5.5.1 两角和与差的正弦、余弦和正切公式 (1)
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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
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答案 \(A\)
解析 \(\sin 59^∘·\cos 89^∘-\cos 59^∘·\sin 89^∘=\sin (59^∘-89^∘ )=\sin (-30^∘ )=-\dfrac{1}{2}\). -
答案 \(\cos 15^{\circ}=\dfrac{\sqrt{6}+\sqrt{2}}{4}\); \(\sin 105^{\circ}=\dfrac{\sqrt{6}+\sqrt{2}}{4}\).
解析 \(\cos 15°=\cos (45°-30°)=\cos 45° \cos 30°+\sin 45° \sin 30°\) \(=\dfrac{\sqrt{2}}{2}\cdot \dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{2}\cdot \dfrac{1}{2}=\dfrac{\sqrt{6}+\sqrt{2}}{4}\);
\(\sin 105° =\sin (60°+45°) =\sin 60° \cos 45° +\cos 60° \sin 45°\)\(=\dfrac{\sqrt{3}}{2}\cdot \dfrac{\sqrt{2}}{2}+\dfrac{1}{2}\cdot \dfrac{\sqrt{2}}{2}=\dfrac{\sqrt{6}+\sqrt{2}}{4}\). -
答案 \(\sqrt{3}\)
解析 \(\dfrac{2 \cos 10^{\circ}-\sin 20^{\circ}}{\sin 70^{\circ}}=\dfrac{2 \cos \left(30^{\circ}-20^{\circ}\right)-\sin 20^{\circ}}{\sin 70^{\circ}}=\dfrac{2\left(\dfrac{\sqrt{3}}{2} \cos 20^{\circ}+\dfrac{1}{2} \sin 20^{\circ}\right)-\sin 20^{\circ}}{\sin 70^{\circ}}\)\(=\dfrac{\sqrt{3} \cos 20^{\circ}}{\sin 70^{\circ}}=\sqrt{3}\). -
答案 \(\sqrt{3}\)
解析 \(\dfrac{1+\tan 15^{\circ}}{1-\tan 15^{\circ}}=\dfrac{\tan 45^{\circ}+\tan 15^{\circ}}{1-\tan 45^{\circ} \cdot \tan 15^{\circ}}=\tan \left(45^{\circ}+15^{\circ}\right)=\tan 60^{\circ}=\sqrt{3}\). -
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\(\sin ^2x+\cos ^2x=1\)、 \(\tan x=\dfrac{\sin x}{\cos x}\) .
【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\(\dfrac{\sqrt{2}}{2}(\sin α+co sα)=\dfrac{4}{5}\),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
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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:\(D\). -
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:\(A\). -
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:\(B\). -
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:\(B\). -
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,
\(\therefore \sin (\alpha+\beta)=\sqrt{1-\cos ^2(\alpha+\beta)}=\dfrac{5 \sqrt{3}}{14}\),
\(∴\sin β=\sin (α+β-α)=\sin (α+β)·\cos α-\cos (α+β)·\sin α\)
\(=\dfrac{5 \sqrt{3}}{14} \times \dfrac{1}{7}-\left(-\dfrac{11}{14}\right) \times \dfrac{4 \sqrt{3}}{7}=\dfrac{\sqrt{3}}{2}\). -
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
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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) \(\sin \left(\dfrac{\pi}{6}+α \right)+\sqrt{3} \cos \left(\dfrac{\pi}{6}+α \right)=2[\dfrac{1}{2} \sin \left(\dfrac{\pi}{6}+α \right)+\dfrac{\sqrt{3}}{2} \cos \left(\dfrac{\pi}{6}+α) \right]\)
\(=2\left[\sin \left(\dfrac{\pi}{6}+α \right) \cos \dfrac{\pi}{3} +\cos \left(\dfrac{\pi}{6}+α \right) \sin \dfrac{\pi}{3} \right]=2 \sin \left(\dfrac{\pi}{6}+α+\dfrac{\pi}{3} \right)\)\(=2 \sin \left(\dfrac{\pi}{2}+α \right) =2 \cos α\). -
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 \(\dfrac{1}{2} \times \dfrac{2 \pi}{\omega}=\pi \Rightarrow \omega=1\). -
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\((-1,2]\). -
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
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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:\(D\). -
答案 \(B\)
解析 \(\cos α=\sqrt{1-\sin ^2α}=\dfrac{4}{5}\),原式\(=\cos α-\sin α=\dfrac{4}{5}-\dfrac{3}{5}=\dfrac{1}{5}\). -
答案 \(D\)
解析 \(\sin B=\dfrac{3 \sqrt{10}}{10}\),\(\sin C=\sin (A+B)=\sin A·\cos B+\cos A·\sin B=\dfrac{2\sqrt{5}}{5}\). -
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:\(D\). -
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:\(B\). -
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\(C\). -
答案 \((\dfrac{\sqrt{3}}{2},1]\)
解析 \(\sin A+\sin C=\sin A+\sin \left(\dfrac{\pi}{3} -A \right)=\sin A+\dfrac{\sqrt{3}}{2} \cos A-\dfrac{1}{2} \sin A\)
\(=\dfrac{1}{2} \sin A+\dfrac{\sqrt{3}}{2} \cos A=\sin \left(A+\dfrac{\pi}{3} \right)\).
\(∵A∈\left(0,\dfrac{\pi}{3} \right )\),\(∴A+\dfrac{\pi}{3} ∈\left(\dfrac{\pi}{3} ,\dfrac{2\pi}{3} \right)\),
\(∴\sin \left(A+\dfrac{\pi}{3} \right)∈\left(\dfrac{\sqrt{3}}{2},1 \right]\). -
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\(∵\dfrac{\pi}{2}<β<π\),\(∴β=\dfrac{3\pi}{4}\). -
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\(\tan (α+β)=3\),\(\tan \left(α+\dfrac{\pi}{4} \right)=2\),
\(\therefore \tan \beta=\tan [(\alpha+\beta)-\alpha]=\dfrac{\tan (\alpha+\beta)-\tan \alpha}{1+\tan (\alpha+\beta) \tan \alpha}=\dfrac{3-\dfrac{1}{3}}{1+3 \times \dfrac{1}{3}}=\dfrac{4}{3}\). -
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\(\left[\dfrac{5\pi}{12},\dfrac{5\pi}{6} \right]\). -
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
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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\(\cos β-\sin β=\dfrac{1}{2}\),\(\sin ^2β+\cos ^2β=1\),
\(∴8 \cos ^2β-4\cos β-3=0\),\(8 \sin ^2β+4\sin β-3=0\),
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:\(C\). -
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:\(B\). -
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
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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:\(-\sqrt{3}\). -
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\(8\).