\dfrac{1-\left(1-2 \sin ^2 \theta\right)}{1+2 \cos ^2 \theta-1}=\tan ^2 \theta=\)右边.
-
证明 左边\(=3+2 \cos ^22α-1-4 \cos 2α=2 \cos ^22α-4 \cos 2α+2\)
\(=2(1-2 \sin ^2α )^2-4(1-2 \sin ^2α )+2 =2-8 \sin ^2α+8 \sin ^4α-4+8 \sin ^2α+2\)
\(=8\sin ^4α=\)右边. -
答案 \(\dfrac{\sqrt{2}}{2}\)
解析 \(\tan \beta=\dfrac{\dfrac{\sin \alpha}{\cos \alpha}-1}{\dfrac{\sin \alpha}{\cos \alpha}+1}=\dfrac{\tan \alpha-\tan \dfrac{\pi}{4}}{1+\tan \alpha \tan \dfrac{\pi}{4}}=\tan \left(\alpha-\dfrac{\pi}{4}\right)\)
因为\(α,β\)都是锐角，所以\(β=α-\dfrac{\pi}{4}\)，
所以 \(\dfrac{\sin \beta}{\sin \alpha-\cos \alpha}=\dfrac{\sin \beta}{\sqrt{2} \sin \left(\alpha-\dfrac{\pi}{4}\right)}=\dfrac{\sqrt{2}}{2}\).
分层练习
【A组---基础题】
1.已知 \(\cos x=\dfrac{2}{3}\)，且\(x\)为第四象限角，则\(\tan \dfrac{x}{2}=\)( )
A. \(-\dfrac{\sqrt{5}}{5}\) \(\qquad \qquad \qquad \qquad\) B.\(\dfrac{\sqrt{5}}{5}\) \(\qquad \qquad \qquad \qquad\) C. \(\dfrac{\sqrt{5}}{2}\) \(\qquad \qquad \qquad \qquad\) D. \(-\dfrac{\sqrt{5}}{3}\)
2.若\(θ∈[\dfrac{\pi}{4},\dfrac{\pi}{2}]\)， \(\sin 2 \theta=\dfrac{3 \sqrt{7}}{8}\)，则\(\sin θ=\) ( )
A．\(\dfrac{3}{5}\) \(\qquad \qquad \qquad \qquad\) B．\(\dfrac{4}{5}\) \(\qquad \qquad \qquad \qquad\) C． \(\dfrac{\sqrt{7}}{4}\) \(\qquad \qquad \qquad \qquad\) D．\(\dfrac{3}{4}\)
3.已知\(\dfrac{3\pi}{2}<α<2π\)，则 \(\sqrt{\dfrac{1+\cos \alpha}{1-\cos \alpha}}+\sqrt{\dfrac{1-\cos \alpha}{1+\cos \alpha}}=\) ( )
A． \(-\dfrac{1}{\sin \alpha}\) \(\qquad \qquad \qquad \qquad\) B． \(\dfrac{1}{\sin \alpha}\) \(\qquad \qquad \qquad \qquad\) C． \(-\dfrac{2}{\sin \alpha}\) \(\qquad \qquad \qquad \qquad\) D． \(\dfrac{2}{\sin \alpha}\)
4.已知 \(\cos \theta=-\dfrac{7}{25}\),\(θ∈(-π,0)\)，则 \(\sin \dfrac{\theta}{2}+\tan \dfrac{\theta}{2}=\)\(\underline{\quad \quad}\)．
5.在\(△ABC\)中，若\(\cos A=\dfrac{1}{3}\)，则 \(\sin ^2 \dfrac{B+C}{2}+\cos 2 A\)等于\(\underline{\quad \quad}\)．
6.化简 \(\sqrt{1+\sin 6}+\sqrt{1-\sin 6}=\)\(\underline{\quad \quad}\) ．
7.化简：\(\sin ^2 2x+2\cos ^2 x\cos 2x=\)\(\underline{\quad \quad}\)．
8.已知\(\sin (α+β)=\dfrac{1}{2}\)，\(\sin (α-β)=\dfrac{1}{3}\)，求证
(1) \(\sin α\cos β=5\cos α\sin β\); (2) \(\tan α=5\tan β\).
9.证明\(\tan (\dfrac{x}{2}+\dfrac{\pi}{4})+\tan (\dfrac{x}{2}-\dfrac{\pi}{4})=2\tan x\).
10.已知 \(\dfrac{1-\tan \theta}{2+\tan \theta}=1\)，求证\(\tan 2θ=-4\tan (θ+\dfrac{\pi}{4})\).
参考答案
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答案 \(-\dfrac{\sqrt{5}}{5}\)
解析 \(\because x\)为第四象限角，\(\therefore \dfrac{x}{2}\)为第二象限角或第四象限角，\(\therefore \tan \dfrac{x}{2}<0\)，
\(\therefore \tan \dfrac{\alpha}{2}=-\sqrt{\dfrac{1-\cos \alpha}{1+\cos \alpha}}=-\dfrac{\sqrt{5}}{5}\). -
答案 \(\dfrac{3}{4}\)
解析 由\(θ∈[\dfrac{\pi}{4},\dfrac{\pi}{2}]\)，得\(2θ∈[\dfrac{\pi}{2},π]\)， \(\cos 2 \theta=-\sqrt{1-\sin ^2 2 \theta}=-\dfrac{1}{8}\)，
\(\therefore \sin \theta=\sqrt{\dfrac{1-\cos 2 \theta}{2}}=\dfrac{3}{4}\)． -
答案 \(C\)
解析 因为\(\dfrac{3\pi}{2}<α<2π\)，所以\(\sin \dfrac{\alpha}{2} >0\),\(\cos \dfrac{\alpha}{2} <0\)，
所以 \(\sqrt{\dfrac{1+\cos \alpha}{1-\cos \alpha}}+\sqrt{\dfrac{1-\cos \alpha}{1+\cos \alpha}}=\sqrt{\dfrac{1+2 \cos ^2 \dfrac{\alpha}{2}-1}{1-1+2 \sin ^2 \dfrac{\alpha}{2}}}+\sqrt{\dfrac{1-1+2 \sin ^2 \dfrac{\alpha}{2}}{1+2 \cos ^2 \dfrac{\alpha}{2}-1}}\)\(=\sqrt{\dfrac{\cos ^2 \dfrac{\alpha}{2}}{\sin ^2 \dfrac{\alpha}{2}}}+\sqrt{\dfrac{\sin ^2 \dfrac{\alpha}{2}}{\cos ^2 \dfrac{\alpha}{2}}}=-\left(\dfrac{\cos \dfrac{\alpha}{2}}{\sin \dfrac{\alpha}{2}}+\dfrac{\sin \dfrac{\alpha}{2}}{\cos \dfrac{\alpha}{2}}\right)=-\dfrac{2}{\sin \alpha}\),
故选：\(C\)． -
答案 \(-\dfrac{32}{15}\)
解析 \(\because θ∈(-π,0)\)，\(\therefore \dfrac{\theta}{2}∈(-\dfrac{\pi}{2} ,0)\)，\(\therefore \sin \dfrac{\theta}{2}<0\)，\(\tan \dfrac{\theta}{2}<0\)
\(\therefore \sin \dfrac{\theta}{2}=-\sqrt{\dfrac{1-\cos \alpha}{2}}=-\dfrac{4}{5}\)， \(\tan \dfrac{\theta}{2}=-\sqrt{\dfrac{1-\cos \alpha}{1+\cos \alpha}}=-\dfrac{4}{3}\)，
\(\therefore \sin \dfrac{\theta}{2}+\tan \dfrac{\theta}{2}=-\dfrac{4}{5}-\dfrac{4}{3}=-\dfrac{32}{15}\). -
答案 \(-\dfrac{1}{9}\)
解析 在\(△ABC\)中， \(\dfrac{B+C}{2}=\dfrac{\pi}{2}-\dfrac{A}{2}\)，
\(\sin ^2 \dfrac{B+C}{2}+\cos 2 A=\sin ^2\left(\dfrac{\pi}{2}-\dfrac{A}{2}\right)+\cos 2 A\)
\(=\cos ^2 \dfrac{A}{2}+\cos 2 A=\dfrac{1+\cos A}{2}+2 \cos ^2 A-1=-\dfrac{1}{9}\)． -
答案 \(-2\cos 3\)
解析 原式 \(=\sqrt{\sin ^2 3+\cos ^2 3+2 \sin 3 \cos 3}+\sqrt{\sin ^2 3+\cos ^2 3-2 \sin 3 \cos 3}\)
\(=|\sin 3+\cos 3|+|\sin 3-\cos 3|\)，
因为\(\dfrac{3\pi}{4}<3<π\)，
所以\(\sin 3+\cos 3=\sqrt{2} \sin (3+\dfrac{\pi}{4})<0\)，\(\sin 3-\cos 3>0\)，
所以原式\(=-(\sin 3+\cos 3)+(\sin 3-\cos 3)=-2\cos 3\). -
答案 \(2\cos ^2 x\)
解析 原式\(=4\sin ^2 x·\cos ^2 x+2\cos ^2 x·\cos 2x=2\cos ^2 x·(2\sin ^2 x+\cos 2x)\)
\(=2\cos ^2 x·(2\sin ^2 x+1-2\sin ^2 x)=2\cos ^2 x\)． -
证明 (1) \(\because \sin (α+β)=\sin α\cos β+\cos α\sin β=\dfrac{1}{2}\)，\(\sin (α-β)=\sin α\cos β-\cos α\sin β=\dfrac{1}{3}\)，
\(\therefore \sin \alpha \cos \beta=\dfrac{5}{12}\), \(\cos \alpha \sin \beta=\dfrac{1}{12}\)，
\(\therefore \sin α\cos β=5\cos α\sin β\).
(2) \(\sin α\cos β=5\cos α\sin β\)
两边同时除以\(\cos α\cos β\)，得\(\tan α=5\tan β\). -
证明 左边\(=\dfrac{\tan \dfrac{x}{2}+1}{1-\tan \dfrac{x}{2}}+\dfrac{\tan \dfrac{x}{2}-1}{1+\tan \dfrac{x}{2}}=\dfrac{4 \tan \dfrac{x}{2}}{1-\tan ^2 \dfrac{x}{2}}=2 \tan x=\)右边.
-
证明 \(\because\dfrac{1-\tan \theta}{2+\tan \theta}=1\),\(\therefore \tan θ=-\dfrac{1}{2}\),
\(\therefore \tan 2 \theta=\dfrac{2 \tan \theta}{1-\tan ^2 \theta}=\dfrac{-1}{1-\dfrac{1}{4}}=-\dfrac{4}{3}\)，
\(-4 \tan \left(\theta+\dfrac{\pi}{4}\right)=-4 \times \dfrac{\tan \theta+1}{1-\tan \theta}=-4 \times \dfrac{-\dfrac{1}{2}+1}{1-\left(-\dfrac{1}{2}\right)}=-\dfrac{4}{3}\)，
\(\therefore \tan 2θ=-4\tan (θ+\dfrac{\pi}{4})\).
【B组---提高题】
1.\(\cos \dfrac{\pi}{7}+\cos \dfrac{3\pi}{7}+\cos \dfrac{5\pi}{7}=\)\(\underline{\quad \quad}\)．
2.已知\(α\) ,\(β\)为锐角，且 \(\alpha-\beta=\dfrac{\pi}{6}\)，那么\(\sin α\sin β\)的取值范围是\(\underline{\quad \quad}\)．
3.证明 \(\dfrac{\tan \alpha \tan 2 \alpha}{\tan 2 \alpha-\tan \alpha}+\sqrt{3}\left(\sin ^2 \alpha-\cos ^2 \alpha\right)=2 \sin \left(2 \alpha-\dfrac{\pi}{3}\right)\).
参考答案
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答案 \(\dfrac{1}{2}\)
解析 \(\cos \dfrac{\pi}{7}+\cos \dfrac{3 \pi}{7}+\cos \dfrac{5 \pi}{7}=\dfrac{1}{\sin \dfrac{\pi}{7}}\left(\sin \dfrac{\pi}{7} \cos \dfrac{\pi}{7}+\sin \dfrac{\pi}{7} \cos \dfrac{3 \pi}{7}+\sin \dfrac{\pi}{7} \cos \dfrac{5 \pi}{7}\right)\)
\(=\dfrac{1}{2 \sin \dfrac{\pi}{7}}\left[\sin \dfrac{2 \pi}{7}+\left(\sin \dfrac{4 \pi}{7}-\sin \dfrac{2 \pi}{7}\right)+\left(\sin \dfrac{6 \pi}{7}-\sin \dfrac{4 \pi}{7}\right)\right]\)
\(=\dfrac{1}{2 \sin \dfrac{\pi}{7}} \sin \dfrac{6 \pi}{7}=\dfrac{1}{2 \sin \dfrac{\pi}{7}} \times \sin \left(\pi-\dfrac{\pi}{7}\right)=\dfrac{1}{2}\)． -
答案 \((0 ,\dfrac{\sqrt{3}}{2})\)
解析 \(\because \alpha-\beta=\dfrac{\pi}{6}\)
\(\therefore \sin \alpha \sin \beta=-\dfrac{1}{2}[\cos (\alpha+\beta)-\cos (\alpha-\beta)]=-\dfrac{1}{2}\left[\cos (\alpha+\beta)-\dfrac{\sqrt{3}}{2}\right]\)
\(=-\dfrac{1}{2}\left[\cos \left(2 \beta+\dfrac{\pi}{6}\right)-\dfrac{\sqrt{3}}{2}\right]\)
\(\because β\)为锐角，即\(0<β<\dfrac{\pi}{3}\)，
\(\therefore \dfrac{\pi}{6}<2 \beta+\dfrac{\pi}{6}<\dfrac{5 \pi}{6}\)，
\(\therefore-\dfrac{\sqrt{3}}{2}<\cos \left(2 \beta+\dfrac{\pi}{6}\right)<\dfrac{\sqrt{3}}{2}\)，
\(\therefore 0<-\dfrac{1}{2}\left[\cos \left(2 \beta+\dfrac{\pi}{6}\right)-\dfrac{\sqrt{3}}{2}\right]<\dfrac{\sqrt{3}}{2}\)，
故 答案为：\((0 ,\dfrac{\sqrt{3}}{2})\). -
证明 左边 \(=\dfrac{\dfrac{\sin \alpha}{\cos \alpha} \cdot \dfrac{\sin 2 \alpha}{\cos 2 \alpha}}{\dfrac{\sin 2 \alpha}{\cos 2 \alpha}-\dfrac{\sin \alpha}{\cos \alpha}}-\sqrt{3} \cos 2 \alpha=\dfrac{\sin \alpha \sin 2 \alpha}{\sin 2 \alpha \cos \alpha-\cos 2 \alpha \sin \alpha}-\sqrt{3} \cos 2 \alpha\)
\(=\sin 2α-\sqrt{3}\cos 2α=2(\dfrac{1}{2} \sin 2α-\dfrac{\sqrt{3}}{2} \sin 2α)=2\sin (2α-\dfrac{\pi}{3})=\)右边.
【C组---拓展题】
- \(\dfrac{1}{\sin 45^{\circ} \sin 46^{\circ}}+\dfrac{1}{\sin 46^{\circ} \sin 47^{\circ}}+\cdots+\dfrac{1}{\sin 89^{\circ} \sin 90^{\circ}}=\)\(\underline{\quad \quad}\)．
参考答案
- 答案 \(\dfrac{1}{\sin 1^{\circ}}\)
解析 \(\dfrac{1}{\sin n^{\circ} \cdot \sin (n+1)^{\circ}}=\dfrac{1}{\sin 1^{\circ}} \cdot \dfrac{\sin \left((n+1)^{\circ}-n^{\circ}\right)}{\sin n^{\circ} \sin (n+1)^{\circ}}\)
\(=\dfrac{1}{\sin ^{\circ}} \cdot \dfrac{\sin (n+1)^{\circ} \cos n^{\circ}-\cos (n+1)^{\circ} \sin n^{\circ}}{\sin ^{\circ} \sin (n+1)^{\circ}}\)
\(=\dfrac{1}{\sin 1^{\circ}} \cdot\left(\dfrac{\cos n^{\circ}}{\sin n^{\circ}}-\dfrac{\cos (n+1)^{\circ}}{\sin (n+1)^{\circ}}\right)\)
\(=\dfrac{1}{\sin 1^{\circ}} \cdot\left(\dfrac{1}{\tan n^{\circ}}-\dfrac{1}{\tan (n+1)^{\circ}}\right)\)，
\(\dfrac{1}{\sin 45^{\circ} \sin 46^{\circ}}+\dfrac{1}{\sin 46^{\circ} \sin 47^{\circ}}+\cdots+\dfrac{1}{\sin 89^{\circ} \sin 90^{\circ}}\)
\(=\dfrac{1}{\sin 1^{\circ}}\left(\dfrac{1}{\tan 45^{\circ}}-\dfrac{1}{\tan 46^{\circ}}+\dfrac{1}{\tan 46^{\circ}}-\dfrac{1}{\tan 47^{\circ}}+\cdots+\dfrac{1}{\tan 89^{\circ}}-\dfrac{\cos 90^{\circ}}{\sin 90^{\circ}}\right)\)
\(=\dfrac{1}{\sin 1^{\circ}}(1-0)=\dfrac{1}{\sin 1^{\circ}}\)．