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模塊導圖
知識剖析
\({\color{Red} {(本專題僅為公式求值、公式變換等鞏固練習，其應用在另一專題講解)}}\)
二倍角的正弦余弦正切公式
① \(\sin 2 \alpha=2 \sin \alpha \cos \alpha\)
②\(\cos 2 \alpha=\cos ^{2} \alpha-\sin ^{2} \alpha=1-2 \sin ^{2} \alpha=2 \cos ^{2} \alpha-1\)
③ \(\tan 2 \alpha=\dfrac{2 \tan \alpha}{1-\tan ^{2} \alpha}\)
(由\(S_{(\alpha \pm \beta)}\)、\(C_{(\alpha \pm \beta)}\)、\(T_{(\alpha \pm \beta)}\)可推導出\(\sin 2 \alpha\)，\(\cos 2 \alpha\)，$ \tan 2 \alpha$的公式)
降冪公式
\(\cos ^{2} \alpha=\dfrac{1+\cos 2 \alpha}{2}\) \(\sin ^{2} \alpha=\dfrac{1-\cos 2 \alpha}{2}\)
(由余弦倍角公式可得)
\({\color{Red} {（以下公式僅供了解）}}\)
半角公式
\(\sin \dfrac{\alpha}{2}=\pm \sqrt{\dfrac{1-\cos \alpha}{2}}\)
\(\cos \dfrac{\alpha}{2}=\pm \sqrt{\dfrac{1+\cos \alpha}{2}}\)
\(\tan \dfrac{\alpha}{2}=\pm \sqrt{\dfrac{1-\cos \alpha}{1+\cos \alpha}}\)
(由降冪公式可得)
萬能公式
\(\sin \alpha=\dfrac{2\tan \dfrac{\alpha}{2}}{1+\tan ^{2} \dfrac{\alpha}{2}}\)
\(\cos \alpha=\dfrac{1-\tan ^{2} \dfrac{\alpha}{2}}{1+\tan ^{2} \dfrac{\alpha}{2}}\)
\(\tan \alpha=\dfrac{2 \tan \dfrac{\alpha}{2}}{1-\tan ^{2} \dfrac{\alpha}{2}}\)
(由倍角公式可得)
積化和公式
\(\begin{aligned} &\sin \alpha \cdot \cos \beta=\dfrac{1}{2}[\sin (\alpha+\beta)+\sin (\alpha-\beta)] \\ &\cos \alpha \cdot \cos \beta=\dfrac{1}{2}[\cos (\alpha+\beta)+\cos (\alpha-\beta)] \\ &\sin \alpha \cdot \sin \beta=\dfrac{1}{2}[\cos (\alpha-\beta)-\cos (\alpha+\beta)] \end{aligned}\)
(由和差公式可得)
和化積公式
\(\sin \alpha+\sin \beta=2 \sin \dfrac{\alpha+\beta}{2} \cos \dfrac{\alpha-\beta}{2}\)
\(\sin \alpha-\sin \beta=2 \cos \dfrac{\alpha+\beta}{2} \sin \dfrac{\alpha-\beta}{2}\)
\(\cos \alpha+\cos \beta=2 \cos \dfrac{\alpha+\beta}{2} \cos \dfrac{\alpha-\beta}{2}\)
\(\cos \alpha-\cos \beta=-2 \sin \dfrac{\alpha+\beta}{2} \sin \dfrac{\alpha-\beta}{2}\)
(由和差公式可得)
經典例題
【題型一】 倍角公式的運用
【典題1】 求值\(\dfrac{\cos 20^{\circ}}{\cos 35^{\circ} \sqrt{1-\sin 20^{\circ}}}=\) \(\underline{\quad{} \quad{}}\)．
【解析】\(\dfrac{\cos 20^{\circ}}{\cos 35^{\circ} \sqrt{1-\sin 20^{\circ}}}\)
\(\begin{aligned} &=\dfrac{\cos ^{2} 10^{\circ}-\sin ^{2} 10}{\cos \left(45^{\circ}-10^{\circ}\right)\left(\cos 10^{\circ}-\sin 10^{\circ}\right)} \\ &=\dfrac{\cos 10^{\circ}+\sin 10^{\circ}}{\cos 45^{\circ} \cos 10^{\circ}+\sin 45^{\circ} \sin 10^{\circ}} \\ &=\dfrac{\cos 10^{\circ}+\sin 10^{\circ}}{\dfrac{\sqrt{2}}{2}\left(\cos 10^{\circ}+\sin 10^{\circ}\right)} \\ &=\sqrt{2} . \end{aligned}\)
【典題2】計算\(4 \cos 50^{\circ}-\tan 40^{\circ}=\)\(\underline{\quad{} \quad{}}\).
【解析】 \(4 \cos 50^{\circ}-\tan 40^{\circ}\)
\(\begin{aligned} &=4 \cos 50^{\circ}-\dfrac{\sin 40^{\circ}}{\cos 40^{\circ}}=\dfrac{4 \cos 50^{\circ} \cos 40^{\circ}-\sin 40^{\circ}}{\cos 40^{\circ}} \\ &=\dfrac{4 \sin 40^{\circ} \cos 40^{\circ}-\sin 40^{\circ}}{\cos 40^{\circ}}=\dfrac{2 \sin 80^{\circ}-\sin 40^{\circ}}{\cos 40^{\circ}} \end{aligned}\)
\(\begin{aligned} &=\dfrac{2 \cos 10^{\circ}-\sin 40^{\circ}}{\cos 40^{\circ}}=\dfrac{2 \cos \left(40^{\circ}-30^{\circ}\right)-\sin 40^{\circ}}{\cos 40^{\circ}} \\ &=\dfrac{\sqrt{3} \cos 40^{\circ}}{\cos 40^{\circ}}=\sqrt{3} \end{aligned}\)
【點撥】
① 正切化弦；
② 注意角度之間的關系，比如互余(\(50^{\circ}\)與\(40^{\circ}\)，\(80^{\circ}\)與\(10^{\circ}\))、倍數關系、角度相差值是特殊值(\(10^{\circ}\)與\(40^{\circ}\)相差\(30^°\)).
【典題3】如果\(\dfrac{1+\tan \alpha}{1-\tan \alpha}=2013\)，那么\(\dfrac{1}{\cos 2 \alpha}+\tan 2 \alpha=\) \(\underline{\quad{} \quad{}}\).
【解析】\(\dfrac{1}{\cos 2 \alpha}+\tan 2 \alpha\)
\(=\dfrac{1}{\cos 2 \alpha}+\dfrac{\sin 2 \alpha}{\cos 2 \alpha}=\dfrac{1+\sin 2 \alpha}{\cos 2 \alpha}\) \({\color{Red} {(化切為弦)}}\)
\(\begin{aligned} &=\dfrac{(\cos \alpha+\sin \alpha)^{2}}{(\cos \alpha+\sin \alpha)(\cos \alpha-\sin \alpha)} \\ &=\dfrac{\cos \alpha+\sin \alpha}{\cos \alpha-\sin \alpha} \\ &=\dfrac{1+\tan \alpha}{1-\tan \alpha}=2013 \end{aligned}\)
【點撥】
① 本題的思路有二，一是先化簡所求式子再利用已知條件，化二倍角為一倍角；二是由已知可求\(\tan \alpha\)，進而可得\(\sinα,\cosα\)，再求\(\tan2α\)與\(\cos2α\)得結果，但數值不好求.
② 化切為弦是常見思路，也可\(\dfrac{1}{\cos 2 \alpha}+\tan 2 \alpha\)\(=\dfrac{\cos ^{2} \alpha+\sin ^{2} \alpha}{\cos ^{2} \alpha-\sin ^{2} \alpha}+\dfrac{2 \tan \alpha}{1-\tan ^{2} \alpha}=\dfrac{1+\tan ^{2} \alpha}{1-\tan ^{2} \alpha}\)\(+\dfrac{2 \tan \alpha}{1-\tan ^{2} \alpha}=\dfrac{(1+\tan \alpha)^{2}}{1-\tan ^{2} \alpha}=\dfrac{1+\tan \alpha}{1-\tan \alpha}=2013\)．方法多樣，多思考.
【典題4】已知\(\sin \left(\dfrac{\pi}{12}-\dfrac{\alpha}{2}\right)=\dfrac{\sqrt{3}}{3}\)，則\(\sin \left(2 \alpha+\dfrac{\pi}{6}\right)\)的值為\(\underline{\quad{} \quad{}}\).
【解析】\(\because \sin \left(\dfrac{\pi}{12}-\dfrac{\alpha}{2}\right)=\dfrac{\sqrt{3}}{3}\)，
\(\therefore \cos \left(\dfrac{\pi}{6}-\alpha\right)=1-2 \sin ^{2}\left(\dfrac{\pi}{12}-\dfrac{\alpha}{2}\right)=\dfrac{1}{3}\)，
\(\therefore \sin \left(2 \alpha+\dfrac{\pi}{6}\right)=\cos \left(\dfrac{\pi}{3}-2 \alpha\right)\)\(=2 \cos ^{2}\left(\dfrac{\pi}{6}-\alpha\right)-1=2 \times\left(\dfrac{1}{3}\right)^{2}-1=-\dfrac{7}{9}\)．
【點撥】\(\dfrac{\alpha}{2}\)與\(2α\)是四倍關系，故可用借助\(α\)進行轉化；解題中多用綜合法與分析法求解.
【典題5】 若\(\alpha \in\left(0, \dfrac{\pi}{2}\right)\)，且\(\cos 2 \alpha=\dfrac{\sqrt{2}}{5} \sin \left(\alpha+\dfrac{\pi}{4}\right)\)，則\(\tanα=\)\(\underline{\quad{} \quad{}}\)．
【解析】\(\because \alpha \in\left(0, \dfrac{\pi}{2}\right)\)，且\(\cos 2 \alpha=\dfrac{\sqrt{2}}{5} \sin \left(\alpha+\dfrac{\pi}{4}\right)\)，
\(\therefore \cos 2 \alpha=\dfrac{\sqrt{2}}{5} \times \dfrac{\sqrt{2}}{2}(\sin \alpha+\cos \alpha)\)\(=\dfrac{1}{5}(\sin \alpha+\cos \alpha)\)，
\(\therefore \cos ^{2} \alpha-\sin ^{2} \alpha\)\(=(\cos \alpha-\sin \alpha)(\sin \alpha+\cos \alpha)=\dfrac{1}{5}(\sin \alpha+\cos \alpha)\)，
\(\therefore \cos \alpha-\sin \alpha=\dfrac{1}{5}\) ① ，
\(∴\)①式兩邊平方可得\(1-2 \sin \alpha \cos \alpha=\dfrac{1}{25}\)，
解得\(2 \sin \alpha \cos \alpha=\dfrac{24}{25}\)，
\(\therefore \dfrac{2 \sin \alpha \cos \alpha}{\sin ^{2} \alpha+\cos ^{2} \alpha}=\dfrac{2 \tan \alpha}{1+\tan ^{2} \alpha}=\dfrac{24}{25}\)，
\({\color{Red} {(巧用\sin ^{2} \alpha+\cos ^{2} \alpha=1，齊次化處理)}}\)
可得\(12 \tan ^{2} \alpha-25 \tan \alpha+12=0\)，解得\(\tan \alpha=\dfrac{3}{4}\)或\(\dfrac{4}{3}\)．
由①可知\(\cos \alpha>\sin \alpha\)，即\(\tanα<1\)，
\({\color{Red} {(注意對最后求值的取舍)}}\)
\(\therefore \tan \alpha=\dfrac{3}{4}\).
【點撥】
本題的處理方法很多，平時要多注意一題多解，提高對公式靈活運用的能力.
比如湊角\(\cos 2 \alpha=\dfrac{\sqrt{2}}{5} \sin \left(\alpha+\dfrac{\pi}{4}\right) \Rightarrow \sin 2\left(\alpha+\dfrac{\pi}{4}\right)=\dfrac{\sqrt{2}}{5} \sin \left(\alpha+\dfrac{\pi}{4}\right)\)；得到\(\cos \alpha-\sin \alpha=\dfrac{1}{5}\)后能求出\(\cosα\)和\(\sinα\)等等.
鞏固練習
1(★)計算\(\dfrac{\sqrt{3}-\tan 12^{\circ}}{\left(2 \cos ^{2} 12^{\circ}-1\right) \sin 12^{\circ}}=\)\(\underline{\quad{} \quad{}}\)．
2(★)已知\(\theta \in\left(0, \dfrac{\pi}{2}\right)\)，\(\sin \theta=\dfrac{\sqrt{5}}{5}\)，則\(\dfrac{\cos 2 \theta}{\tan \theta}=\)\(\underline{\quad{} \quad{}}\).
3(★)若\(\tan \alpha+\dfrac{1}{\tan \alpha}=3\)，則\(\cos4α=\)\(\underline{\quad{} \quad{}}\).
4(★★) 設\(\tan \alpha=\dfrac{1}{2}\)，\(\cos (\pi+\beta)=-\dfrac{4}{5}(\beta \in(0, \pi))\)，則\(\tan (2 \alpha-\beta)\)的值為\(\underline{\quad{} \quad{}}\).
5(★★)已知\(\alpha \in(0, \pi)\)，且\(3 \cos 2 \alpha-8 \cos \alpha=5\)，則\(\sinα=\) \(\underline{\quad{} \quad{}}\).
6(★★) 已知\(\alpha \in\left(0, \dfrac{\pi}{2}\right)\)，若\(\sin 2 \alpha-2 \cos 2 \alpha=2\)，則\(\sinα=\)\(\underline{\quad{} \quad{}}\).
7(★★)已知\(\alpha \in\left(\dfrac{\pi}{2}, \pi\right)\)，\(\tan 2 \alpha=\dfrac{3}{4}\)，則\(\sin 2 \alpha+\cos ^{2} \alpha=\)\(\underline{\quad{} \quad{}}\).
8(★★)已知\(\sinα=2\sinβ\)，\(\tanα=3\tanβ\)，則\(\cos2α=\)\(\underline{\quad{} \quad{}}\).
參考答案
- \(8\)
- \(\dfrac{6}{5}\)
- \(\dfrac{1}{9}\)
- \(\dfrac{7}{24}\)
- \(\dfrac{\sqrt{5}}{3}\)
- \(\dfrac{2 \sqrt{5}}{5}\)
- \(-\dfrac{1}{2}\)
- \(-\dfrac{1}{4}\)或\(1\)
【題型二】 降冪公式的運用
【典題1】 在\(∆ ABC\)中，若\(3 \cos ^{2} \dfrac{A-B}{2}+5 \sin ^{2} \dfrac{A+B}{2}=4\)，求\(\tan A \tan B\).
【解析】在\(∆ ABC\)中，若\(3 \cos ^{2} \dfrac{A-B}{2}+5 \sin ^{2} \dfrac{A+B}{2}=4\)
\(\therefore 3 \times \dfrac{1+\cos (A-B)}{2}+5 \times \dfrac{1-\cos (A+B)}{2}=4\)
即\(\dfrac{3}{2} \cos (A-B)-\dfrac{5}{2} \cos (A+B)=0\)
即\(3(\cos A \cos B+\sin A \sin B)=5(\cos A \cos B-\sin A \sin B)\)，
即\(2 \cos A \cos B=8 \sin A \sin B\) ，
\(\therefore \tan A \tan B=\dfrac{1}{4}\)
【點撥】式子中出現“平方”形式，想到降冪公式\(\cos ^{2} \alpha=\dfrac{1+\cos 2 \alpha}{2}\)、\(\sin ^{2} \alpha=\dfrac{1-\cos 2 \alpha}{2}\).
鞏固練習
1(★★)若\(\cos 2 \theta=\dfrac{1}{4}\)，則\(\sin ^{2} \theta+2 \cos ^{2} \theta\)的值為 \(\underline{\quad{} \quad{}}\).
2(★★)已知\(\tan \theta\)是方程\(x^{2}-6 x+1=0\)的一根，則\(\cos^{2}\left(\theta+\dfrac{\pi}{4}\right)=\)\(\underline{\quad{} \quad{}}\).
3(★★)已知\(\dfrac{\cos 2 \alpha}{\sin \alpha+\cos \alpha}=\dfrac{\sqrt{2}}{4}\)，則\(\cos ^{2}\left(\dfrac{3}{4} \pi+\alpha\right)\)的值是\(\underline{\quad{} \quad{}}\).
參考答案
- \(\dfrac{13}{8}\)
- \(\dfrac{1}{3}\)
- \(\dfrac{7}{8}\)
【題型三】角的變換
【典題1】若\(\sin \left(\theta+\dfrac{\pi}{8}\right)=\dfrac{1}{3}\)，則\(\sin \left(2 \theta-\dfrac{\pi}{4}\right)=\)\(\underline{\quad{} \quad{}}\).
【解析】\(\because 2 \theta-\dfrac{\pi}{4}=2\left(\theta+\dfrac{\pi}{8}\right)-\dfrac{\pi}{2}\)，
\(\therefore \sin \left(2 \theta-\dfrac{\pi}{4}\right)=\sin \left[2\left(\theta+\dfrac{\pi}{8}\right)-\dfrac{\pi}{2}\right]=-\cos 2\left(\theta+\dfrac{\pi}{8}\right)\)\(=-\left[1-2 \sin ^{2}\left(\theta+\dfrac{\pi}{8}\right)\right]=-\dfrac{7}{9}\)．
【點撥】因為已知角\(\theta+\dfrac{\pi}{8}\)和所求角\(2 \theta-\dfrac{\pi}{4}\)中\(θ\)的系數是\(2\)倍的關系，故想到\(2\left(\theta+\dfrac{\pi}{8}\right)\)與\(2 \theta-\dfrac{\pi}{4}\)的差\(\dfrac{\pi}{2}\)是特殊角為關鍵，則有\(2 \theta-\dfrac{\pi}{4}=2\left(\theta+\dfrac{\pi}{8}\right)-\dfrac{\pi}{2}\).
【典題2】 已知\(\sin \left(\alpha+\dfrac{3 \pi}{4}\right)=\dfrac{4}{5}\)，\(\cos \left(\dfrac{\pi}{4}-\beta\right)=\dfrac{3}{5}\)，且\(-\dfrac{\pi}{4}<\alpha<\dfrac{\pi}{4}\) ，\(\dfrac{\pi}{4}<\beta<\dfrac{3 \pi}{4}\)，求\(\cos2(α－β)\)的值．
【解析】 由\(-\dfrac{\pi}{4}<\alpha<\dfrac{\pi}{4}\)得，\(\dfrac{\pi}{2}<\alpha+\dfrac{3}{4} \pi<\pi\)，
\({\color{Red} {(注意角度的范圍)}}\)
所以\(\cos \left(\alpha+\dfrac{3}{4} \pi\right)=-\sqrt{1-\sin ^{2}\left(\alpha+\dfrac{3}{4} \pi\right)}=-\dfrac{3}{5}\)，
由\(\dfrac{\pi}{4}<\beta<\dfrac{3}{4} \pi\)得，\(-\dfrac{\pi}{2}<\dfrac{\pi}{4}-\beta<0\)，
所以\(\sin \left(\dfrac{\pi}{4}-\beta\right)=-\sqrt{1-\cos ^{2}\left(\dfrac{\pi}{4}-\beta\right)}=-\dfrac{4}{5}\)，
所以\(\cos \left[\left(\alpha+\dfrac{3}{4} \pi\right)+\left(\dfrac{\pi}{4}-\beta\right)\right]\)
\(=\cos \left(\alpha+\dfrac{3}{4} \pi\right) \cos \left(\dfrac{\pi}{4}-\beta\right)-\sin \left(\alpha+\dfrac{3}{4} \pi\right) \sin \left(\dfrac{\pi}{4}-\beta\right)\)
\(=\left(-\dfrac{3}{5}\right) \times \dfrac{3}{5}-\dfrac{4}{5} \times\left(-\dfrac{4}{5}\right)=\dfrac{7}{25}\)
即\(-\cos (\alpha-\beta)=\dfrac{7}{25}\)，
所以\(\cos 2(\alpha-\beta)=2 \cos ^{2}(\alpha-\beta)-1\)\(=2 \times\left(-\dfrac{7}{25}\right)^{2}-1=-\dfrac{527}{625}\)
【點撥】 本題關鍵在於發現兩個已知角之和\(\left(\alpha+\dfrac{3}{4} \pi\right)+\left(\dfrac{\pi}{4}-\beta\right)=\pi+\alpha-\beta\)與所求角\(2(α－β)\)之間差個特殊角\(π\)存在兩倍的關系.
【總結】
① 當已知角只有一個時，可已知角與所求角的和或差的值是否為一固定特殊角，或看已知角(所求角)的\(2\)倍與所求角(已知角)和或差的值是否為一固定特殊角；
當已知角有兩個時，主要看兩個已知角的和或差形式與所求角的關系；
特殊角為\(0\)、\(\dfrac{\pi}{3}\) 、\(\dfrac{\pi}{4}\)、 \(\dfrac{\pi}{6}\) 、\(π\)等.
② 常見的角變換有：\(\alpha=2 \cdot \dfrac{\alpha}{2}\)，\(\alpha=(\alpha+\beta)-\beta=\beta-(\alpha+\beta)\)，\(\dfrac{\pi}{4}+\alpha=\dfrac{\pi}{2}-\left(\dfrac{\pi}{4}-\alpha\right)\)，
\(\beta=\dfrac{1}{2}[(\alpha+\beta)-(\alpha-\beta)]^{2}\)]等.
③ 在運用和差角公式和倍角公式時，要注意“整體思想”的運用.
鞏固練習
1(★★)若\(\cos \left(\alpha+\dfrac{\pi}{12}\right)=\dfrac{\sqrt{2}}{3}\)，則\(\sin \left(\dfrac{\pi}{3}-2 \alpha\right)\)的值為\(\underline{\quad{} \quad{}}\)．
2(★★)已知\(\cos \left(\alpha+\dfrac{\pi}{6}\right)=\dfrac{3}{5}\)，\(\alpha \in\left(0, \dfrac{\pi}{2}\right)\)，則\(\cos \left(2 \alpha+\dfrac{7 \pi}{12}\right)=\) \(\underline{\quad{} \quad{}}\)．
3(★★) 已知\(\cos \left(\theta+\dfrac{\pi}{6}\right)=-\dfrac{\sqrt{3}}{3}\)，則\(\sin \left(\dfrac{\pi}{6}-2 \theta\right)=\)\(\underline{\quad{} \quad{}}\)．
4(★★) 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\(β=\)\(\underline{\quad{} \quad{}}\)．
5(★★★) 已知\(\dfrac{\pi}{2}<\beta<\alpha<\dfrac{3 \pi}{4}\)，且\(\cos (\alpha-\beta)=\dfrac{12}{13}\)，\(\sin (\alpha+\beta)=-\dfrac{3}{5}\)，求\(\cos2α\)的值．
6(★★★)設\(0<x_{1}<x_{2}<\pi\)，若\(\sin \left(2 x_{1}-\dfrac{\pi}{3}\right)=\sin \left(2 x_{2}-\dfrac{\pi}{3}\right)=\dfrac{3}{5}\)，求\(\cos(x_1-x_2)\).
參考答案
- \(-\dfrac{5}{9}\)
- \(-\dfrac{31 \sqrt{2}}{50}\)
- \(-\dfrac{1}{3}\)
- \(\dfrac{\pi}{4}\)
- \(-\dfrac{33}{65}\)
- \(\dfrac{3}{5}\)
【題型四】簡單的三角恆等變換\({\color{Red} {(選學內容)}}\)
【典題1】 若\(\alpha \in(0, \pi)\)，且\(\sinα+2\cosα=2\)，則\(\tan \dfrac{\alpha}{2}\)等於 \(\underline{\quad{} \quad{}}\).
【解析】\(∵α∈(0 ,π)\)，\(\therefore \dfrac{\alpha}{2} \in\left(0, \dfrac{\pi}{2}\right)\)，
設\(\tan \dfrac{\alpha}{2}=x\)，\(x>0\)，
\(\because \sin \alpha=\dfrac{2 \tan \dfrac{\alpha}{2}}{1+\tan ^{2} \dfrac{\alpha}{2}}=\dfrac{2 x}{1+x^{2}}\)，
\(\cos \alpha=\dfrac{1-\tan ^{2} \dfrac{\alpha}{2}}{1+\tan ^{2} \dfrac{\alpha}{2}}=\dfrac{1-x^{2}}{1+x^{2}}\)，
\(\therefore \sin \alpha+2 \cos \alpha=\dfrac{2 x}{1+x^{2}}+2 \cdot \dfrac{1-x^{2}}{1+x^{2}}=\dfrac{2 x+2-2 x^{2}}{1+x^{2}}=2\)，
即\(x+1-x^{2}=1+x^{2}\)，解得\(x=\dfrac{1}{2}\).
【點撥】本題利用萬能公式，也可利用\(\sinα+2\cosα=2\)求出\(\sinα,\cosα\)，再求\(\tanα\)得到\(\tan \dfrac{a}{2}\).
【典題2】在\(△ABC\)中，\(B=\dfrac{\pi}{4}\)，則\(\sin A\sin C\)的最大值是\(\underline{\quad{} \quad{}}\).
【解析】\({\color{Red} {方法一 兩角和差公式、二倍角公式}}\)
\(\sin A \sin C=\sin A \sin (\pi-A-B)\)
\(=\sin A \sin \left(\dfrac{3 \pi}{4}-A\right)\)
\(=\sin A\left(\dfrac{\sqrt{2}}{2} \cos A+\dfrac{\sqrt{2}}{2} \sin A\right)\)
\(=\dfrac{\sqrt{2}}{4} \sin 2 A-\dfrac{\sqrt{2}}{4} \cos 2 A+\dfrac{\sqrt{2}}{4}\)
\(=\dfrac{1}{2} \sin \left(2 A-\dfrac{\pi}{4}\right)+\dfrac{\sqrt{2}}{4}\)
\(\because 0<A<\dfrac{3 \pi}{4}\)，
\(\therefore-\dfrac{\pi}{4}<2 A-\dfrac{\pi}{4}<\dfrac{5 \pi}{4}\)
\(∴\)當\(2 A-\dfrac{\pi}{4}=\dfrac{\pi}{2}\)，即\(A=\dfrac{3 \pi}{8}\)時，
\(\sin A\sin C\)取得最大值\(\dfrac{2+\sqrt{2}}{4}\)．
\({\color{Red} {方法二 積化和差}}\)
\(\sin A \sin C=\dfrac{1}{2}[\cos (A-C)-\cos (A+C)]\)
\(=\dfrac{1}{2}\left[\cos (A-C)-\cos \dfrac{3 \pi}{4}\right]\)
\(=\dfrac{1}{2}\left[\cos (A-C)+\dfrac{\sqrt{2}}{2}\right]\)
\(\because-1 \leq \cos (A-C) \leq 1\)
\(\therefore-\dfrac{2-\sqrt{2}}{4} \leq \dfrac{1}{2}\left[\cos (A-C)+\dfrac{\sqrt{2}}{2}\right] \leq \dfrac{2+\sqrt{2}}{4}\).
當\(A-C=0\)，即\(A=C=\dfrac{3 \pi}{8}\)時，
\(\sin A\sin C\)取得最大值\(\dfrac{2+\sqrt{2}}{4}\)．
【點撥】掌握積化和差公式，對於處理含涉及\(\sin A\sin B\)，\(\cos A \cos B\)，\(\sin A \cos B\)的題目較為有利.
鞏固練習
1(★★)\(\sin ^{2} 20^{\circ}+\cos 80^{\circ} \cos 40^{\circ}=\)\(\underline{\quad{} \quad{}}\)．
2(★★) \(\dfrac{\sin \left(\alpha+30^{\circ}\right)-\sin \left(\alpha-30^{\circ}\right)}{\cos \alpha}\)的值為\(\underline{\quad{} \quad{}}\)．
3(★★)已知\(θ\)為第二象限角，\(25 \sin ^{2} \theta+\sin \theta-24=0,\)，則\(\sin \dfrac{\theta}{2}\)的值為\(\underline{\quad{} \quad{}}\).
4(★★)若\(\sin \alpha=-\dfrac{3}{5}\)，\(α\)是第三象限角，則\(\dfrac{1-\tan \dfrac{\alpha}{2}}{1+\tan \dfrac{\alpha}{2}}=\)\(\underline{\quad{} \quad{}}\).
5(★★)已知\(\cos \alpha+\cos \beta=\dfrac{1}{2}\)，則\(\cos \dfrac{\alpha+\beta}{2} \cos \dfrac{\alpha-\beta}{2}\)的值為\(\underline{\quad{} \quad{}}\).
6(★★★) 已知\(α ,β\)為銳角，且\(\alpha-\beta=\dfrac{\pi}{6}\)，那么\(\sin \alpha \sin \beta\)的取值范圍是\(\underline{\quad{} \quad{}}\)．
7(★★★) \(\cos \dfrac{\pi}{7}+\cos \dfrac{3 \pi}{7}+\cos \dfrac{5 \pi}{7}=\) \(\underline{\quad{} \quad{}}\)．
答案
- \(\dfrac{1}{4}\)
- \(1\)
- \(\pm \dfrac{4}{5}\)
- \(-2\)
- \(\dfrac{1}{4}\)
- \(\left(0, \dfrac{\sqrt{3}}{2}\right)\)
- \(\dfrac{1}{2}\)