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模塊導圖
知識剖析
兩角和差的正弦，余弦與正切公式
(理解公式的推導，體會其方法，而不死背公式)
① 余弦兩角和差公式
推導如下
如圖，設單位圓與\(x\)軸的正半軸相交於點\(A(1 ,0)\)，以\(x\)軸為非負半軸為始邊作角\(α\)，\(β\)，\(α-β\)，它們的終邊分別與單位圓相較於點\(P_1 (\cosα ,\sinα)\)，\(A_{1}(\cos \beta, \sin \beta)\)，\(P(\cos (\alpha-\beta), \sin (\alpha-\beta))\)，連接\(A_1 P_1\)，\(AP\)，若把扇形\(OAP\)繞點\(O\)旋轉\(β\)角，則點\(A\)，\(P\)分別與$A_1 $ ,\(P_1\)重合.根據圓的旋轉對稱性可知，\(\widehat{A P}\)與\(\overline{A_{1} P_{1}}\)重合，從而\(\widehat{A P}=\widehat{A_{1} P_{1}}\)，所以\(AP=A_1 P_1\).
根據兩點間的距離公式，得
\([\cos (\alpha-\beta)-1]^{2}+\sin ^{2}(\alpha-\beta)=(\cos \alpha-\cos \beta)^{2}+(\sin \alpha-\sin \beta)^{2}\)
化簡得
\(\cos (\alpha-\beta)=\cos \alpha \cos \beta+\sin \alpha \sin \beta\)
而
\(\cos (\alpha+\beta)=\cos [\alpha-(-\beta)]=\cos \alpha \cos \beta-\sin \alpha \sin \beta\)
② 正弦兩角和差公式
推導如下
\(\begin{aligned} \sin (\alpha+\beta) &=\cos \left[\left(\dfrac{\pi}{2}-\alpha\right)-\beta\right] \\ &=\cos \left(\dfrac{\pi}{2}-\alpha\right) \cos \beta+\sin \left(\dfrac{\pi}{2}-\alpha\right) \sin \beta \\ &=\sin \alpha \cos \beta+\cos \alpha \sin \beta \end{aligned}\)
\(\begin{aligned} \sin (\alpha-\beta) &=\cos \left[\left(\dfrac{\pi}{2}-\alpha\right)+\beta\right] \\ &=\cos \left(\dfrac{\pi}{2}-\alpha\right) \cos \beta-\sin \left(\dfrac{\pi}{2}-\alpha\right) \sin \beta \\ &=\sin \alpha \cos \beta-\cos \alpha \sin \beta \end{aligned}\)
③ 正切兩角和差公式
(由\(S_{(\alpha \pm \beta)} 、 C_{(\alpha \pm \beta)}\)可推導正切的和差角公式)
對公式中\(α\)、\(β\)的理解，他們可表示為一個數字、一個字母，甚至一個式子
\({\color{Red}{Eg}}\)：① \(\sin 75^{\circ}=\sin \left(45^{\circ}+30^{\circ}\right)\)\(=\sin 45^{\circ} \cos 30^{\circ}+\cos 45^{\circ} \sin 30^{\circ}=\dfrac{\sqrt{6}+\sqrt{2}}{4}\)
對應公式\(\sin (\alpha \pm \beta)=\sin \alpha \cos \beta \pm \cos \alpha \sin \beta\)，把\(α\)看成數字\(45°\) ,\(β\)看成數字\(30^°\)；
② \(\cos \left(x+\dfrac{\pi}{3}\right)=\cos x \cdot \cos \dfrac{\pi}{3}-\sin x \cdot \sin \dfrac{\pi}{3}\)
對應公式\(\cos (α+β)=\cos α \cos β-\sin α \sin β\)，把\(α\)看成字母\(x\)，\(β\)看成數字\(\dfrac{\pi}{3}\)；
③\(\tan \dfrac{\pi}{4}=\tan \left[\left(x+\dfrac{\pi}{8}\right)+\left(\dfrac{\pi}{8}-x\right)\right]\)\(=\dfrac{\tan \left(x+\dfrac{\pi}{8}\right)+\tan \left(\dfrac{\pi}{8}-x\right)}{1-\tan \left(x+\dfrac{\pi}{8}\right) \tan \left(\dfrac{\pi}{8}-x\right)}\)，
對應公式\(\tan (\alpha+\beta)=\dfrac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}\)，把\(α、β\)分別看成式子\(x+\dfrac{\pi}{8}\)，\(x-\dfrac{\pi}{8}\).
對應公式的運用，注意整體變換的思想.
輔助角公式
\(a \sin x+b \cos x=\sqrt{a^{2}+b^{2}} \sin (x+\varphi)\)
其中\(\tan \varphi=\dfrac{b}{a}\).
熟記兩個特殊角的化簡過程
\((1) a:b=1:1\)型，配\(\dfrac{\pi}{4}\)
\((2)a: b=\sqrt{3}: 1\)型，配\(\dfrac{\pi}{6}\)或\(\dfrac{\pi}{3}\)
經典例題
【題型一】和差角公式的基本運用
【典題1】 計算\(\sin 25^{\circ} \sin 70^{\circ}-\cos 155^{\circ} \sin 20^{\circ}=\)\(\underline{\quad \quad}\) .
【解析】 \(\sin 25^{\circ} \sin 70^{\circ}-\cos 155^{\circ} \sin 20^{\circ}\) \({\color{Red} {(大角化小角)}}\)
\(\begin{aligned} &=\sin 25^{\circ} \cos 20^{\circ}+\cos 25^{\circ} \sin 20^{\circ} \\ &=\sin \left(25^{\circ}+20^{\circ}\right) \\ &=\sin 45^{\circ} \\ &=\dfrac{\sqrt{2}}{2} \end{aligned}\)
【典題2】求\(\tan 27^{\circ}+\tan 33^{\circ}+\sqrt{3} \tan 27^{\circ} \tan 33^{\circ}\)
【解析】 \(\because \tan (27+33)^{\circ}=\tan 60^{\circ}=\sqrt{3}\)
\(\therefore \dfrac{\tan 27^{\circ}+\tan 33^{\circ}}{1-\tan 27^{\circ} \tan 33^{\circ}}=\sqrt{3}\)
\(\therefore \tan 27^{\circ}+\tan 33^{\circ}=\sqrt{3}-\sqrt{3} \tan 27^{\circ} \tan 33^{\circ}\)
\(\therefore \tan 27^{\circ}+\tan 33^{\circ}+\sqrt{3} \tan 27^{\circ} \tan 33^{\circ}=\sqrt{3}\)
【點撥】由\(\tan (\alpha+\beta)=\dfrac{\tan \alpha+\tan \beta}{1-\tan \alpha \tan \beta}\)可得
\(\tan \alpha+\tan \beta=\tan (\alpha+\beta)(1-\tan \alpha \tan \beta)\)
\(\tan \alpha+\tan \beta+\tan \alpha \tan \beta \tan (\alpha+\beta)=\tan (\alpha+\beta)\)
【典題3】 若\(\alpha, \beta \in\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)\)，且\(\tan \alpha\)，\(\tan \beta\)是方程\(x^{2}+4 \sqrt{3} x+5=0\)的兩個根，則\(α+β=\)\(\underline{\quad \quad}\) .
【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\(\tan \alpha<0\)，\(\tan \beta=5<0\)，
\(\therefore \alpha, \beta \in\left(-\dfrac{\pi}{2}, 0\right)\)，則\(\alpha+\beta \in(-\pi, 0)\)，
\(\therefore \alpha+\beta=-\dfrac{2 \pi}{3}\)．
【點撥】注意考慮角度的范圍.
【典題4】已知\(\sin \alpha-\sin \beta=-\dfrac{1}{3}\)，\(\cos \alpha+\cos \beta=\dfrac{1}{2}\)，則\(\cos (\alpha+\beta)=\)\(\underline{\quad \quad}\) ．
【解析】已知兩等式分別平方得\((\sin \alpha-\sin \beta)^{2}=\sin ^{2} \alpha-2 \sin \alpha \sin \beta+\sin ^{2} \beta=\dfrac{1}{9}\)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\(\cos \alpha \cos \beta-\sin \alpha \sin \beta=-\dfrac{59}{72},\)，
則\(\cos (\alpha+\beta)=\cos \alpha \cos \beta-\sin \alpha \sin \beta=-\dfrac{59}{72}\)
【典題5】 設\(0<\beta<\alpha<\dfrac{\pi}{2}\)，\(\tan (a-\beta)+\tan \beta=\dfrac{1}{\cos \beta}\)，則( )
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】由題意知，\(\tan (\alpha-\beta)+\tan \beta=\dfrac{1}{\cos \beta}\)，
即\(\dfrac{\sin (\alpha-\beta)}{\cos (\alpha-\beta)}+\dfrac{\sin \beta}{\cos \beta}=\dfrac{1}{\cos \beta}\)，
\({\color{Red} { (正切化弦)}}\)
等式兩邊同乘以\(\cos (\alpha-\beta) \cos \beta\)，
得\(\sin (\alpha-\beta) \cos \beta+\cos (\alpha-\beta) \sin \beta=\cos (\alpha-\beta)\)，
所以\(\sin \alpha=\cos (\alpha-\beta)\)，
即\(\cos \left(\dfrac{\pi}{2}-\alpha\right)=\cos (\alpha-\beta)\)；
\({\color{Red} {(化為同一函數名)}}\)
又\(0<\beta<\alpha<\dfrac{\pi}{2}\)，
所以\(\dfrac{\pi}{2}-\alpha \in\left(0, \dfrac{\pi}{2}\right)\)，\(0<\alpha-\beta<\dfrac{\pi}{2}\)，
\({\color{Red} {(注意角度的范圍限制)}}\)
所以\(\dfrac{\pi}{2}-\alpha=\alpha-\beta\)，所以\(2 \alpha-\beta=\dfrac{\pi}{2}\)．
故選：\(B\)．
【點撥】遇到含正切與正弦余弦的等式，可采取“切化弦”的方法.
【典題6】 在\(△ABC\)中，\(\tan A+\tan B+\sqrt{3}=\sqrt{3} \tan A \tan B\)，\(\sin A \cos B=\dfrac{\sqrt{3}}{4}\)，則\(△ABC\)的形狀為\(\underline{\quad \quad}\) ．
【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\(\sin A \cos B=\dfrac{\sqrt{3}}{4}\)，
\(\therefore \sin C=\sin (A+B)\)\(=\sin A \cos B+\cos A \sin B=\dfrac{\sqrt{3}}{2}\)，
\(\therefore \cos A \sin B=\dfrac{\sqrt{3}}{4}\)，
\(\therefore \sin (A-B)=\sin A \cos B-\cos A \sin B=0\)
\(∴A=B\)，
\(∴△ABC\)為等邊三角形.
【點撥】在三角形\(\triangle ABC\)中，\(\sin C=\sin (A+B)\)，\(\cos C=-\cos (A+B)\).
鞏固練習
1(★) \(\sin 80^{\circ} \cos 50^{\circ}+\cos 140^{\circ} \sin 10^{\circ}=\)\(\underline{\quad \quad}\) ．
2(★)若\(\sin \alpha=\dfrac{3}{5}\)，且\(\alpha \in\left(\dfrac{\pi}{2}, \pi\right)\)，則\(\tan \left(\alpha+\dfrac{\pi}{4}\right)=\)\(\underline{\quad \quad}\) ．
3(★)已知：\(α ,β\)均為銳角，\(\tan \alpha=\dfrac{1}{2}\)，\(\tan \beta=\dfrac{1}{3}\)，則\(α+β=\)\(\underline{\quad \quad}\) ．
4 (★★)在\(△ABC\)中，\(\cos A+\sin A=\dfrac{1}{5}\)，則\(\tan \left(A-\dfrac{\pi}{4}\right)=\) \(\underline{\quad \quad}\) ．
5(★★★)設\(α=70^°\)，若\(\beta \in\left(0, \dfrac{\pi}{2}\right)\)，且\(\tan \alpha=\dfrac{1+\sin \beta}{\cos \beta}\)，則\(β=\)\(\underline{\quad \quad}\) ．
6(★★★)設\(\alpha, \beta \in\left(0, \dfrac{\pi}{2}\right)\)，\(\sin \alpha \cos \beta=3 \sin \beta \cos \alpha\)，則\(α-β\)的最大值為\(\underline{\quad \quad}\) ．
7(★★★)已知銳角\(α,β\)滿足\(\alpha-\beta=\dfrac{\pi}{3}\)，則\(\dfrac{1}{\cos \alpha \cdot \cos \beta}+\dfrac{1}{\sin \alpha \cdot \sin \beta}\)的最小值為\(\underline{\quad \quad}\) ．
答案
- \(\dfrac{1}{2}\)
- \(\dfrac{1}{7}\)
- \(\dfrac{\pi}{4}\)
- \(7\)
- \(50^{\circ}\)
- \(\dfrac{\pi}{6}\)
- \(8\)
【題型二】角的變換
【典題1】 若\(\sin \left(\alpha+\dfrac{\pi}{5}\right)=-\dfrac{1}{3}\)，\(α∈(0 ,π)\)，則\(\cos \left(\dfrac{\pi}{20}-\alpha\right)=\) \(\underline{\quad \quad}\) ．
【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，
\({\color{Red} {(注意角度的范圍)}}\)
\(\therefore \cos \left(\alpha+\dfrac{\pi}{5}\right)=-\dfrac{2 \sqrt{2}}{3}\)，
則\(\cos \left(\dfrac{\pi}{20}-\alpha\right)=\cos \left[\dfrac{\pi}{4}-\left(\alpha+\dfrac{\pi}{5}\right)\right]\)\(=\dfrac{\sqrt{2}}{2} \times\left(-\dfrac{2 \sqrt{2}}{3}\right)+\dfrac{\sqrt{2}}{2} \times\left(-\dfrac{1}{3}\right)=\dfrac{-4-\sqrt{2}}{6}\)．
【點撥】
① 因為已知角\(\alpha+\dfrac{\pi}{5}\)和所求角\(\dfrac{\pi}{20}-\alpha\)中\(α\)的系數是相反數，故想到兩角和\(\alpha+\dfrac{\pi}{5}+\dfrac{\pi}{20}-\alpha=\dfrac{\pi}{4}\)是特殊角為關鍵，則有\(\dfrac{\pi}{20}-\alpha=\dfrac{\pi}{4}-\left(\alpha+\dfrac{\pi}{5}\right)\).
② 在角的變換中，要注意已知角與所求角之間的和差是否為定值.
【典題2】若\(\sin 2 \alpha=\dfrac{\sqrt{5}}{5}\)，\(\sin (\beta-\alpha)=\dfrac{\sqrt{10}}{10}\)，且\(\alpha \in\left[\dfrac{\pi}{4}, \pi\right]\)，\(\beta \in\left[\pi, \dfrac{3 \pi}{2}\right]\)，則\(α+β\)的值是\(\underline{\quad \quad}\) .
【解析】 \({\color{Red} {(找到已知角2α、β-α與所求角α+β之間的關系α+β=2α+(β-α))}}\)
則\(\cos (\alpha+\beta)=\cos [2 \alpha+(\beta-\alpha)]\)
\(=\cos 2 \alpha \cdot \cos (\beta-\alpha)-\sin 2 \alpha \cdot \sin (\beta-\alpha)\)
\({\color{Red} {(求\sin (α+β)也ok，還要求\cos 2α，\cos (β-α))}}\)
\(\because \alpha \in\left[\dfrac{\pi}{4}, \pi\right]\)，\(\therefore 2 \alpha \in\left[\dfrac{\pi}{2}, 2 \pi\right]\)，
又\(0<\sin 2 \alpha=\dfrac{\sqrt{5}}{5}<\dfrac{1}{2}\)，\(\therefore 2 \alpha \in\left(\dfrac{5 \pi}{6}, \pi\right)\)，
\(\therefore \cos 2 \alpha=-\sqrt{1-\sin ^{2} 2 \alpha}=-\dfrac{2 \sqrt{5}}{5}\)；
\(\because 2 \alpha \in\left(\dfrac{5 \pi}{6}, \pi\right) \Rightarrow \alpha \in\left(\dfrac{5 \pi}{12}, \dfrac{\pi}{2}\right), \beta \in\left[\pi, \dfrac{3 \pi}{2}\right],\)，
\(\therefore \beta-\alpha \in\left(\dfrac{\pi}{2}, \dfrac{13 \pi}{12}\right)\)，
\(\therefore \cos (\beta-\alpha)=-\sqrt{1-\sin ^{2}(\beta-\alpha)}=-\dfrac{3 \sqrt{10}}{10}\)，
\({\color{Red} {(確定2α與β-α的范圍，以確定\cos2α和\cos(β-α)的正負號)}}\)
\(\therefore \cos (\alpha+\beta)=-\dfrac{2 \sqrt{5}}{5} \times\left(-\dfrac{3 \sqrt{10}}{10}\right)-\dfrac{\sqrt{5}}{5} \times \dfrac{\sqrt{10}}{10}=\dfrac{\sqrt{2}}{2}\)，
又\(\alpha \in\left(\dfrac{5 \pi}{12}, \dfrac{\pi}{2}\right), \quad \beta \in\left[\pi, \dfrac{3 \pi}{2}\right]\)，
\(\therefore(\alpha+\beta) \in\left(\dfrac{17 \pi}{12}, 2 \pi\right)\)，
\(\therefore \alpha+\beta=\dfrac{7 \pi}{4}\).
【典題3】已知\(\alpha, \beta \in\left(0, \dfrac{\pi}{2}\right)\)，\(\sin (2 \alpha+\beta)=2 \sin \beta\)，則\(\tan \beta\)的最大值為\(\underline{\quad \quad}\) .
【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\(\tan (\alpha+\beta)=\dfrac{\tan \alpha+\tan \beta}{1-\tan \alpha \cdot \tan \beta}=3 \tan \alpha\)，
化簡整理得\(\tan \beta=\dfrac{2 \tan \alpha}{1+3 \tan ^{2} \alpha}\)\(=\dfrac{2}{\dfrac{1}{\tan \alpha}+3 \tan \alpha} \leq \dfrac{2}{2 \sqrt{\dfrac{1}{\tan \alpha} \cdot 3 \tan \alpha}}=\dfrac{\sqrt{3}}{3}\)，
當且\(\dfrac{1}{\tan \alpha}=3 \tan \alpha\)，即\(\tan \alpha=\dfrac{\sqrt{3}}{3}\)，等號成立，\(\tan \beta\)取得最大值\(\dfrac{\sqrt{3}}{3}\)．
鞏固練習
1(★★)已知\(0<\alpha<\beta<\dfrac{\pi}{2}\)，且\(\cos (\alpha-\beta)=\dfrac{63}{65}\)，\(\sin \beta=\dfrac{12}{13}\)，則\(\sin \alpha=\)\(\underline{\quad \quad}\) ．
2(★★)若\(α ,β∈(0 ,π)\)，\(\cos \left(\alpha-\dfrac{\beta}{2}\right)=-\dfrac{12}{13}\)，\(\sin \left(\dfrac{\alpha}{2}-\beta\right)=\dfrac{4}{5}\)，則\(\sin \dfrac{\alpha+\beta}{2}=\)\(\underline{\quad \quad}\) ．
3(★★)若\(0<\alpha<\dfrac{\pi}{2}\)，\(-\dfrac{\pi}{2}<\beta<0\)，\(\cos \left(\dfrac{\pi}{4}+\alpha\right)=\dfrac{1}{3}\)，\(\cos \left(\dfrac{\pi}{4}-\dfrac{\beta}{2}\right)=\dfrac{\sqrt{3}}{3}\)，則\(\cos \left(\alpha+\dfrac{\beta}{2}\right)=\)\(\underline{\quad \quad}\) ．
4 (★★)已知\(\cos \alpha=\dfrac{2 \sqrt{5}}{5}\)，\(\tan (\alpha-\beta)=-\dfrac{1}{3}\)，\(α ,β\)均為銳角，則\(β=\)\(\underline{\quad \quad}\) ．
5(★★)已知\(\cos \alpha=\dfrac{2 \sqrt{5}}{5}\)，\(\cos (\beta-\alpha)=\dfrac{3 \sqrt{10}}{10}\)，且\(0<\alpha<\beta<\dfrac{\pi}{2}\)，則\(β\)的值\(\underline{\quad \quad}\) ．
6 (★★)若\(\sin 2 \alpha=\dfrac{\sqrt{5}}{5}\)，\(\sin (\beta-\alpha)=\dfrac{\sqrt{10}}{10}\)，且\(\alpha \in\left[\dfrac{\pi}{4}, \pi\right]\)，\(\beta \in\left[\pi, \dfrac{3 \pi}{2}\right]\)，則\(α+β\)的值是\(\underline{\quad \quad}\) ．
答案
- \(\dfrac{4}{5}\)
- \(\dfrac{63}{65}\)
- \(\dfrac{5 \sqrt{3}}{9}\)
- \(\dfrac{\pi}{4}\)
- \(\dfrac{\pi}{4}\)
- \(\dfrac{7 \pi}{4}\)
【題型三】輔助角公式的運用
【典題1】 若\(\dfrac{\pi}{4}<\alpha<\beta<\dfrac{\pi}{2}\)，\(\sin \alpha+\cos \alpha=a\)，\(\sin \beta+\cos \beta=b\) ,則\(a，b\)的大小關系是\(\underline{\quad \quad}\) ．
【解析】化簡可得\(a=\sin \alpha+\cos \alpha=\sqrt{2} \sin \left(\alpha+\dfrac{\pi}{4}\right)\)，\(b=\sin \beta+\cos \beta=\sqrt{2} \sin \left(\beta+\dfrac{\pi}{4}\right)\)，
\(\because \dfrac{\pi}{4}<\alpha<\beta<\dfrac{\pi}{2}\) ,
\(\therefore \dfrac{\pi}{2}<\alpha+\dfrac{\pi}{4}<\beta+\dfrac{\pi}{4}<\dfrac{3 \pi}{4},\)，
由正弦函數的單調性可知\(a>b\).
【點撥】熟記\(\sin x \pm \cos x=\sqrt{2} \sin \left(x \pm \dfrac{\pi}{4}\right)\).
【典題2】 設當\(x=\theta\)時，函數\(f(x)=2 \sin x+\cos x\)取得最小值，則\(\cos \left(\theta+\dfrac{\pi}{4}\right)=\)\(\underline{\quad \quad}\) ．
【解析】函對於數\(f(x)=2 \sin x+\cos x=\sqrt{5} \sin (x+\varphi)\)，
其中\(\cos \varphi=\dfrac{2}{\sqrt{5}}, \sin \varphi=\dfrac{1}{\sqrt{5}}\)，\(φ\)為銳角．
當\(x=θ\)時，函數取得最小值，\(\therefore \sqrt{5} \sin (\theta+\varphi)=-\sqrt{5}\)，
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\(\cos \left(\theta+\dfrac{\pi}{4}\right)=\cos \left(-\dfrac{\pi}{4}-\varphi+2 k \pi\right)=\cos \left(\varphi+\dfrac{\pi}{4}\right)\)
\(=\dfrac{\sqrt{2}}{2} \cos \varphi-\dfrac{\sqrt{2}}{2} \sin \varphi=\dfrac{\sqrt{2}}{2}\left(\dfrac{2}{\sqrt{5}}-\dfrac{1}{\sqrt{5}}\right)=\dfrac{\sqrt{10}}{10}\).
【點撥】
① 輔助角公式\(a \sin x+b \cos x=\sqrt{a^{2}+b^{2}} \sin (x+\varphi)\)，要理解其中\(φ\)的含義，\(\tan \varphi=\dfrac{b}{a}\);
② 涉及到三角函數\(f(x)=a \sin x+b \cos x\)的性質問題(比如單調性、對稱性、最值等)，往往要通過輔助角公式把函數\(y=f(x)\)轉化為\(f(x)=A\sin(ωx+φ)\)的形式.
【典題3】 已知函數\(f(x)=2 \sin x-a \cos x\)圖象的一條對稱軸為\(x=-\dfrac{\pi}{6}\)，\(f(x_1)+f(x_2)=0\)，且函數\(f(x)\)在\((x_1 ,x_2)\)上單調，則\(|3x_1+2x_2 |\)的最小值為\(\underline{\quad \quad}\) .
【解析】由題意\(f(x)=2 \sin x-a \cos x=\sqrt{4+a^{2}} \sin (x+\theta)\)，\(θ\)為輔助角，
因為對稱軸\(x=-\dfrac{\pi}{6}\)，
所以\(f\left(-\dfrac{\pi}{6}\right)=-1-\dfrac{\sqrt{3}}{2} a\)，
即\(\sqrt{4+a^{2}}=\left|-1-\dfrac{\sqrt{3}}{2} a\right|\)，
\({\color{Red} {(三角函數對稱軸對應的y值是最值) }}\)
解得\(a=2 \sqrt{3}\)，
所以\(f(x)=4 \sin \left(x-\dfrac{\pi}{3}\right)\)，對稱軸方程為\(x=-\dfrac{\pi}{6}+k \pi(k \in Z)\)，
又因為\(f(x)\)在\((x_1 ,x_2)\)上具有單調性，且\(f(x_1)+f(x_2)=0\)，
設\(A(x_1 ,f(x_1))\)，\(B(x_2 ,f(x_2))\)，
則線段\(AB\)的中點為函數\(f(x)\)的對稱中心，
所以\(x_{1}+x_{2}=2 k \pi+\dfrac{2 \pi}{3}(k \in Z)\)，
\(\left|3 x_{1}+2 x_{2}\right|=\left|2\left(x_{1}+x_{2}\right)+x_{1}\right|=\left|4 k \pi+\dfrac{4 \pi}{3}+x_{1}\right|\)
顯然當\(k=0\)，\(x_{1}=-\dfrac{\pi}{6}\)時，即\(x_{1}=-\dfrac{\pi}{6}, x_{2}=\dfrac{5 \pi}{6}\)時取最小值\(\dfrac{7 \pi}{6}\).
(結合函數圖像分析)
鞏固練習
1(★★)已知函數\(f(x)=|\sqrt{3} \sin \omega x-\cos \omega x|(\omega>0)\)的最小正周期為\(π\)，則\(ω=\)\(\underline{\quad \quad}\) ．
2(★★)\(A ,B ,C\)是\(△ABC\)的內角，其中\(B=\dfrac{2 \pi}{3}\)，則\(\sin A+\sin C\)的取值范圍是\(\underline{\quad \quad}\) ．
3(★★)若函數\(f(x)=\sin 2 x-\sqrt{3} \cos 2 x\)在\([0 ,t]\)上的值域為\([-\sqrt{3}, 2]\)，則\(t\)的取值范圍為\(\underline{\quad \quad}\) ．
4(★★★) 已知函數\(f(x)=\sin \omega x+\cos \omega x(\omega>0)\)在\(\left(\dfrac{\pi}{6}, \dfrac{5 \pi}{12}\right)\)上僅有\(1\)個最值，且是最大值，則實數\(ω\)的取值范圍為\(\underline{\quad \quad}\) ．
5(★★★)已知函數\(f(x)=2 \sin \left(\omega x+\dfrac{\pi}{6}\right)+a\cos\omega x(a>0, \omega>0)\)對任意的\(x_1 ,x_2∈R\)，都有\(f\left(x_{1}\right)+f\left(x_{2}\right) \leq 4 \sqrt{3}\)，若\(f(x)\)在\([0 ,π]\)上的值域為\([3,2 \sqrt{3}]\)，則實數\(ω\)的取值范圍為\(\underline{\quad \quad}\) ．
答案
- \(1\)
- \(\left(\dfrac{\sqrt{3}}{2}, 1\right]\)
- \(\left[\dfrac{5 \pi}{12}, \dfrac{5 \pi}{6}\right]\)
- \(\left(\dfrac{3}{5}, \dfrac{3}{2}\right)\)
- \(\left[\dfrac{1}{6}, \dfrac{1}{3}\right]\)