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模块导图
知识剖析
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
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\(P_1 (\cosα ,\sinα)\),\(A_{1}(\cos \beta, \sin \beta)\),\(P(\cos (\alpha-\beta), \sin (\alpha-\beta))\),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\(\overline{A_{1} P_{1}}\)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
\([\cos (\alpha-\beta)-1]^{2}+\sin ^{2}(\alpha-\beta)=(\cos \alpha-\cos \beta)^{2}+(\sin \alpha-\sin \beta)^{2}\)
化简得
\(\cos (\alpha-\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta\)
而
\(\cos (\alpha+\beta)=\cos [\alpha-(-\beta)]=\cos \alpha \cos \beta-\sin \alpha \sin \beta\)
② 正弦两角和差公式
推导如下
\(\begin{aligned} \sin (\alpha+\beta) &=\cos \left[\left(\dfrac{\pi}{2}-\alpha\right)-\beta\right] \\ &=\cos \left(\dfrac{\pi}{2}-\alpha\right) \cos \beta+\sin \left(\dfrac{\pi}{2}-\alpha\right) \sin \beta \\ &=\sin \alpha \cos \beta+\cos \alpha \sin \beta \end{aligned}\)
\(\begin{aligned} \sin (\alpha-\beta) &=\cos \left[\left(\dfrac{\pi}{2}-\alpha\right)+\beta\right] \\ &=\cos \left(\dfrac{\pi}{2}-\alpha\right) \cos \beta-\sin \left(\dfrac{\pi}{2}-\alpha\right) \sin \beta \\ &=\sin \alpha \cos \beta-\cos \alpha \sin \beta \end{aligned}\)
③ 正切两角和差公式
(由\(S_{(\alpha \pm \beta)} 、 C_{(\alpha \pm \beta)}\)可推导正切的和差角公式)
对公式中\(α\)、\(β\)的理解,他们可表示为一个数字、一个字母,甚至一个式子
\({\color{Red}{Eg}}\):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\(\sin (\alpha \pm \beta)=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta\),把\(α\)看成数字\(45°\) ,\(β\)看成数字\(30^°\);
② \(\cos \left(x+\dfrac{\pi}{3}\right)=\cos x \cdot \cos \dfrac{\pi}{3}-\sin x \cdot \sin \dfrac{\pi}{3}\)
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\(\tan (\alpha+\beta)=\dfrac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}\),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
\(a \sin x+b \cos x=\sqrt{a^{2}+b^{2}} \sin (x+\varphi)\)
其中\(\tan \varphi=\dfrac{b}{a}\).
熟记两个特殊角的化简过程
\((1) a:b=1:1\)型,配\(\dfrac{\pi}{4}\)
\((2)a: b=\sqrt{3}: 1\)型,配\(\dfrac{\pi}{6}\)或\(\dfrac{\pi}{3}\)
经典例题
【题型一】和差角公式的基本运用
【典题1】 计算\(\sin 25^{\circ} \sin 70^{\circ}-\cos 155^{\circ} \sin 20^{\circ}=\)\(\underline{\quad \quad}\) .
【解析】 \(\sin 25^{\circ} \sin 70^{\circ}-\cos 155^{\circ} \sin 20^{\circ}\) \({\color{Red} {(大角化小角)}}\)
\(\begin{aligned} &=\sin 25^{\circ} \cos 20^{\circ}+\cos 25^{\circ} \sin 20^{\circ} \\ &=\sin \left(25^{\circ}+20^{\circ}\right) \\ &=\sin 45^{\circ} \\ &=\dfrac{\sqrt{2}}{2} \end{aligned}\)
【典题2】求\(\tan 27^{\circ}+\tan 33^{\circ}+\sqrt{3} \tan 27^{\circ} \tan 33^{\circ}\)
【解析】 \(\because \tan (27+33)^{\circ}=\tan 60^{\circ}=\sqrt{3}\)
\(\therefore \dfrac{\tan 27^{\circ}+\tan 33^{\circ}}{1-\tan 27^{\circ} \tan 33^{\circ}}=\sqrt{3}\)
\(\therefore \tan 27^{\circ}+\tan 33^{\circ}=\sqrt{3}-\sqrt{3} \tan 27^{\circ} \tan 33^{\circ}\)
\(\therefore \tan 27^{\circ}+\tan 33^{\circ}+\sqrt{3} \tan 27^{\circ} \tan 33^{\circ}=\sqrt{3}\)
【点拨】由\(\tan (\alpha+\beta)=\dfrac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}\)可得
\(\tan \alpha+\tan \beta=\tan (\alpha+\beta)(1-\tan \alpha \tan \beta)\)
\(\tan \alpha+\tan \beta+\tan \alpha \tan \beta \tan (\alpha+\beta)=\tan (\alpha+\beta)\)
【典题3】 若\(\alpha, \beta \in\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)\),且\(\tan \alpha\),\(\tan \beta\)是方程\(x^{2}+4 \sqrt{3} x+5=0\)的两个根,则\(α+β=\)\(\underline{\quad \quad}\) .
【解析】由已知可得\(\tan \alpha+\tan \beta=-4 \sqrt{3}\),\(\tan \alpha \cdot \tan \beta=5\),
\(\therefore \tan (\alpha+\beta)=\dfrac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta}=\dfrac{-4 \sqrt{3}}{1-5}=\sqrt{3}\).
\(\because \alpha, \beta \in\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)\),且\(\tan \alpha<0\),\(\tan \beta=5<0\),
\(\therefore \alpha, \beta \in\left(-\dfrac{\pi}{2}, 0\right)\),则\(\alpha+\beta \in(-\pi, 0)\),
\(\therefore \alpha+\beta=-\dfrac{2 \pi}{3}\).
【点拨】注意考虑角度的范围.
【典题4】已知\(\sin \alpha-\sin \beta=-\dfrac{1}{3}\),\(\cos \alpha+\cos \beta=\dfrac{1}{2}\),则\(\cos (\alpha+\beta)=\)\(\underline{\quad \quad}\) .
【解析】已知两等式分别平方得\((\sin \alpha-\sin \beta)^{2}=\sin ^{2} \alpha-2 \sin \alpha \sin \beta+\sin ^{2} \beta=\dfrac{1}{9}\)①
\((\cos \alpha+\cos \beta)^{2}=\cos ^{2} \alpha+2 \cos \alpha \cos \beta+\cos ^{2} \beta=\dfrac{1}{4}\)②
①+②得:\(2+2(\cos \alpha \cos \beta-\sin \alpha \sin \beta)=\dfrac{13}{36}\)
即\(\cos \alpha \cos \beta-\sin \alpha \sin \beta=-\dfrac{59}{72},\),
则\(\cos (\alpha+\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta=-\dfrac{59}{72}\)
【典题5】 设\(0<\beta<\alpha<\dfrac{\pi}{2}\),\(\tan (a-\beta)+\tan \beta=\dfrac{1}{\cos \beta}\),则( )
A.\(2 \alpha+\beta=\dfrac{\pi}{2}\) \(\qquad \qquad\)B.\(2 \alpha-\beta=\dfrac{\pi}{2}\) \(\qquad \qquad\)C.\(a+2 \beta=\dfrac{\pi}{2}\) \(\qquad \qquad\)D.\(\alpha-2 \beta=\dfrac{\pi}{2}\)
【解析】由题意知,\(\tan (\alpha-\beta)+\tan \beta=\dfrac{1}{\cos \beta}\),
即\(\dfrac{\sin (\alpha-\beta)}{\cos (\alpha-\beta)}+\dfrac{\sin \beta}{\cos \beta}=\dfrac{1}{\cos \beta}\),
\({\color{Red} { (正切化弦)}}\)
等式两边同乘以\(\cos (\alpha-\beta) \cos \beta\),
得\(\sin (\alpha-\beta) \cos \beta+\cos (\alpha-\beta) \sin \beta=\cos (\alpha-\beta)\),
所以\(\sin \alpha=\cos (\alpha-\beta)\),
即\(\cos \left(\dfrac{\pi}{2}-\alpha\right)=\cos (\alpha-\beta)\);
\({\color{Red} {(化为同一函数名)}}\)
又\(0<\beta<\alpha<\dfrac{\pi}{2}\),
所以\(\dfrac{\pi}{2}-\alpha \in\left(0, \dfrac{\pi}{2}\right)\),\(0<\alpha-\beta<\dfrac{\pi}{2}\),
\({\color{Red} {(注意角度的范围限制)}}\)
所以\(\dfrac{\pi}{2}-\alpha=\alpha-\beta\),所以\(2 \alpha-\beta=\dfrac{\pi}{2}\).
故选:\(B\).
【点拨】遇到含正切与正弦余弦的等式,可采取“切化弦”的方法.
【典题6】 在\(△ABC\)中,\(\tan A+\tan B+\sqrt{3}=\sqrt{3} \tan A \tan B\),\(\sin A \cos B=\dfrac{\sqrt{3}}{4}\),则\(△ABC\)的形状为\(\underline{\quad \quad}\) .
【解析】\(\because \tan A+\tan B+\sqrt{3}=\sqrt{3} \tan A \tan B\),
\(\therefore \tan (A+B)=\dfrac{\tan A+\tan B}{1-\tan A \tan B}\)\(=\dfrac{\sqrt{3} \cdot(\tan A \tan B-1)}{1-\tan A \tan B}\)\(=-\sqrt{3}=-\tan C\),
\(\therefore \tan C=\sqrt{3}\),\(\therefore C=\dfrac{\pi}{3}\),\(A+B=\dfrac{2 \pi}{3}\).
又\(\sin A \cos B=\dfrac{\sqrt{3}}{4}\),
\(\therefore \sin C=\sin (A+B)\)\(=\sin A \cos B+\cos A \sin B=\dfrac{\sqrt{3}}{2}\),
\(\therefore \cos A \sin B=\dfrac{\sqrt{3}}{4}\),
\(\therefore \sin (A-B)=\sin A \cos B-\cos A \sin B=0\)
\(∴A=B\),
\(∴△ABC\)为等边三角形.
【点拨】在三角形\(\triangle ABC\)中,\(\sin C=\sin (A+B)\),\(\cos C=-\cos (A+B)\).
巩固练习
1(★) \(\sin 80^{\circ} \cos 50^{\circ}+\cos 140^{\circ} \sin 10^{\circ}=\)\(\underline{\quad \quad}\) .
2(★)若\(\sin \alpha=\dfrac{3}{5}\),且\(\alpha \in\left(\dfrac{\pi}{2}, \pi\right)\),则\(\tan \left(\alpha+\dfrac{\pi}{4}\right)=\)\(\underline{\quad \quad}\) .
3(★)已知:\(α ,β\)均为锐角,\(\tan \alpha=\dfrac{1}{2}\),\(\tan \beta=\dfrac{1}{3}\),则\(α+β=\)\(\underline{\quad \quad}\) .
4 (★★)在\(△ABC\)中,\(\cos A+\sin A=\dfrac{1}{5}\),则\(\tan \left(A-\dfrac{\pi}{4}\right)=\) \(\underline{\quad \quad}\) .
5(★★★)设\(α=70^°\),若\(\beta \in\left(0, \dfrac{\pi}{2}\right)\),且\(\tan \alpha=\dfrac{1+\sin \beta}{\cos \beta}\),则\(β=\)\(\underline{\quad \quad}\) .
6(★★★)设\(\alpha, \beta \in\left(0, \dfrac{\pi}{2}\right)\),\(\sin \alpha \cos \beta=3 \sin \beta \cos \alpha\),则\(α-β\)的最大值为\(\underline{\quad \quad}\) .
7(★★★)已知锐角\(α,β\)满足\(\alpha-\beta=\dfrac{\pi}{3}\),则\(\dfrac{1}{\cos \alpha \cdot \cos \beta}+\dfrac{1}{\sin \alpha \cdot \sin \beta}\)的最小值为\(\underline{\quad \quad}\) .
答案
- \(\dfrac{1}{2}\)
- \(\dfrac{1}{7}\)
- \(\dfrac{\pi}{4}\)
- \(7\)
- \(50^{\circ}\)
- \(\dfrac{\pi}{6}\)
- \(8\)
【题型二】角的变换
【典题1】 若\(\sin \left(\alpha+\dfrac{\pi}{5}\right)=-\dfrac{1}{3}\),\(α∈(0 ,π)\),则\(\cos \left(\dfrac{\pi}{20}-\alpha\right)=\) \(\underline{\quad \quad}\) .
【解析】\(\because \alpha+\dfrac{\pi}{5}+\dfrac{\pi}{20}-\alpha=\dfrac{\pi}{4}\), \(\therefore \dfrac{\pi}{20}-\alpha=\dfrac{\pi}{4}-\left(\alpha+\dfrac{\pi}{5}\right)\),
\(∵α∈(0 ,π)\),\(\therefore \alpha+\dfrac{\pi}{5} \in\left(\dfrac{\pi}{5}, \dfrac{6 \pi}{5}\right)\),
又\(\sin \left(\alpha+\dfrac{\pi}{5}\right)=-\dfrac{1}{3}<0\),
即\(\alpha+\dfrac{\pi}{5}\)在第三象限,
\({\color{Red} {(注意角度的范围)}}\)
\(\therefore \cos \left(\alpha+\dfrac{\pi}{5}\right)=-\dfrac{2 \sqrt{2}}{3}\),
则\(\cos \left(\dfrac{\pi}{20}-\alpha\right)=\cos \left[\dfrac{\pi}{4}-\left(\alpha+\dfrac{\pi}{5}\right)\right]\)\(=\dfrac{\sqrt{2}}{2} \times\left(-\dfrac{2 \sqrt{2}}{3}\right)+\dfrac{\sqrt{2}}{2} \times\left(-\dfrac{1}{3}\right)=\dfrac{-4-\sqrt{2}}{6}\).
【点拨】
① 因为已知角\(\alpha+\dfrac{\pi}{5}\)和所求角\(\dfrac{\pi}{20}-\alpha\)中\(α\)的系数是相反数,故想到两角和\(\alpha+\dfrac{\pi}{5}+\dfrac{\pi}{20}-\alpha=\dfrac{\pi}{4}\)是特殊角为关键,则有\(\dfrac{\pi}{20}-\alpha=\dfrac{\pi}{4}-\left(\alpha+\dfrac{\pi}{5}\right)\).
② 在角的变换中,要注意已知角与所求角之间的和差是否为定值.
【典题2】若\(\sin 2 \alpha=\dfrac{\sqrt{5}}{5}\),\(\sin (\beta-\alpha)=\dfrac{\sqrt{10}}{10}\),且\(\alpha \in\left[\dfrac{\pi}{4}, \pi\right]\),\(\beta \in\left[\pi, \dfrac{3 \pi}{2}\right]\),则\(α+β\)的值是\(\underline{\quad \quad}\) .
【解析】 \({\color{Red} {(找到已知角2α、β-α与所求角α+β之间的关系α+β=2α+(β-α))}}\)
则\(\cos (\alpha+\beta)=\cos [2 \alpha+(\beta-\alpha)]\)
\(=\cos 2 \alpha \cdot \cos (\beta-\alpha)-\sin 2 \alpha \cdot \sin (\beta-\alpha)\)
\({\color{Red} {(求\sin (α+β)也ok,还要求\cos 2α,\cos (β-α))}}\)
\(\because \alpha \in\left[\dfrac{\pi}{4}, \pi\right]\),\(\therefore 2 \alpha \in\left[\dfrac{\pi}{2}, 2 \pi\right]\),
又\(0<\sin 2 \alpha=\dfrac{\sqrt{5}}{5}<\dfrac{1}{2}\),\(\therefore 2 \alpha \in\left(\dfrac{5 \pi}{6}, \pi\right)\),
\(\therefore \cos 2 \alpha=-\sqrt{1-\sin ^{2} 2 \alpha}=-\dfrac{2 \sqrt{5}}{5}\);
\(\because 2 \alpha \in\left(\dfrac{5 \pi}{6}, \pi\right) \Rightarrow \alpha \in\left(\dfrac{5 \pi}{12}, \dfrac{\pi}{2}\right), \beta \in\left[\pi, \dfrac{3 \pi}{2}\right],\),
\(\therefore \beta-\alpha \in\left(\dfrac{\pi}{2}, \dfrac{13 \pi}{12}\right)\),
\(\therefore \cos (\beta-\alpha)=-\sqrt{1-\sin ^{2}(\beta-\alpha)}=-\dfrac{3 \sqrt{10}}{10}\),
\({\color{Red} {(确定2α与β-α的范围,以确定\cos2α和\cos(β-α)的正负号)}}\)
\(\therefore \cos (\alpha+\beta)=-\dfrac{2 \sqrt{5}}{5} \times\left(-\dfrac{3 \sqrt{10}}{10}\right)-\dfrac{\sqrt{5}}{5} \times \dfrac{\sqrt{10}}{10}=\dfrac{\sqrt{2}}{2}\),
又\(\alpha \in\left(\dfrac{5 \pi}{12}, \dfrac{\pi}{2}\right), \quad \beta \in\left[\pi, \dfrac{3 \pi}{2}\right]\),
\(\therefore(\alpha+\beta) \in\left(\dfrac{17 \pi}{12}, 2 \pi\right)\),
\(\therefore \alpha+\beta=\dfrac{7 \pi}{4}\).
【典题3】已知\(\alpha, \beta \in\left(0, \dfrac{\pi}{2}\right)\),\(\sin (2 \alpha+\beta)=2 \sin \beta\),则\(\tan \beta\)的最大值为\(\underline{\quad \quad}\) .
【解析】\(\because \alpha, \beta \in\left(0, \dfrac{\pi}{2}\right)\),\(\sin (2 \alpha+\beta)=2 \sin \beta\),
\(\therefore \sin [(\alpha+\beta)+\alpha]=2 \sin [(\alpha+\beta)-\alpha]\),
\(\therefore \sin (\alpha+\beta) \cos \alpha+\cos (\alpha+\beta) \sin \alpha\)\(=2[\sin (\alpha+\beta) \cos \alpha-\cos (\alpha+\beta) \sin \alpha]\),
即\(3 \cos (\alpha+\beta) \sin \alpha=\sin (\alpha+\beta) \cos \alpha\),
\(\therefore \tan (\alpha+\beta)=3 \tan \alpha\),
即\(\tan (\alpha+\beta)=\dfrac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta}=3 \tan \alpha\),
化简整理得\(\tan \beta=\dfrac{2 \tan \alpha}{1+3 \tan ^{2} \alpha}\)\(=\dfrac{2}{\dfrac{1}{\tan \alpha}+3 \tan \alpha} \leq \dfrac{2}{2 \sqrt{\dfrac{1}{\tan \alpha} \cdot 3 \tan \alpha}}=\dfrac{\sqrt{3}}{3}\),
当且\(\dfrac{1}{\tan \alpha}=3 \tan \alpha\),即\(\tan \alpha=\dfrac{\sqrt{3}}{3}\),等号成立,\(\tan \beta\)取得最大值\(\dfrac{\sqrt{3}}{3}\).
巩固练习
1(★★)已知\(0<\alpha<\beta<\dfrac{\pi}{2}\),且\(\cos (\alpha-\beta)=\dfrac{63}{65}\),\(\sin \beta=\dfrac{12}{13}\),则\(\sin \alpha=\)\(\underline{\quad \quad}\) .
2(★★)若\(α ,β∈(0 ,π)\),\(\cos \left(\alpha-\dfrac{\beta}{2}\right)=-\dfrac{12}{13}\),\(\sin \left(\dfrac{\alpha}{2}-\beta\right)=\dfrac{4}{5}\),则\(\sin \dfrac{\alpha+\beta}{2}=\)\(\underline{\quad \quad}\) .
3(★★)若\(0<\alpha<\dfrac{\pi}{2}\),\(-\dfrac{\pi}{2}<\beta<0\),\(\cos \left(\dfrac{\pi}{4}+\alpha\right)=\dfrac{1}{3}\),\(\cos \left(\dfrac{\pi}{4}-\dfrac{\beta}{2}\right)=\dfrac{\sqrt{3}}{3}\),则\(\cos \left(\alpha+\dfrac{\beta}{2}\right)=\)\(\underline{\quad \quad}\) .
4 (★★)已知\(\cos \alpha=\dfrac{2 \sqrt{5}}{5}\),\(\tan (\alpha-\beta)=-\dfrac{1}{3}\),\(α ,β\)均为锐角,则\(β=\)\(\underline{\quad \quad}\) .
5(★★)已知\(\cos \alpha=\dfrac{2 \sqrt{5}}{5}\),\(\cos (\beta-\alpha)=\dfrac{3 \sqrt{10}}{10}\),且\(0<\alpha<\beta<\dfrac{\pi}{2}\),则\(β\)的值\(\underline{\quad \quad}\) .
6 (★★)若\(\sin 2 \alpha=\dfrac{\sqrt{5}}{5}\),\(\sin (\beta-\alpha)=\dfrac{\sqrt{10}}{10}\),且\(\alpha \in\left[\dfrac{\pi}{4}, \pi\right]\),\(\beta \in\left[\pi, \dfrac{3 \pi}{2}\right]\),则\(α+β\)的值是\(\underline{\quad \quad}\) .
答案
- \(\dfrac{4}{5}\)
- \(\dfrac{63}{65}\)
- \(\dfrac{5 \sqrt{3}}{9}\)
- \(\dfrac{\pi}{4}\)
- \(\dfrac{\pi}{4}\)
- \(\dfrac{7 \pi}{4}\)
【题型三】辅助角公式的运用
【典题1】 若\(\dfrac{\pi}{4}<\alpha<\beta<\dfrac{\pi}{2}\),\(\sin \alpha+\cos \alpha=a\),\(\sin \beta+\cos \beta=b\) ,则\(a,b\)的大小关系是\(\underline{\quad \quad}\) .
【解析】化简可得\(a=\sin \alpha+\cos \alpha=\sqrt{2} \sin \left(\alpha+\dfrac{\pi}{4}\right)\),\(b=\sin \beta+\cos \beta=\sqrt{2} \sin \left(\beta+\dfrac{\pi}{4}\right)\),
\(\because \dfrac{\pi}{4}<\alpha<\beta<\dfrac{\pi}{2}\) ,
\(\therefore \dfrac{\pi}{2}<\alpha+\dfrac{\pi}{4}<\beta+\dfrac{\pi}{4}<\dfrac{3 \pi}{4},\),
由正弦函数的单调性可知\(a>b\).
【点拨】熟记\(\sin x \pm \cos x=\sqrt{2} \sin \left(x \pm \dfrac{\pi}{4}\right)\).
【典题2】 设当\(x=\theta\)时,函数\(f(x)=2 \sin x+\cos x\)取得最小值,则\(\cos \left(\theta+\dfrac{\pi}{4}\right)=\)\(\underline{\quad \quad}\) .
【解析】函对于数\(f(x)=2 \sin x+\cos x=\sqrt{5} \sin (x+\varphi)\),
其中\(\cos \varphi=\dfrac{2}{\sqrt{5}}, \sin \varphi=\dfrac{1}{\sqrt{5}}\),\(φ\)为锐角.
当\(x=θ\)时,函数取得最小值,\(\therefore \sqrt{5} \sin (\theta+\varphi)=-\sqrt{5}\),
即\(\sin (\theta+\varphi)=-1\),
故可令\(\theta+\varphi=-\dfrac{\pi}{2}+2 k \pi(k \in Z)\),即\(\theta=-\dfrac{\pi}{2}-\varphi+2 k \pi\),
故\(\cos \left(\theta+\dfrac{\pi}{4}\right)=\cos \left(-\dfrac{\pi}{4}-\varphi+2 k \pi\right)=\cos \left(\varphi+\dfrac{\pi}{4}\right)\)
\(=\dfrac{\sqrt{2}}{2} \cos \varphi-\dfrac{\sqrt{2}}{2} \sin \varphi=\dfrac{\sqrt{2}}{2}\left(\dfrac{2}{\sqrt{5}}-\dfrac{1}{\sqrt{5}}\right)=\dfrac{\sqrt{10}}{10}\).
【点拨】
① 辅助角公式\(a \sin x+b \cos x=\sqrt{a^{2}+b^{2}} \sin (x+\varphi)\),要理解其中\(φ\)的含义,\(\tan \varphi=\dfrac{b}{a}\);
② 涉及到三角函数\(f(x)=a \sin x+b \cos x\)的性质问题(比如单调性、对称性、最值等),往往要通过辅助角公式把函数\(y=f(x)\)转化为\(f(x)=A\sin(ωx+φ)\)的形式.
【典题3】 已知函数\(f(x)=2 \sin x-a \cos x\)图象的一条对称轴为\(x=-\dfrac{\pi}{6}\),\(f(x_1)+f(x_2)=0\),且函数\(f(x)\)在\((x_1 ,x_2)\)上单调,则\(|3x_1+2x_2 |\)的最小值为\(\underline{\quad \quad}\) .
【解析】由题意\(f(x)=2 \sin x-a \cos x=\sqrt{4+a^{2}} \sin (x+\theta)\),\(θ\)为辅助角,
因为对称轴\(x=-\dfrac{\pi}{6}\),
所以\(f\left(-\dfrac{\pi}{6}\right)=-1-\dfrac{\sqrt{3}}{2} a\),
即\(\sqrt{4+a^{2}}=\left|-1-\dfrac{\sqrt{3}}{2} a\right|\),
\({\color{Red} {(三角函数对称轴对应的y值是最值) }}\)
解得\(a=2 \sqrt{3}\),
所以\(f(x)=4 \sin \left(x-\dfrac{\pi}{3}\right)\),对称轴方程为\(x=-\dfrac{\pi}{6}+k \pi(k \in Z)\),
又因为\(f(x)\)在\((x_1 ,x_2)\)上具有单调性,且\(f(x_1)+f(x_2)=0\),
设\(A(x_1 ,f(x_1))\),\(B(x_2 ,f(x_2))\),
则线段\(AB\)的中点为函数\(f(x)\)的对称中心,
所以\(x_{1}+x_{2}=2 k \pi+\dfrac{2 \pi}{3}(k \in Z)\),
\(\left|3 x_{1}+2 x_{2}\right|=\left|2\left(x_{1}+x_{2}\right)+x_{1}\right|=\left|4 k \pi+\dfrac{4 \pi}{3}+x_{1}\right|\)
显然当\(k=0\),\(x_{1}=-\dfrac{\pi}{6}\)时,即\(x_{1}=-\dfrac{\pi}{6}, x_{2}=\dfrac{5 \pi}{6}\)时取最小值\(\dfrac{7 \pi}{6}\).
(结合函数图像分析)
巩固练习
1(★★)已知函数\(f(x)=|\sqrt{3} \sin \omega x-\cos \omega x|(\omega>0)\)的最小正周期为\(π\),则\(ω=\)\(\underline{\quad \quad}\) .
2(★★)\(A ,B ,C\)是\(△ABC\)的内角,其中\(B=\dfrac{2 \pi}{3}\),则\(\sin A+\sin C\)的取值范围是\(\underline{\quad \quad}\) .
3(★★)若函数\(f(x)=\sin 2 x-\sqrt{3} \cos 2 x\)在\([0 ,t]\)上的值域为\([-\sqrt{3}, 2]\),则\(t\)的取值范围为\(\underline{\quad \quad}\) .
4(★★★) 已知函数\(f(x)=\sin \omega x+\cos \omega x(\omega>0)\)在\(\left(\dfrac{\pi}{6}, \dfrac{5 \pi}{12}\right)\)上仅有\(1\)个最值,且是最大值,则实数\(ω\)的取值范围为\(\underline{\quad \quad}\) .
5(★★★)已知函数\(f(x)=2 \sin \left(\omega x+\dfrac{\pi}{6}\right)+a\cos\omega x(a>0, \omega>0)\)对任意的\(x_1 ,x_2∈R\),都有\(f\left(x_{1}\right)+f\left(x_{2}\right) \leq 4 \sqrt{3}\),若\(f(x)\)在\([0 ,π]\)上的值域为\([3,2 \sqrt{3}]\),则实数\(ω\)的取值范围为\(\underline{\quad \quad}\) .
答案
- \(1\)
- \(\left(\dfrac{\sqrt{3}}{2}, 1\right]\)
- \(\left[\dfrac{5 \pi}{12}, \dfrac{5 \pi}{6}\right]\)
- \(\left(\dfrac{3}{5}, \dfrac{3}{2}\right)\)
- \(\left[\dfrac{1}{6}, \dfrac{1}{3}\right]\)
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