【变式1】计算:(1)$\frac{\cos 27° - \sin 57° \sin 30°}{\cos 57°} =$；
(2)(一题多解)$\tan 23° + \tan 37° + \sqrt{3}\tan 23° · \tan 37° =$。

(1)$\frac{\cos 27° - \sin 57° \sin 30°}{\cos 57°} = \frac{\cos(57° - 30°) - \sin 57° \sin 30°}{\cos 57°} = \frac{\cos 57° \cos 30° + \sin 57° \sin 30° - \sin 57° \sin 30°}{\cos 57°} = \cos 30° = \frac{\sqrt{3}}{2}$。
(2)方法1:$\tan 23° + \tan 37° + \sqrt{3}\tan 23° · \tan 37° = \tan(23° + 37°)(1 - \tan 23° \tan 37°) + \sqrt{3}\tan 23° \tan 37° = \tan 60°(1 - \tan 23° \tan 37°) + \sqrt{3}\tan 23° \tan 37° = \sqrt{3} - \sqrt{3}\tan 23° \tan 37° + \sqrt{3}\tan 23° \tan 37° = \sqrt{3}$。
方法2:因为$\tan(23° + 37°) = \frac{\tan 23° + \tan 37°}{1 - \tan 23° \tan 37°}$，
所以$\sqrt{3} = \frac{\tan 23° + \tan 37°}{1 - \tan 23° \tan 37°}$，
所以$\sqrt{3} - \sqrt{3}\tan 23° \tan 37° = \tan 23° + \tan 37°$，
所以$\tan 23° + \tan 37° + \sqrt{3}\tan 23° \tan 37° = \sqrt{3}$。

【典例2】已知$\frac{π}{4}＜α＜\frac{3π}{4},0＜β＜\frac{π}{4},\cos(\frac{π}{4}+α)=-\frac{3}{5},\sin(\frac{3π}{4}+β)=\frac{5}{13}$,则$\sin(α+β)=（ ）$

A.$\frac{16}{65}$
B.$-\frac{16}{65}$
C.$-\frac{63}{65}$
D.$\frac{63}{65}$
解题指导 由题意可知,$\sin(\frac{π}{4}+α)=\frac{4}{5},\cos(\frac{3π}{4}+β)=-\frac{12}{13}$,进而根据$\sin(α+β)=-\sin[(\frac{π}{4}+α)+(\frac{3π}{4}+β)]$计算即可.
解析 由$\frac{π}{4}＜α＜\frac{3π}{4},0＜β＜\frac{π}{4}$,得$\frac{π}{2}＜\frac{π}{4}+α＜π,\frac{3π}{4}＜\frac{3π}{4}+β＜π$.
因为$\cos(\frac{π}{4}+α)=-\frac{3}{5},\sin(\frac{3π}{4}+β)=\frac{5}{13}$,
所以$\sin(\frac{π}{4}+α)=\frac{4}{5},\cos(\frac{3π}{4}+β)=-\frac{12}{13}$,
所以$\sin(α+β)=-\sin[(\frac{π}{4}+α)+(\frac{3π}{4}+β)]=-[\sin(\frac{π}{4}+α)\cos(\frac{3π}{4}+β)+\cos(\frac{π}{4}+α)\sin(\frac{3π}{4}+β)]=-[\frac{4}{5}×(-\frac{12}{13})+(-\frac{3}{5})×\frac{5}{13}]=\frac{63}{65}$.
答案 D

D

由$\frac{π}{4}＜α＜\frac{3π}{4},0＜β＜\frac{π}{4}$，可得$\frac{π}{2}＜\frac{π}{4}+α＜π$，$\frac{3π}{4}＜\frac{3π}{4}+β＜π$。
由同角三角函数基本关系，结合已知$\cos(\frac{π}{4}+α)=-\frac{3}{5}$，得$\sin(\frac{π}{4}+α)=\frac{4}{5}$；结合已知$\sin(\frac{3π}{4}+β)=\frac{5}{13}$，得$\cos(\frac{3π}{4}+β)=-\frac{12}{13}$。
利用角的变换：$(\frac{π}{4}+α)+(\frac{3π}{4}+β)=π+α+β$，因此$\sin(α+β)=-\sin[(\frac{π}{4}+α)+(\frac{3π}{4}+β)]$，代入两角和的正弦公式展开计算：
$\sin(α+β)=-[\sin(\frac{π}{4}+α)\cos(\frac{3π}{4}+β)+\cos(\frac{π}{4}+α)\sin(\frac{3π}{4}+β)]=-[\frac{4}{5}×(-\frac{12}{13})+(-\frac{3}{5})×\frac{5}{13}]=\frac{63}{65}$。

【典例3】已知$α \in (0,\dfrac{π}{2})$，且$\tan(α+\dfrac{π}{4})=\dfrac{4}{3}$.
(1)求$\sin2α$的值；
(2)(一题多解)求$\dfrac{\cos2α}{\sin(α+\dfrac{π}{4})}$的值.
解题指导 (1)利用两角和的正切公式求解$\tanα$，利用弦切互化结合平方关系求$\sinα$，$\cosα$，进而计算$\sin2α$.
(2)方法1:应用余弦二倍角公式结合正弦、余弦齐次式，将$\tanα$代入可求$\cos2α$，再通过$\tan(α+\dfrac{π}{4})=\dfrac{4}{3}$确定$\sin(α+\dfrac{π}{4})$的值，最终求出分式的结果.
答案 解:(1)由$\tan(α+\dfrac{π}{4})=\dfrac{\tanα +1}{1-\tanα}=\dfrac{4}{3}$，得$\tanα=\dfrac{1}{7}$，
则$\begin{cases}\cosα=7\sinα,\\\sin^2α+\cos^2α=1,\\\cosα＞0,\sinα＞0,\end{cases}$ 解得$\begin{cases}\sinα=\dfrac{\sqrt{2}}{10},\\\cosα=\dfrac{7\sqrt{2}}{10},\end{cases}$
所以$\sin2α=2\sinα\cosα=2×\dfrac{\sqrt{2}}{10}×\dfrac{7\sqrt{2}}{10}=\dfrac{28}{100}=\dfrac{7}{25}$.
(2)(一题多解)方法1:$\cos2α=\dfrac{\cos^2α-\sin^2α}{\sin^2α+\cos^2α}=\dfrac{1-\tan^2α}{\tan^2α+1}=\dfrac{1-(\dfrac{1}{7})^2}{(\dfrac{1}{7})^2+1}=\dfrac{\dfrac{48}{49}}{\dfrac{50}{49}}=\dfrac{24}{25}$.
由$α\in(0,\dfrac{π}{2})$可得，$α+\dfrac{π}{4}\in(\dfrac{π}{4},\dfrac{3π}{4})$.
因为$\tan(α+\dfrac{π}{4})=\dfrac{4}{3}＞0$，所以$α+\dfrac{π}{4}\in(\dfrac{π}{4},\dfrac{π}{2})$.
由$\begin{cases}4\cos(α+\dfrac{π}{4})=3\sin(α+\dfrac{π}{4}),\\\sin^2(α+\dfrac{π}{4})+\cos^2(α+\dfrac{π}{4})=1,\\\cos(α+\dfrac{π}{4})＞0,\sin(α+\dfrac{π}{4})＞0,\end{cases}$
解得$\sin(α+\dfrac{π}{4})=\dfrac{4}{5}$，
所以$\dfrac{\cos2α}{\sin(α+\dfrac{π}{4})}=\dfrac{\dfrac{24}{25}}{\dfrac{4}{5}}=\dfrac{24}{25}×\dfrac{5}{4}=\dfrac{6}{5}$.
方法2:$\dfrac{\cos2α}{\sin(α+\dfrac{π}{4})}=\dfrac{\cos^2α-\sin^2α}{\dfrac{\sqrt{2}}{2}(\sinα+\cosα)}=\sqrt{2}(\cosα-\sinα)=\sqrt{2}×(\dfrac{7\sqrt{2}}{10}-\dfrac{\sqrt{2}}{10})=\dfrac{6}{5}$.

解：
(1) 由两角和的正切公式：
$\tan(α+\frac{π}{4})=\frac{\tanα + 1}{1-\tanα}=\frac{4}{3}$
整理得$3(\tanα+1)=4(1-\tanα)$，解得$\tanα=\frac{1}{7}$。
由$\tanα=\frac{\sinα}{\cosα}=\frac{1}{7}$，得$\cosα=7\sinα$，
结合$α\in(0,\frac{π}{2})$，可知$\sinα＞0,\cosα＞0$，且$\sin^2α+\cos^2α=1$，
代入得$50\sin^2α=1$，解得：
$\begin{cases}\sinα=\frac{\sqrt{2}}{10}\\\cosα=\frac{7\sqrt{2}}{10}\end{cases}$
因此：
$\sin2α=2\sinα\cosα=2×\frac{\sqrt{2}}{10}×\frac{7\sqrt{2}}{10}=\frac{7}{25}$
(2) 方法1：
利用同角三角函数齐次性化简：
$\cos2α=\frac{\cos^2α-\sin^2α}{\sin^2α+\cos^2α}=\frac{1-\tan^2α}{\tan^2α+1}=\frac{1-(\frac{1}{7})^2}{(\frac{1}{7})^2+1}=\frac{24}{25}$
由$α\in(0,\frac{π}{2})$，得$α+\frac{π}{4}\in(\frac{π}{4},\frac{3π}{4})$，
又$\tan(α+\frac{π}{4})=\frac{4}{3}＞0$，故$α+\frac{π}{4}\in(\frac{π}{4},\frac{π}{2})$，即$\sin(α+\frac{π}{4})＞0$。
由$\tan(α+\frac{π}{4})=\frac{\sin(α+\frac{π}{4})}{\cos(α+\frac{π}{4})}=\frac{4}{3}$，结合$\sin^2(α+\frac{π}{4})+\cos^2(α+\frac{π}{4})=1$，解得$\sin(α+\frac{π}{4})=\frac{4}{5}$。
因此：
$\frac{\cos2α}{\sin(α+\frac{π}{4})}=\frac{\frac{24}{25}}{\frac{4}{5}}=\frac{6}{5}$
方法2：
由二倍角公式和两角和的正弦公式化简：
$\frac{\cos2α}{\sin(α+\frac{π}{4})}=\frac{\cos^2α-\sin^2α}{\frac{\sqrt{2}}{2}(\sinα+\cosα)}=\sqrt{2}(\cosα-\sinα)$
将$\sinα=\frac{\sqrt{2}}{10}$，$\cosα=\frac{7\sqrt{2}}{10}$代入得：
$\sqrt{2}×(\frac{7\sqrt{2}}{10}-\frac{\sqrt{2}}{10})=\frac{6}{5}$
综上，(1) $\sin2α=\frac{7}{25}$；(2) $\frac{\cos2α}{\sin(α+\frac{π}{4})}=\frac{6}{5}$。