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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

两角和差的正弦，余弦与正切公式

(理解公式的推导，体会其方法，而不死背公式)
1 余弦两角和差公式
$$\cos (α±β)=\cos α\cdot \cos β∓ \sin α\cdot \sin β$$
推导如下
如图，设单位圆与\(x\)轴的正半轴相交于点\(A(1 ,0)\)，以\(x\)轴为非负半轴为始边作角\(α\)，\(β\)，\(α-β\)，它们的终边分别与单位圆相较于点\(P_1 (\cos α ,\sin α)\)，\(A_1 (\cos β ,\sin β)\)，\(P(\cos ⁡(α-β) ,\sin ⁡(α-β) )\)，连接\(A_1 P_1\)，\(AP\)，若把扇形\(OAP\)绕点\(O\)旋转\(β\)角，则点\(A\)，\(P\)分别与\(A_1\) ,\(P_1\)重合.根据圆的旋转对称性可知， \(\widehat{A P}\)与 \(\widehat{A_1 P_1}\)重合，从而 \(\widehat{A P}=\widehat{A_1 P_1},\)，所以\(AP=A_1 P_1\).

根据两点间的距离公式，得
\([\cos ⁡(α-β)-1]^2+ \sin ^2⁡ (α-β)=(\cos α-\cos β) ^2+(\sin α-\sin β)^2\)
化简得 \(\cos ⁡(α-β)=\cos α\cdot \cos β+ \sin α\cdot \sin β\)
而 \(\cos (α+β)=\cos [α-(-β)]=\cos α\cdot \cos β- \sin α\cdot \sin β\)
 

2 正弦两角和差公式
$$\sin (α±β)=\sin α\cdot \cos β± \cos α\cdot \sin β$$
推导如下
\(\sin ⁡(α+β)=\cos \left[\dfrac{\pi}{2}-(α+β) \right] =\cos \left[\left(\dfrac{\pi}{2}-α\right)-β\right]\)
\(=\cos \left(\dfrac{\pi}{2}-α\right)\cos β+\sin\left(\dfrac{\pi}{2}-α\right)\sin β=\sin α\cos β+ \cos α\sin β\)
\(\sin (\alpha-\beta)=\cos \left[\dfrac{\pi}{2}-(\alpha-\beta)\right]=\cos \left[\left(\dfrac{\pi}{2}-\alpha\right)+\beta\right]\)
\(=\cos \left(\dfrac{\pi}{2}-α\right)\cos β-\sin \left(\dfrac{\pi}{2}-α\right)\sin β=\sin α\cos β- \cos α\sin β\)
 

3 正切两角和差公式
$$\tan (\alpha \pm \beta)=\dfrac{\tan \alpha \pm \tan \beta}{1 \mp \tan \alpha \tan \beta}$$
(由 \(S_{(\alpha \pm \beta)}\)、 \(C_{(\alpha \pm \beta)}\)可推导正切的和差角公式)
 

4 对公式的理解
公式中\(α\)、\(β\)的理解，他们可表示为一个数字、一个字母，甚至一个式子
Eg
① \(\sin 75^{\circ}=\sin \left(45^{\circ}+30^{\circ}\right)=\sin 45^{\circ} \cos 30^{\circ}+\cos 45^{\circ} \sin 30^{\circ}=\dfrac{\sqrt{6}+\sqrt{2}}{4}\)
对应公式\(\sin (α±β)=\sin α \cos β± \cos α \sin β\)，把\(α\)看成数字\(45^{\circ}\) ,\(β\)看成数字\(30^{\circ}\)；
② \(\cos ⁡(x+\dfrac{\pi}{3} )=\cos x\cdot \cos \dfrac{\pi}{3}-\sin x\cdot \sin \dfrac{\pi}{3}\)
对应公式\(\cos (α+β)=\cos α \cos β-\sin α \sin β\)，把\(α\)看成字母\(x\)， \(β\)看成数字\(\dfrac{\pi}{3}\) ；
③\(\tan \dfrac{\pi}{4}=\tan \left[\left(x+\dfrac{\pi}{8}\right)+\left(\dfrac{\pi}{8}-x\right)\right]=\dfrac{\tan \left(x+\dfrac{\pi}{8}\right)+\tan \left(\dfrac{\pi}{8}-x\right)}{1-\tan \left(x+\dfrac{\pi}{8}\right) \tan \left(\dfrac{\pi}{8}-x\right)}\)，
对应公式 \(\tan (\alpha \pm \beta)=\dfrac{\tan \alpha \pm \tan \beta}{1 \mp \tan \alpha \tan \beta}\)，把\(α\)、\(β\)分别看成式子\(x+\dfrac{\pi}{8}\) 、\(x-\dfrac{\pi}{8}\) .
对应公式的运用，注意整体变换的思想.
 

辅助角公式

\(a:b=\sqrt{3}:1\)型，配\(\dfrac{\pi}{6}\)或\(\dfrac{\pi}{3}\)

基本方法

【题型1】 给角求值问题

【典题1】 化简下列各式：
  (1) \(\sin 70^∘·\sin 65^∘-\sin 20^∘·\sin 25^∘\)；
  (2) \(\sin (54^∘-x)·\cos (36^∘+x)+\cos (54^∘-x)·\sin (36^∘+x)\)；
  (3) \(\dfrac{\sqrt{3}-\tan 15^{\circ}}{1+\sqrt{3} \tan 15^{\circ}}\)；
  (4) \(\tan 23^∘+\tan 37^∘+\sqrt{3} \tan 23^∘·\tan 37^∘\)．
解析 (1)原式\(=\sin 70^∘·\cos 25^∘-\cos 70^∘·\sin 25^∘=\sin ⁡(70^∘-25^∘ )=\sin 45^∘=\dfrac{\sqrt{2}}{2}\)．
(2)原式\(=\sin ⁡[(54^∘-x)+(36^∘+x)]=\sin 90^∘=1\)．
(3)原式\(=\dfrac{\tan 60^{\circ}-\tan 15^{\circ}}{1+\tan 60^{\circ} \tan 15^{\circ}}=\tan 45^{\circ}=1\)．
(4)原式\(=\tan (23^∘+37^∘ )(1-\tan 23^∘·\tan 37^∘ )+\sqrt{3} \tan 23^∘·\tan 37^∘\)
\(=\sqrt{3}-\sqrt{3} \tan 23^∘·\tan 37^∘+\sqrt{3} \tan 23^∘·\tan 37^∘=\sqrt{3}\)．
点拨 第(1)问，使用两角和差公式时，注意函数名与角度的变换；
第(2)问，使用公式时，注意“整体思想”，把\(54^∘-x\)看成\(α\)，\(36^∘+x\)看成\(β\)；
第(4)问是对 \(\tan (\alpha \pm \beta)=\dfrac{\tan \alpha \pm \tan \beta}{1 \mp \tan \alpha \tan \beta}\)的变式\(\tan α± \tan β=\tan (α±β)(1 ∓ \tan α \tan β)\)的运用.
 

【典题2】 \(\dfrac{\sin 47^{\circ}-\sin 17^{\circ} \cos 30^{\circ}}{\cos 17^{\circ}}=\) (　　)
 A．\(-\dfrac{\sqrt{3}}{2}\) \(\qquad \qquad \qquad \qquad\) B．\(-\dfrac{\sqrt{1}}{2}\) \(\qquad \qquad \qquad \qquad\) C．\(\dfrac{\sqrt{1}}{2}\) \(\qquad \qquad \qquad \qquad\) D．\(\dfrac{\sqrt{3}}{2}\)
解析 原式\(=\dfrac{\sin \left(17^{\circ}+30^{\circ}\right)-\sin 17^{\circ} \cos 30^{\circ}}{\cos 17^{\circ}}\)
\(=\dfrac{\sin 17^{\circ} \cos 30^{\circ}+\cos 17^{\circ} \sin 30^{\circ}-\sin 17^{\circ} \cos 30^{\circ}}{\cos 17^{\circ}}\)
\(=\dfrac{\cos 17^{\circ} \sin 30^{\circ}}{\cos 17^{\circ}}\)
\(=\sin 30^∘=\dfrac{\sqrt{1}}{2}\)．
点拨 注意观察角度之间和、差或倍数等的关系.
 

【巩固练习】

1.\(\sin 59^∘·\cos 89^∘-\cos 59^∘·\sin 89^∘\)的值为(　　)
 A．\(-\dfrac{1}{2}\) \(\qquad \qquad \qquad \qquad\) B．\(\dfrac{1}{2}\) \(\qquad \qquad \qquad \qquad\) C．\(-\dfrac{\sqrt{3}}{2}\) \(\qquad \qquad \qquad \qquad\) D．\(-\sqrt{3}\)
 

2.求\(\cos 15°\)，\(\sin ⁡ 105°\)的值.
 
 

3.求\(\dfrac{2 \cos 10^{\circ}-\sin 20^{\circ}}{\sin 70^{\circ}}\) 的值.
 
 

4.求\(\dfrac{1+\tan 15^{\circ}}{1-\tan 15^{\circ}}\)的值.
 
 

5.求值：\(\dfrac{\sin 7^{\circ}+\cos 15^{\circ} \cdot \sin 8^{\circ}}{\cos 7^{\circ}-\sin 15^{\circ} \cdot \sin 8^{\circ}}\)．
 
 

参考答案

1. 答案 \(A\)
   解析 \(\sin 59^∘·\cos 89^∘-\cos 59^∘·\sin 89^∘=\sin (59^∘-89^∘ )=\sin (-30^∘ )=-\dfrac{1}{2}\)．

2. 答案 \(\cos 15^{\circ}=\dfrac{\sqrt{6}+\sqrt{2}}{4}\)； \(\sin 105^{\circ}=\dfrac{\sqrt{6}+\sqrt{2}}{4}\).
   解析 \(\cos 15°=\cos ⁡(45°-30°)=\cos ⁡ 45° \cos 30°+\sin ⁡ 45° \sin ⁡ 30°\) \(=\dfrac{\sqrt{2}}{2}\cdot \dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{2}\cdot \dfrac{1}{2}=\dfrac{\sqrt{6}+\sqrt{2}}{4}\)；
   \(\sin ⁡ 105° =\sin ⁡ (60°+45°) =\sin ⁡ 60° \cos ⁡ 45° +\cos ⁡ 60° \sin ⁡ 45°\)\(=\dfrac{\sqrt{3}}{2}\cdot \dfrac{\sqrt{2}}{2}+\dfrac{1}{2}\cdot \dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{6}+\sqrt{2}}{4}\).

3. 答案 \(\sqrt{3}\)
   解析 \(\dfrac{2 \cos 10^{\circ}-\sin 20^{\circ}}{\sin 70^{\circ}}=\dfrac{2 \cos \left(30^{\circ}-20^{\circ}\right)-\sin 20^{\circ}}{\sin 70^{\circ}}=\dfrac{2\left(\dfrac{\sqrt{3}}{2} \cos 20^{\circ}+\dfrac{1}{2} \sin 20^{\circ}\right)-\sin 20^{\circ}}{\sin 70^{\circ}}\)\(=\dfrac{\sqrt{3} \cos 20^{\circ}}{\sin 70^{\circ}}=\sqrt{3}\).

4. 答案 \(\sqrt{3}\)
   解析 \(\dfrac{1+\tan 15^{\circ}}{1-\tan 15^{\circ}}=\dfrac{\tan 45^{\circ}+\tan 15^{\circ}}{1-\tan 45^{\circ} \cdot \tan 15^{\circ}}=\tan \left(45^{\circ}+15^{\circ}\right)=\tan 60^{\circ}=\sqrt{3}\).

5. 答案 \(2-\sqrt{3}\)
   解析 原式 \(=\dfrac{\sin \left(15^{\circ}-8^{\circ}\right)+\cos 15^{\circ} \cdot \sin 8^{\circ}}{\cos \left(15^{\circ}-8^{\circ}\right)-\sin 15^{\circ} \cdot \sin 8^{\circ}}=\dfrac{\sin 15^{\circ} \cdot \cos 8^{\circ}}{\cos 15^{\circ} \cdot \cos 8^{\circ}}=\tan 15^{\circ}\)
   \(=\tan \left(45^{\circ}-30^{\circ}\right)=\dfrac{\tan 45^{\circ}-\tan 30^{\circ}}{1+\tan 45^{\circ} \tan 30^{\circ}}=\dfrac{1-\dfrac{\sqrt{3}}{3}}{1+\dfrac{\sqrt{3}}{3}}=2-\sqrt{3}\)．
    

【题型2】 给值求值问题

【典题1】 已知\(\sin ⁡α=\dfrac{4}{5}\)，\(α∈(\dfrac{\pi}{2},π)\)，\(\cos β=-\dfrac{5}{13}\)，\(β\)是第三象限角，求\(\cos (α+β)\)，\(\tan (α+β)\)的值．
解析 由\(\sin α=\dfrac{4}{5}\)，\(α∈(\dfrac{\pi}{2},π)\)，得 \(\cos \alpha=-\sqrt{1-\sin ^2 \alpha}=-\sqrt{1-\left(\dfrac{4}{5}\right)^2}=-\dfrac{3}{5}\)．
又由\(\cos β=-\dfrac{5}{13}\)，\(β\)为第三象限角，得 \(\sin \beta=-\sqrt{1-\cos ^2 \beta}=-\sqrt{1-\left(-\dfrac{5}{13}\right)^2}=-\dfrac{12}{13}\)，
\(\therefore \cos (\alpha+\beta)=\cos \alpha \cdot \cos \beta-\sin \alpha \cdot \sin \beta=\left(-\dfrac{3}{5}\right) \times\left(-\dfrac{5}{13}\right)-\dfrac{4}{5} \times\left(-\dfrac{12}{13}\right)=\dfrac{63}{65}\)，
\(\sin (\alpha+\beta)=\sin \alpha \cdot \cos \beta+\cos \alpha \cdot \sin \beta=\dfrac{4}{5} \times\left(-\dfrac{5}{13}\right)+\left(-\dfrac{3}{5}\right) \times\left(-\dfrac{12}{13}\right)=\dfrac{16}{65}\)，
\(\therefore \tan (\alpha+\beta)=\dfrac{\sin (\alpha+\beta)}{\cos (\alpha+\beta)}=\dfrac{16}{63}\)．
也可由\(\cos α=-\dfrac{3}{5}\)，\(\sin α=\dfrac{4}{5}\)，得\(\tan α=-\dfrac{4}{3}\)．
由\(\sin β=-\dfrac{12}{13}\)，\(\cos β=-\dfrac{5}{13}\)，得 \(\tan \beta=\dfrac{12}{5}\)，
\(\therefore \tan (\alpha+\beta)=\dfrac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=\dfrac{16}{63}\)．
点拨 注意结合三角形函数的关系\(\sin ^2⁡x+\cos ^2⁡x=1\)、 \(\tan x=\dfrac{\sin x}{\cos x}\) .
 

【典题2】 已知\(\sin \left(\alpha+\dfrac{\pi}{4}\right)=\dfrac{4}{5}\)，且\(α+\dfrac{\pi}{4}\)为第二象限角，求\(\cos α\)的值.
解析 依题意可得\(\cos \left(\alpha+\dfrac{\pi}{4}\right)=-\dfrac{3}{5}\)，
\(\therefore \cos \alpha=\cos \left[\left(\alpha+\dfrac{\pi}{4}\right)-\dfrac{\pi}{4}\right]=\cos \left(\alpha+\dfrac{\pi}{4}\right) \cos \dfrac{\pi}{4}-\sin \left(\alpha+\dfrac{\pi}{4}\right) \sin \dfrac{\pi}{4}\)
\(=-\dfrac{3}{5} \times \dfrac{\sqrt{2}}{2}+\dfrac{4}{5} \times \dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{2}}{10}\).
点拨 本题注意到了\(α+\dfrac{\pi}{4}\)与\(α\)的关系：\(α=\left(\alpha+\dfrac{\pi}{4}\right)-\dfrac{\pi}{4}\)，解题过程挺顺畅的，若把\(\sin \left(\alpha+\dfrac{\pi}{4}\right)\)展开得到\(\dfrac{\sqrt{2}}{2}(\sin ⁡α+co s⁡α)=\dfrac{4}{5}\)，也可求\(\cos α\)，但略显麻烦.
 

【典题3】 已知\(\cos ⁡(α-β)=-\dfrac{12}{13}\)，\(\cos (α+β)=\dfrac{12}{13}\)，且\(α-β∈\left(\dfrac{\pi}{2},π \right)\)，\(α+β∈\left(\dfrac{3\pi}{2},2π \right)\)，求角\(β\)的值．
解析 由\(α-β∈\left(\dfrac{\pi}{2},π \right)\)，且\(\cos ⁡(α-β)=-\dfrac{12}{13}\)，得\(\sin (α-β)=\dfrac{5}{13}\)．
由\(α+β∈\left(\dfrac{3\pi}{2},2π \right)\)，且\(\cos (α+β)=\dfrac{12}{13}\),得\(\sin (α+β)=-\dfrac{5}{13}\)．
\(∴\cos ⁡2β=\cos ⁡[(α+β)-(α-β)]=\cos ⁡(α+β) ·\cos ⁡(α-β)+\sin ⁡(α+β)·\sin ⁡(α-β)\)
\(=-\dfrac{12}{13}×\dfrac{12}{13}+\left(-\dfrac{5}{13} \right)×\dfrac{5}{13}=-1\)．
又\(α-β∈\left(\dfrac{\pi}{2},π \right)\)，\(α+β∈\left(\dfrac{3\pi}{2},2π \right)\)，\(∴2β∈\left(\dfrac{\pi}{2},\dfrac{3\pi}{2} \right)\)，
\(∴2β=π\)，\(∴β=\dfrac{\pi}{2}\)．
点拨
1.注意已知角\(α-β\)、\(α+β\)与所求角\(β\)之间的关系：\(2β=(α+β)-(α-β)\)，往它们的和、差或倍数的关系去思考；
2.\(\cos ⁡2β=-1\)得\(2β=(2k+1)π\)，注意角度的取值范围.
 

【巩固练习】

1.若\(\sin α=\dfrac{3}{5}\)，且\(α∈\left(\dfrac{\pi}{2},π \right)\)，则\(\tan \left(α+\dfrac{\pi}{4} \right)=\)(　　)
 A．\(-\dfrac{3}{4}\) \(\qquad \qquad \qquad \qquad\) B．\(\dfrac{3}{4}\) \(\qquad \qquad \qquad \qquad\) C．\(7\) \(\qquad \qquad \qquad \qquad\) D．\(\dfrac{1}{7}\)
 

2.已知\(α∈(0,π)\)，\(\cos \left(α+\dfrac{\pi}{6} \right)=\dfrac{3}{5}\)，则\(\sin α\)的值为(　　)
 A．\(\dfrac{4 \sqrt{3}-3}{10}\) \(\qquad \qquad \qquad\) B．\(\dfrac{3 \sqrt{3}-4}{10}\) \(\qquad \qquad \qquad\) C．\(\dfrac{7}{10}\) \(\qquad \qquad \qquad\) D． \(\dfrac{2 \sqrt{3}}{5}\)
 

3.已知\(\sin θ+\sin \left(θ+\dfrac{\pi}{3} \right)=1\)，则\(\sin \left(θ+\dfrac{\pi}{6} \right)=\)(　　)
 A．\(\dfrac{1}{2}\) \(\qquad \qquad \qquad \qquad\) B．\(\dfrac{\sqrt{3}}{3}\) \(\qquad \qquad \qquad \qquad\) C．\(\dfrac{2}{3}\) \(\qquad \qquad \qquad \qquad\) D．\(\dfrac{\sqrt{2}}{2}\)
 

4.在\(△ABC\)中，已知\(\cos B=\dfrac{4}{5}\)，\(\sin C=\dfrac{5}{13}\)，\(\cos A=\)(　　)
 A．\(\dfrac{63}{65}\) \(\qquad \qquad \qquad \qquad\) B．\(-\dfrac{33}{65}\) \(\qquad \qquad \qquad \qquad\) C．\(\dfrac{63}{65}\)或\(-\dfrac{33}{65}\) \(\qquad \qquad \qquad \qquad\) D．以上答案都不对
 
5.已知\(α\)，\(β\)是锐角，且 \(\sin \alpha=\dfrac{4 \sqrt{3}}{7}\)， \(\cos (\alpha+\beta)=-\dfrac{11}{14}\)，求\(\sin β\)的值．
 
 

6.已知\(\tan α=2\)，\(\tan β=3\)，且\(α\)，\(β\)都是锐角，求\(α+β\)的值．
 
 

参考答案

1. 答案 \(D\)
   解析 若\(\sin α=\dfrac{3}{5}\)，且\(α∈\left(\dfrac{\pi}{2},π \right)\)，
   则 \(\cos \alpha=-\sqrt{1-\sin ^2 \alpha}=-\sqrt{1-\left(\dfrac{3}{5}\right)^2}=-\dfrac{4}{5}\)，
   所以 \(\tan \alpha=\dfrac{\sin \alpha}{\cos \alpha}=\dfrac{\dfrac{3}{5}}{-\dfrac{4}{5}}=-\dfrac{3}{4}\)，
   所以 \(\tan \left(\alpha+\dfrac{\pi}{4}\right)=\dfrac{\tan \alpha+\tan \dfrac{\pi}{4}}{1-\tan \alpha \tan \dfrac{\pi}{4}}=\dfrac{-\dfrac{3}{4}+1}{1-\left(-\dfrac{3}{4}\right) \times 1}=\dfrac{1}{7}\)．
   故选：\(D\)．

2. 答案 \(A\)
   解析 \(∵α∈(0,π)\),\(\cos \left(α+\dfrac{\pi}{6} \right)=\dfrac{3}{5}\)，\(∴\sin \left(α+\dfrac{\pi}{6} \right)=\dfrac{4}{5}\)，
   \(\therefore \sin \alpha=\sin \left[\left(\alpha+\dfrac{\pi}{6}\right)-\dfrac{\pi}{6}\right]=\dfrac{4}{5} \times \dfrac{\sqrt{3}}{2}-\dfrac{1}{2} \times \dfrac{3}{5}=\dfrac{4 \sqrt{3}-3}{10}\)．
   故选：\(A\)．

3. 答案 \(B\)
   解析 \(∵\sin θ+\sin \left(θ+\dfrac{\pi}{3} \right)=1\)，\(∴\sin θ+\dfrac{1}{2} \sin θ+\dfrac{\sqrt{3}}{2} \cos θ=1\)，
   即\(\dfrac{3}{2} \sin θ+\dfrac{\sqrt{3}}{2} \cos θ=1\)，得\(\sqrt{3}\left(\dfrac{1}{2} \cos θ+\dfrac{\sqrt{3}}{2} \sin θ \right)=1\)，
   即\(\sqrt{3} \sin \left(θ+\dfrac{\pi}{6} \right)=1\)，得\(\sin \left(θ+\dfrac{\pi}{6} \right)=\dfrac{\sqrt{3}}{3}\)，
   故选：\(B\)．

4. 答案 \(B\)
   解析 在\(△ABC\)中，由\(\cos B=\dfrac{4}{5}\)，得 \(\sin B=\sqrt{1-\cos ^2 B}=\dfrac{3}{5}\)，
   又\(\sin C=\dfrac{5}{13}\)，且\(\sin B>\sin C\)，
   \(∴\)角\(C\)为锐角，得 \(\cos C=\sqrt{1-\sin ^2 C}=\dfrac{12}{13}\)，
   \(\therefore \cos A=-\cos (B+C)=-[\cos B \cos C-\sin B \sin C]=-\left[\dfrac{4}{5} \times \dfrac{12}{13}-\dfrac{3}{5} \times \dfrac{5}{13}\right]=-\dfrac{33}{65}\)．
   故选：\(B\)．

5. 答案 \(\dfrac{\sqrt{3}}{2}\)
   解析 \(∵α\)是锐角，且 \(\sin \alpha=\dfrac{4 \sqrt{3}}{7}\)，
   \(\therefore \cos \alpha=\sqrt{1-\sin ^2 \alpha}=\sqrt{1-\left(\dfrac{4 \sqrt{3}}{7}\right)^2}=\dfrac{1}{7}\)．
   又 \(\cos (\alpha+\beta)=-\dfrac{11}{14}\)，\(α\)，\(β\)均为锐角，
   \(\therefore \sin (\alpha+\beta)=\sqrt{1-\cos ^2(\alpha+\beta)}=\dfrac{5 \sqrt{3}}{14}\)，
   \(∴\sin ⁡β=\sin ⁡(α+β-α)=\sin ⁡(α+β)·\cos ⁡α-\cos ⁡(α+β)·\sin ⁡α\)
   \(=\dfrac{5 \sqrt{3}}{14} \times \dfrac{1}{7}-\left(-\dfrac{11}{14}\right) \times \dfrac{4 \sqrt{3}}{7}=\dfrac{\sqrt{3}}{2}\)．

6. 答案 \(135°\)
   解析 \(\tan (\alpha+\beta)=\dfrac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=\dfrac{2+3}{1-2 \times 3}=-1\)．
   又\(α\)，\(β\)均为锐角，\(∴0^∘<α+β<180^∘\)，
   \(∴α+β=135°\)．
    

【题型3】辅助角公式

【典题1】 化下列代数式为一个角的三角函数
  (1) \(\sqrt{3} \sin ⁡α+\cos ⁡α\)； \(\qquad \qquad\) (2)\(2 \sin \left(\dfrac{\pi}{4}-α \right)-2\cos \left(\dfrac{\pi}{4}-α \right)\)；
解析 (1)\(\sqrt{3} \sin ⁡α+\cos ⁡α=2\left(\dfrac{\sqrt{3}}{2} \sin ⁡α+\dfrac{1}{2} \cos ⁡α \right)\)
\(=2\left(\sin ⁡α \cos ⁡ \dfrac{\pi}{6} +\cos ⁡α \sin ⁡ \dfrac{\pi}{6} \right)=2 \sin ⁡ \left(α+\dfrac{\pi}{6} \right)\)；
(2) \(2 \sin \left(\dfrac{\pi}{4}-α \right)-2 \cos \left(\dfrac{\pi}{4}-α \right)=2\left[\sin \left(\dfrac{\pi}{4}-α \right)-2 \cos \left(\dfrac{\pi}{4}-α \right) \right]\)
\(=4\left[\dfrac{\sqrt{2}}{2} \sin \left(\dfrac{\pi}{4}-α \right)-\dfrac{\sqrt{2}}{2} \cos \left(\dfrac{\pi}{4}-α \right) \right]=4\left[\sin \left(\dfrac{\pi}{4}-α \right) \cos ⁡ \dfrac{\pi}{4} -\cos ⁡ \left(\dfrac{\pi}{4}-α \right) \sin ⁡ \dfrac{\pi}{4} \right]\)
\(=4 \sin ⁡\left(\dfrac{\pi}{4}-α-\dfrac{\pi}{4} \right) =4 \sin ⁡ (-α) =-4 \sin ⁡α\).
点拨 注意辅助角公式常见的特殊角情况
\(a:b=1:1\)型，配\(\dfrac{\pi}{4}\)：\(\sin x±\cos x=\sqrt{2} \sin \left(x±\dfrac{\pi}{4}\right)\)
\(a:b=\sqrt{3}:1\)型，配\(\dfrac{\pi}{6}\)或\(\dfrac{\pi}{3}\)：\(\sin x±\sqrt{3} \cos x=2 \sin \left(x±\dfrac{\pi}{3} \right)\)，\(\sqrt{3} \sin x±\cos x=2 \sin \left(x±\dfrac{\pi}{6} \right)\).
 

【典题2】 已知函数\(f(x)=\sqrt{3} \sin 3ωx-\cos 3ωx(ω>0)\)图象的相邻两条对称轴之间的距离为\(\dfrac{\pi}{2}\)，则\(f\left(\dfrac{\pi}{3} \right)=\) \(\underline{\quad \quad}\)．
解析 \(f(x)=\sqrt{3} \sin 3ωx-\cos 3ωx=2\sin \left(3ωx-\dfrac{\pi}{6} \right)\)，
又\(∵\)函数\(f(x)\)图象的相邻两条对称轴之间的距离为\(\dfrac{\pi}{2}\)，
\(\therefore \dfrac{2 \pi}{3 \omega}=\pi\)，解得： \(\omega=\dfrac{2}{3}\)，
\(\therefore f\left(\dfrac{\pi}{3}\right)=2 \sin \left(3 \times \dfrac{2}{3} \times \dfrac{\pi}{3}-\dfrac{\pi}{6}\right)=2\)．
点拨 处理函数\(f(x)=a\sin x+b\cos x\)的基本性质，先转化为\(f(x)=A\sin (x+φ)\)的形式.
 

【巩固练习】

1.化下列代数式为一个角的三角函数
  (1)\(\sin ⁡α+\cos ⁡α\)；\(\qquad \qquad\) (2) \(\sin \left(\dfrac{\pi}{6}+α \right)+\sqrt{3} \cos \left(\dfrac{\pi}{6}+α \right)\)；

2.已知函数\(f(x)=\left|\sqrt{3} \sin ωx-\cos ωx \right|(ω>0)\)的最小正周期为\(π\)，则\(ω=\) \(\underline{\quad \quad}\) .
 

3.求\(f(x)=\sin ⁡x-\sqrt{3} \cos ⁡x\)在\(\left[-\dfrac{\pi}{2},π \right]\)的值域 .
 
 

4.已知\(f(x)=\sin ⁡(π-2x)+\sin \left(\dfrac{\pi}{2}+2x \right)\)．
(1)化简\(f(x)\)并求函数\(f(x)\)图象的对称轴方程；
(2)当\(x∈\left[\dfrac{\pi}{4},\dfrac{3\pi}{4} \right]\)时，求函数\(f(x)\)的最大值和最小值．
 
 

参考答案

1. 答案 (1) \(\sin ⁡ \left(α+\dfrac{\pi}{4}\right)\)；(2) \(2 \cos ⁡α\)
   解析 (1)\(\sin ⁡α+\cos ⁡α=\sqrt{2} \left(\dfrac{\sqrt{2}}{2} \sin ⁡α+\dfrac{\sqrt{2}}{2} \cos ⁡α \right)\)
   \(=\sqrt{2} \left(\sin ⁡α \cos ⁡ \dfrac{\pi}{4} +\cos ⁡α \sin ⁡ \dfrac{\pi}{4} \right)=\sqrt{2} \sin ⁡ \left(α+\dfrac{\pi}{4} \right)\)；
   (2) \(\sin ⁡\left(\dfrac{\pi}{6}+α \right)+\sqrt{3} \cos ⁡\left(\dfrac{\pi}{6}+α \right)=2[\dfrac{1}{2} \sin ⁡\left(\dfrac{\pi}{6}+α \right)+\dfrac{\sqrt{3}}{2} \cos ⁡\left(\dfrac{\pi}{6}+α) \right]\)
   \(=2\left[\sin \left(\dfrac{\pi}{6}+α \right) \cos ⁡\dfrac{\pi}{3} +\cos ⁡ \left(\dfrac{\pi}{6}+α \right) \sin ⁡\dfrac{\pi}{3} \right]=2 \sin ⁡ \left(\dfrac{\pi}{6}+α+\dfrac{\pi}{3} \right)\)\(=2 \sin ⁡ \left(\dfrac{\pi}{2}+α \right) =2 \cos ⁡α\).

2. 答案 \(1\)
   解析 因为函数\(f(x)=\left|\sqrt{3} \sin ωx-\cos ωx \right|=\left|2\sin \left(ωx-\dfrac{\pi}{6} \right) \right|\)；
   故其最小正周期为 \(\dfrac{1}{2} \times \dfrac{2 \pi}{\omega}=\pi \Rightarrow \omega=1\).

3. 答案 \((-1,2]\)
   解析 \(f(x)=\sin ⁡x-\sqrt{3} \cos ⁡x=2 \sin ⁡ \left(x-\dfrac{\pi}{3} \right)\)
   \(∵-\dfrac{\pi}{2}<x<π\)，\(∴-\dfrac{5\pi}{6}<x-\dfrac{\pi}{3} <\dfrac{2\pi}{3}\)，
   \(∴-\dfrac{1}{2}<\sin ⁡ \left(x-\dfrac{\pi}{3} \right) ≤1\)，\(∴-1<2 \sin ⁡ \left(x-\dfrac{\pi}{3} \right) ≤2\)，
   即函数\(y=f(x)\)的值域为\((-1,2]\).

4. 答案 (1)\(f(x)=\sqrt{2} \sin ⁡\left(2x+\dfrac{\pi}{4} \right)\)，\(x=\dfrac{k\pi}{2}+\dfrac{\pi}{8}\) ,\(k∈Z\)；
   (2) \(f(x)_{\max} =1\)， \(f(x)_{\min}=-\sqrt{2}\).
   解析 (1)由已知可得\(f(x)=\sin ⁡2x+\cos ⁡2x=\sqrt{2} \sin ⁡\left(2x+\dfrac{\pi}{4} \right)\)，
   令\(2x+\dfrac{\pi}{4}=kπ+\dfrac{\pi}{2}\),\(k∈Z\)，解得\(x=\dfrac{k\pi}{2}+\dfrac{\pi}{8}\),\(k∈Z\)，即为函数\(f(x)\)的对称轴方程；
   (2)当\(x∈\left[\dfrac{\pi}{4},\dfrac{3\pi}{4} \right]\)时，\(2x+\dfrac{\pi}{4}∈\left[\dfrac{3\pi}{4},\dfrac{7\pi}{4} \right]\)，
   所以当\(2x+\dfrac{\pi}{4}=\dfrac{3\pi}{4}\)，即\(x=\dfrac{\pi}{4}\)时， \(f(x) \max =\sqrt{2} \times \dfrac{\sqrt{2}}{2}=1\)，
   当\(2x+\dfrac{\pi}{4}=\dfrac{3\pi}{2}\)，即\(x=\dfrac{5\pi}{8}\) 时， \(f(x) \min =\sqrt{2} \times(-1)=-\sqrt{2}\)．
    

分层练习

【A组—基础题】

1.\(\sin 80^{\circ} \cos 50^{\circ}+\cos 140^{\circ} \sin 10^{\circ}=\)(　　)
 A．\(-\dfrac{\sqrt{3}}{2}\) \(\qquad \qquad \qquad \qquad\) B．\(\dfrac{\sqrt{3}}{2}\) \(\qquad \qquad \qquad \qquad\) C．\(-\dfrac{1}{2}\) \(\qquad \qquad \qquad \qquad\) D．\(\dfrac{1}{2}\)
 

2.设\(α∈\left(0,\dfrac{\pi}{2} \right)\)，若\(\sin ⁡α=\dfrac{3}{5}\)，则\(\sqrt{2} \cos \left(α+\dfrac{\pi}{4} \right)=\) (　　)
 A．\(\dfrac{7}{5}\) \(\qquad \qquad \qquad \qquad\) B．\(\dfrac{1}{5}\) \(\qquad \qquad \qquad \qquad\) C．\(-\dfrac{7}{5}\) \(\qquad \qquad \qquad \qquad\) D．\(-\dfrac{1}{5}\)
 

3.在\(△ABC\)中，\(A=\dfrac{\pi}{4}\)， \(\cos B=\dfrac{\sqrt{10}}{10}\)，则\(\sin C=\)(　　)
 A．\(-\dfrac{\sqrt{5}}{5}\) \(\qquad \qquad \qquad \qquad\) B．\(\dfrac{\sqrt{5}}{5}\) \(\qquad \qquad \qquad \qquad\) C． \(-\dfrac{2\sqrt{5}}{5}\) \(\qquad \qquad \qquad \qquad\) D． \(\dfrac{2\sqrt{5}}{5}\)
 

4.已知\(0<α<β<\dfrac{\pi}{2}\)，且 \(\cos (\alpha-\beta)=\dfrac{63}{65}\), \(\sin \beta=\dfrac{12}{13}\)，则\(\sin α=\)(　　)
 A．\(-\dfrac{3}{5}\) \(\qquad \qquad \qquad \qquad\) B．\(\dfrac{3}{5}\) \(\qquad \qquad \qquad \qquad\) C．\(-\dfrac{4}{5}\) \(\qquad \qquad \qquad \qquad\) D．\(\dfrac{4}{5}\)
 

5.已知：\(α\),\(β\)均为锐角，\(\tan α=\dfrac{1}{2}\)，\(\tan β= \dfrac{1}{3}\)，则\(α+β=\)(　　)
 A．\(\dfrac{\pi}{6}\) \(\qquad \qquad \qquad \qquad\) B．\(\dfrac{\pi}{4}\) \(\qquad \qquad \qquad \qquad\) C．\(\dfrac{\pi}{3}\) \(\qquad \qquad \qquad \qquad\) D．\(\dfrac{5\pi}{12}\)
 

6.若\(0<α<\dfrac{\pi}{2}\)，\(-\dfrac{\pi}{2}<β<0\)，\(\cos \left(\dfrac{\pi}{4}+α \right)= \dfrac{1}{3}\) ，\(\cos \left(\dfrac{\pi}{4}-\dfrac{β}{2} \right)=\dfrac{\sqrt{3}}{3}\)，则\(\cos \left(α+\dfrac{β}{2} \right)=\) (　　)
 A．\(\dfrac{\sqrt{3}}{3}\) \(\qquad \qquad \qquad \qquad\) B．\(-\dfrac{\sqrt{3}}{3}\) \(\qquad \qquad \qquad \qquad\) C． \(\dfrac{5 \sqrt{3}}{9}\) \(\qquad \qquad \qquad \qquad\) D． \(-\dfrac{\sqrt{6}}{9}\)
 

7.\(A\),\(B\),\(C\)是\(△ABC\)的内角，其中\(B=\dfrac{2\pi}{3}\) ，则\(\sin A+\sin C\)的取值范围\(\underline{\quad \quad}\) .
 

8.已知\(\tan ⁡α= \dfrac{1}{3}\)，\(\tan ⁡(β-α)=-2\)，且\(\dfrac{\pi}{2}<β<π\)，则\(β=\)\(\underline{\quad \quad}\)．
 

9.已知\(\tan (α+β)=3\)，\(\tan \left(α+\dfrac{\pi}{4} \right)=2\)，那么\(\tan β=\)\(\underline{\quad \quad}\)　　．
 

10.若函数\(f(x)=\sin 2x-\sqrt{3} \cos 2x\)在\([0,t]\)上的值域为\([-\sqrt{3},2]\)，则\(t\)的取值范围为\(\underline{\quad \quad}\)．
 

11.已知函数\(f(x)=\sin \left(2x+\dfrac{\pi}{6} \right)+\sin \left(2x-\dfrac{\pi}{6} \right)+\cos ⁡2x-1\)．
  (1)求\(f(x)\)的最小正期；
  (2)当\(x∈\left[0,\dfrac{\pi}{4} \right]\)时，求\(f(x)\)的单调区间；
  (3)在(2)的件下，求\(f(x)\)的最小值，以及取得最小值时相应自变量\(x\)的取值范围.
 
 
 

参考答案

1. 答案 \(D\)
   解析 \(\sin 80^{\circ} \cos 50^{\circ}+\cos 140^{\circ} \sin 10^{\circ}=\cos 10^{\circ} \cos 50^{\circ}-\sin 50^{\circ} \sin 10^{\circ}\)\(=\cos (50^{\circ}+10^{\circ})=\cos 60^{\circ}=\dfrac{1}{2}\)．
   故选：\(D\)．

2. 答案 \(B\)
   解析 \(\cos α=\sqrt{1-\sin ^2⁡α}=\dfrac{4}{5}\)，原式\(=\cos α-\sin α=\dfrac{4}{5}-\dfrac{3}{5}=\dfrac{1}{5}\)．

3. 答案 \(D\)
   解析 \(\sin B=\dfrac{3 \sqrt{10}}{10}\)，\(\sin C=\sin ⁡(A+B)=\sin A·\cos B+\cos A·\sin B=\dfrac{2\sqrt{5}}{5}\)．

4. 答案 \(D\)
   解析 \(∵\)已知\(0<α<β<\dfrac{\pi}{2}\)，且 \(\cos (\alpha-\beta)=\dfrac{63}{65}\), \(\sin \beta=\dfrac{12}{13}\)，
   \(∴α-β∈\left(-\dfrac{\pi}{2},0 \right)\)， \(\sin (\alpha-\beta)=-\sqrt{1-\cos ^2(\alpha-\beta)}=-\dfrac{16}{65}\)， \(\cos \beta=\sqrt{1-\sin ^2 \beta}=\dfrac{5}{13}\)．
   \(∴\sin α=\sin [(α-β)+β]=\sin (α-β)\cos β+\cos (α-β)\sin β\)
   \(=-\dfrac{16}{65} \cdot \dfrac{5}{13}+\dfrac{63}{65} \cdot \dfrac{12}{13}=\dfrac{676}{845}=\dfrac{4}{5}\)，
   故选：\(D\)．

5. 答案 \(B\)
   解析 由于\(α\),\(β\)均为锐角，\(\tan α=\dfrac{1}{2}\)，\(\tan β= \dfrac{1}{3}\)，
   所以\(0<α+β<\dfrac{\pi}{2}+\dfrac{\pi}{2}=π\)．
   所以 \(\tan (\alpha+\beta)=\dfrac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}=\dfrac{\dfrac{1}{2}+\dfrac{1}{3}}{1-\dfrac{1}{6}}=1\)，所以\(α+β=\dfrac{\pi}{4}\)．
   故选：\(B\)．

6. 答案 \(C\)
   解析 \(∵\cos (\dfrac{\pi}{4}+α)= \dfrac{1}{3}\) ，\(0<α<\dfrac{\pi}{2}\)， \(\therefore \sin \left(\dfrac{\pi}{4}+\alpha\right)=\dfrac{2 \sqrt{2}}{3}\)，
   又\(∵\cos (\dfrac{\pi}{4}-\dfrac{β}{2})=\dfrac{\sqrt{3}}{3}\)，\(-\dfrac{\pi}{2}<β<0\)，
   \(\therefore \sin \left(\dfrac{\pi}{4}-\dfrac{\beta}{2}\right)=\dfrac{\sqrt{6}}{3}\)，
   \(\therefore \cos \left(\alpha+\dfrac{\beta}{2}\right)=\cos \left[\left(\dfrac{\pi}{4}+\alpha\right)-\left(\dfrac{\pi}{4}-\dfrac{\beta}{2}\right)\right]\)
   \(=\cos ⁡ \left(\dfrac{\pi}{4}+α \right)\cdot \cos \left(\dfrac{\pi}{4}-\dfrac{β}{2} \right) +\sin \left(\dfrac{\pi}{4}+α \right) \sin \left(\dfrac{\pi}{4}-\dfrac{β}{2} \right)\)
   \(=\dfrac{1}{3} \times \dfrac{\sqrt{3}}{3}+\dfrac{2 \sqrt{2}}{3} \times \dfrac{\sqrt{6}}{3}=\dfrac{5 \sqrt{3}}{9}\)，故选\(C\)．

7. 答案 \((\dfrac{\sqrt{3}}{2},1]\)
   解析 \(\sin A+\sin C=\sin A+\sin \left(\dfrac{\pi}{3} -A \right)=\sin A+\dfrac{\sqrt{3}}{2} \cos A-\dfrac{1}{2} \sin A\)
   \(=\dfrac{1}{2} \sin A+\dfrac{\sqrt{3}}{2} \cos A=\sin \left(A+\dfrac{\pi}{3} \right)\)．
   \(∵A∈\left(0,\dfrac{\pi}{3} \right )\)，\(∴A+\dfrac{\pi}{3} ∈\left(\dfrac{\pi}{3} ,\dfrac{2\pi}{3} \right)\)，
   \(∴\sin \left(A+\dfrac{\pi}{3} \right)∈\left(\dfrac{\sqrt{3}}{2},1 \right]\).

8. 答案 \(\dfrac{3\pi}{4}\)
   解析 \(\tan \beta=\tan [\alpha+(\beta-\alpha)]=\dfrac{\tan \alpha+\tan (\beta-\alpha)}{1-\tan \alpha \cdot \tan (\beta-\alpha)}=\dfrac{\dfrac{1}{3}-2}{1+\dfrac{2}{3}}=-1\)．
   又\(∵\dfrac{\pi}{2}<β<π\)，\(∴β=\dfrac{3\pi}{4}\)．

9. 答案 \(\dfrac{4}{3}\)
   解析 \(∵\tan \left(α+\dfrac{\pi}{4} \right)=2\)， \(\therefore \dfrac{1+\tan \alpha}{1-\tan \alpha}=2\)，解得\(\tan α= \dfrac{1}{3}\)；
   又\(\tan (α+β)=3\)，\(\tan \left(α+\dfrac{\pi}{4} \right)=2\)，
   \(\therefore \tan \beta=\tan [(\alpha+\beta)-\alpha]=\dfrac{\tan (\alpha+\beta)-\tan \alpha}{1+\tan (\alpha+\beta) \tan \alpha}=\dfrac{3-\dfrac{1}{3}}{1+3 \times \dfrac{1}{3}}=\dfrac{4}{3}\)．

10. 答案 \(\left [\dfrac{5\pi}{12},\dfrac{5\pi}{6} \right]\)
    解析 \(f(x)=\sin 2x-\sqrt{3} \cos 2x=2\sin \left(2x-\dfrac{\pi}{3} \right)\)
    当\(x=0\)时，函数值是\(-\sqrt{3}\)；当\(x=\dfrac{5\pi}{12}\)时，函数值是\(2\)；当\(x=\dfrac{5\pi}{6}\)时，函数值是\(-\sqrt{3}\)；
    又函数在\(\left[0,\dfrac{5\pi}{12} \right]\)上增，在\(\left[\dfrac{5\pi}{12},\dfrac{5\pi}{6} \right]\)上减，可得\(t\)的取值范围\(\left[\dfrac{5\pi}{12},\dfrac{5\pi}{6} \right]\)．

11. 答案 (1) \(π\)；(2) 单调递增区间为\(\left[0,\dfrac{\pi}{6} \right]\)，单调递减区间为\(\left[\dfrac{\pi}{6},\dfrac{\pi}{4} \right]\)；(3)在\(x=0\)时取得最小值\(0\).
    解析 (1)\(f(x)=\sin ⁡2x\cos ⁡\dfrac{\pi}{6}+\cos ⁡2x\sin ⁡\dfrac{\pi}{6}+\sin ⁡2x\cos ⁡\dfrac{\pi}{6}-\cos ⁡2x\sin ⁡\dfrac{\pi}{6}+\cos ⁡2x-1\)
    \(=\sqrt{3} \sin ⁡2x+\cos ⁡2x-1=2\sin ⁡\left(2x+\dfrac{\pi}{6} \right)-1\)，
    \(T=\dfrac{2 \pi}{|\omega|}=\dfrac{2 \pi}{2}=\pi\)，故\(f(x)\)的最小正周期为\(π\)．
    (2)先求出增区间，即：令\(-\dfrac{\pi}{2}+2kπ≤2x+\dfrac{\pi}{6}≤\dfrac{\pi}{2}+2kπ,(k∈Z)\)，
    解得\(x∈\left[-\dfrac{\pi}{3} +kπ,\dfrac{\pi}{6}+kπ \right]\)，
    \(∴f(x)\)的单调递增区间为\(\left[-\dfrac{\pi}{3} +kπ,\dfrac{\pi}{6}+kπ \right]\)，
    \(f(x)\)的单调递减区间为\(\left[\dfrac{\pi}{6}+kπ,\dfrac{2\pi}{3} +kπ \right]\)，
    所以在区间\(\left[0,\dfrac{\pi}{4} \right]\)上，当\(x∈\left[0,\dfrac{\pi}{6} \right]\)时，函数\(f(x)\)单调递增，
    当\(x∈\left[\dfrac{\pi}{6},\dfrac{\pi}{4} \right]\)时，函数\(f(x)\)单调递减；
    所以\(f(x)\)的单调递增区间为\(\left[0,\dfrac{\pi}{6} \right]\)，单调递减区间为\(\left[\dfrac{\pi}{6},\dfrac{\pi}{4} \right]\)．
    (3)由(2)所得到的单调性可得\(f(0)=2\sin ⁡\dfrac{\pi}{6}-1=0\)，\(f\left(\dfrac{\pi}{4} \right)=2\sin \left(\dfrac{\pi}{2}+\dfrac{\pi}{6} \right)-1=\sqrt{3}\),
    所以\(f(x)\)在\(x=0\)时取得最小值\(0\)．
     

【B组—提高题】

1.已知\(α\),\(β∈(0,2π)\)，且满足\(\sin α-\cos α=\dfrac{1}{2}\)，\(\cos β-\sin β=\dfrac{1}{2}\)，则\(\sin (α+β)=\)(　　)
 A．\(1\) \(\qquad \qquad \qquad \qquad\) B．\(-\dfrac{\sqrt{2}}{2}\)或\(1\) \(\qquad \qquad \qquad \qquad\) C．\(-\dfrac{3}{4}\)或\(1\) \(\qquad \qquad \qquad \qquad\) D．\(1\)或\(-1\)
 

2.设\(α\),\(β∈\left(0,\dfrac{\pi}{2} \right)\)，\(\sin α\cos β=3\sin β\cos α\)，则\(α-β\)的最大值为(　　)
 A． \(\dfrac{\pi}{12}\)\(\qquad \qquad \qquad \qquad\) B．\(\dfrac{\pi}{6}\)\(\qquad \qquad \qquad \qquad\) C．\(\dfrac{\pi}{4}\) \(\qquad \qquad \qquad \qquad\) D．\(\dfrac{\pi}{3}\)
 

3.设当\(x=θ\)时，函数\(f(x)=\sin x+3\cos x\)取得最大值，则\(\cos \left(θ-\dfrac{\pi}{4} \right)=\) \(\underline{\quad \quad}\)．
 

参考答案

1. 答案 \(C\)
   解析 \(∵\sin α-\cos α=\dfrac{1}{2}\)，\(\sin ^2 α+\cos ^2 α=1\)，
   \(∴8\sin ^2 α-4\sin α-3=0\)，\(8\cos ^2 α+4\cos α-3=0\)
   又\(\cos β-\sin β=\dfrac{1}{2}\)，\(\sin ^2⁡β+\cos ^2⁡β=1\)，
   \(∴8 \cos ^2⁡β-4\cos β-3=0\)，\(8 \sin ^2⁡β+4\sin β-3=0\)，
   ①若\(\sin α=\cos β\)，则\(α+β=\dfrac{\pi}{2}\)或\(\dfrac{5\pi}{2}\)，此时\(\sin (α+β)=1\)，
   ②若\(\sin α≠\cos β\)，
   则\(\sin α\)，\(\cos β\)是方程\(8x^2-4x-3=0\)的根，故\(\sin α\cos β=-\dfrac{3}{8}\)，
   同时\(\cos α\)，\(\sin β\)是方程\(8x^2+4x-3=0\)的根，故\(\cos α\sin β=-\dfrac{3}{8}\)，
   故\(\sin (α+β)=\sin α\cos β+\cos α\sin β=-\dfrac{3}{4}\)，
   故\(\sin (α+β)\)的值是\(1\)或\(-\dfrac{3}{4}\)，
   故选：\(C\)．

2. 答案 \(B\)
   解析 由\(\sin α\cos β=3\sin β\cos α\)可得\(\tan α=3\tan β\)，
   \(∵α\),\(β∈\left(0,\dfrac{\pi}{2} \right)\)，
   所以 \(\tan (\alpha-\beta)=\dfrac{\tan \alpha-\tan \beta}{1+\tan \alpha \tan \beta}=\dfrac{2 \tan \beta}{1+3 \tan ^2 \beta}=\dfrac{2}{3 \tan \beta+\dfrac{1}{\tan \beta}}\)\(\leq \dfrac{2}{2 \sqrt{3 \tan \beta \cdot \dfrac{1}{\tan \beta}}}=\dfrac{\sqrt{3}}{3}\)，
   当且仅当\(3 \tan \beta=\dfrac{1}{\tan \beta}\)即\(\tan β=\dfrac{\sqrt{3}}{3}\)，\(\tan α=\sqrt{3}\)时取等号，
   此时\(α-β\)取得最大值\(\dfrac{\pi}{6}\)．
   故选：\(B\)．

3. 答案 \(\dfrac{2 \sqrt{5}}{5}\)
   解析 \(∵\)当\(x=θ\)时，函数 \(f(x)=\sin x+3 \cos x=\sqrt{10}\left(\dfrac{1}{\sqrt{10}} \sin x+\dfrac{3}{\sqrt{10}} \cos x\right)\)取得最大值，
   \(\therefore \cos \theta=\dfrac{3}{\sqrt{10}}\)， \(\sin \theta=\dfrac{1}{\sqrt{10}}\)，
   \(\therefore \sin \theta+3 \cos \theta=\sqrt{1+9}=\sqrt{10}\)，
   则 \(\cos \left(\theta-\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{2}(\cos \theta+\sin \theta) \dfrac{\sqrt{2}}{2} \times \dfrac{4}{\sqrt{10}}=\dfrac{2 \sqrt{5}}{5}\)．
    

【C组—拓展题】

1.在钝角三角形\(ABC\)中， \(\dfrac{1}{\tan A}+\dfrac{1}{\tan B}+2 \tan C=0\)，则\(\tan C\)的最大值是\(\underline{\quad \quad}\)．
 
 

2.已知锐角\(α\),\(β\)满足\(α-β=\dfrac{\pi}{3}\)，则 \(\dfrac{1}{\cos \alpha \cos \beta}+\dfrac{1}{\sin \alpha \sin \beta}\)的最小值为\(\underline{\quad \quad}\)．
 
 

参考答案

1. 答案 \(-\sqrt{3}\)
   解析 在钝角三角形\(ABC\)中， \(\dfrac{1}{\tan A}+\dfrac{1}{\tan B}+2 \tan C=0\)，
   可得 \(\dfrac{\tan A+\tan B}{\tan A \tan B}=-2 \tan C\)，
   即\(\tan A+\tan B=-2\tan A\tan B\tan C\)，
   则\(\tan (A+B)(1-\tan A\tan B)=-2\tan A\tan B\tan C\)，
   即\(-\tan C(1-\tan A\tan B)=-2\tan A\tan B\tan C\)，
   所以\(\tan A\tan B= \dfrac{1}{3}\)，
   则\(\tan C=-\tan (A+B)=-\dfrac{\tan A+\tan B}{1-\tan A \tan B}=-\dfrac{3}{2}(\tan A+\tan B)\)
   \(\leq-\dfrac{3}{2} \times 2 \sqrt{\tan A \tan B}=-\sqrt{3}\)．
   当且仅当\(\tan A=\tan B\)时，取等号，故\(\tan C\)的最大值是\(-\sqrt{3}\)．
   故答案为：\(-\sqrt{3}\)．

2. 答案 \(8\)
   解析 因为锐角\(α\),\(β\)满足\(α-β=\dfrac{\pi}{3}\)，
   所以\(\cos (α-β)=\cos α\cos β+\sin α\sin β=\dfrac{1}{2}\)，
   令\(x=\cos α\cos β\)，\(y=\sin α\sin β\)，
   则\(x+y=\dfrac{1}{2}\)，
   由题意得\(x>0\)，\(y>0\)，
   则 \(\dfrac{1}{\cos \alpha \cos \beta}+\dfrac{1}{\sin \alpha \sin \beta}=\dfrac{1}{x}+\dfrac{1}{y}=2(x+y)\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)\(=2\left(2+\dfrac{y}{x}+\dfrac{x}{y}\right) \geq 2\left(2+2 \sqrt{\dfrac{x}{y} \cdot \dfrac{y}{x}}\right)=8\)，
   当且仅当\(x=y\)时取等号，此时 \(\dfrac{1}{\cos \alpha \cos \beta}+\dfrac{1}{\sin \alpha \sin \beta}\)的最小值\(8\)．