${\color{Red} 分式裂项}$ 分式裂项
一、基本原理
裂消原理：$\cfrac{大－小}{小*大}＝\cfrac{1}{小}－\cfrac{1}{大}$
相消规律：对称性
相邻项，首尾各剩下1项；
隔一项，首尾各剩下2项；
隔两项，首尾各剩下3项；
二、八大裂项模型：
${\color{Red} \quad ①\quad a_n=\cfrac{1}{n(n+1)} =\cfrac{1}{n} -\cfrac{1}{n+1} } \quad \qquad$
${\color{Green}\quad ②\quad a_n=\cfrac{1}{(2n-1)(2n+1)} =\cfrac{1}{2} \cfrac{(2n+1)-(2n-1)}{(2n-1)(2n+1)}=\cfrac{1}{2}\cdot (\cfrac{1}{2n-1}-\cfrac{1}{2n+1})}\quad \qquad$
${\color{Red} \quad ③\quad a_n=\cfrac{3}{(6n-1)(6n+5)} =\cfrac{1}{2} \cfrac{(6n+5)-(6n-1)}{(6n-1)(6n+5)}=\cfrac{1}{2} \cfrac{6}{(6n-1)(6n+5)}}\quad \qquad$
${\color{Green}\quad ④\quad a_n=\cfrac{2^{n+1}}{(2^n+1)(2^{n+1}+1)} =\cfrac{2[(2^{n+1}+1)-(2^n+1)]}{(2^n+1)(2^{n+1}+1)}}\quad $
${\color{Red} \quad ⑤\quad a_n=\cfrac{n+1}{4n^2(n+2)^2} =\cfrac{1}{4} \cdot \cfrac{1}{4} \cdot \cfrac{(n+2)^2-n^2}{n^2(n+2)^2}}\quad $

${\color{Green}\quad ⑥\quad a_n=(-1)^n\cfrac{2n+1}{n(n+1)}=(-1)^n\cdot\cfrac{n+1+n}{n(n+1)} }\quad 裂项要出现负号，这里有-1$
${\color{Red} \quad ⑦\quad a_n=\cfrac{1}{n(n+1)(n+2)} =\cfrac{1}{2} \cfrac{(n-2)-n}{n(n+1)(n+2)}=\cfrac{1}{2} [\cfrac{1}{n(n+1)}-\cfrac{1}{(n+1)(n+2)} ]}\quad$当中间n+1项是透明的
${\color{Green}\quad ⑧\quad a_n=\cfrac{n}{(n+1)!}=\cfrac{(n+1)-1}{1\times 2\times 3\times4 \times \cdots \cdots (n+1)} ＝\cfrac{1}{n!}-\cfrac{1}{(n+1)!} }\quad$ 只看首尾项
实战：

(1)设$b_n=(2n-1)a_n\quad S_n表示b_n的前n项和，S_n=2n$
$①n=1时，S_1=b_1=2;$
$②n\ge 2时，S_n-S_{n-1}=b_n=2n-2(n-1)=2,故b_n$是常数数列。
$b_n=(2n-1)a_n=2\Rightarrow a_n=\cfrac{2}{2n-1}$
(2)$\cfrac{a_n}{2n+1} =\cfrac{2}{(2n-1)(2n+1)}=\cfrac{1}{2n-1}-\cfrac{1}{2n+1}$
设前n项和为$T_n={\color{Green} \cfrac{1}{1}} -\cfrac{1}{3}+\cfrac{1}{3} -\cfrac{1}{5}+\cfrac{1}{5} -\cfrac{1}{7}+\cdots \cdots+$
$\cfrac{1}{2n-3}-\cfrac{1}{2n-1}+\cfrac{1}{ 2n-1}-{\color{Green} \cfrac{1}{2n+1}} =\cfrac{2n}{2n+1}$

$n\ge 2时，a_n=S_n-S_{n-1}=\sqrt{S_n} +\sqrt{S_{n-1}} \Rightarrow (\sqrt{S_n})^2 -(\sqrt{S_{n-1}})^2 =\sqrt{S_n} +\sqrt{S_{n-1}}$
$\sqrt{S_n} -\sqrt{S_{n-1}}=1\Rightarrow \sqrt{S_n}$ 是公差为1的首项为$\sqrt{a_1} $的等差数列。
$\sqrt{S_n} =n\Rightarrow S_n=n^2\quad a_n=n^2-(n-1)^2=2n-1$
$b_n=(-1)^n\cfrac{n}{a_na_{n+1}}= (-1)^n\cfrac{n}{(2n-1)(2n+1)}=\cfrac{1}{4} \times (-1)^n\cfrac{2n+1+2n-1}{(2n-1)(2n+1)}$
$b_n=\cfrac{1}{4} (-1)^n(\cfrac{1}{2n-1}+\cfrac{1}{2n+1} )$
$T_{2n}=\cfrac{1}{4} [{\color{Green}-(\cfrac{1}{1} } +\cfrac{1}{3})+(\cfrac{1}{3} +\cfrac{1}{5})-(\cfrac{1}{5}-\cfrac{1}{7})+\dots +\cfrac{1}{4n-1}+{\color{Green}\cfrac{1}{4n+1} } ]$
$T_{2n}=-\cfrac{n}{4n+1}$

${\color{Red} PART\quad TWO\quad 根式裂项}$
一、模型总结：
${\color{Red} \quad ②\quad a_n=\cfrac{1}{\sqrt{n}+\sqrt{n+1} } =\cfrac{\sqrt{n+1}-\sqrt{n} }{(\sqrt{n+1}+\sqrt{n} )(\sqrt{n+1}-\sqrt{n} )} =\sqrt{n+1}-\sqrt{n} }$
${\color{Green}\quad ②\quad a_n=\cfrac{1}{\sqrt{n}(n+1)+n\sqrt{n+1} } =\cfrac{1}{\sqrt{n}\sqrt{n+1}(\sqrt{n+1} +\sqrt{n} ) } =\cfrac{1}{\sqrt{n} \sqrt{n+1} }(\sqrt{n+1}-\sqrt{n})}$
${\color{Green} \quad ②\quad a_n=\cfrac{1}{\sqrt{n} } -\cfrac{1}{\sqrt{n+1}} } $


${\color{Green} a_n=\cfrac{1}{\sqrt{n} } -\cfrac{1}{\sqrt{n+1}} } $
$S_n=\cfrac{1}{\sqrt{1} } -\cfrac{1}{\sqrt{n+1}} $
$S_{2024}=1-\cfrac{1}{\sqrt{2024+1}} ,45^2=2025,2^2=3+12^2+....+45^2,45-2+1=44个$

${\color{Red} 第三部分、平方递推裂项} $
一、模型总结：
${\color{Green} ①a_{n+1}=a_n^2-a_n+1,(a_{n+1}=a_n^2两边对数求通项，但此式求不了通项。} $
$a_{n+1}-a_n=(a_n^2-1)^2,a_1\ne 1时，$是递增数列。
$a_{n+1}=a_n^2-a_n+1\Rightarrow a_{n+1}=a_n(a_n-1)+1\Rightarrow a_{n+1}-1=a_n(a_n-1)\Rightarrow 取倒\cfrac{1}{a_{n+1}-1}=\cfrac{1}{a_n(a_n-1)} $
$\cfrac{1}{a_{n+1}-1}=\cfrac{1}{a_n(a_n-1)} =\cfrac{1}{a_n-1} -\cfrac{1}{a_n} \Rightarrow \cfrac{1}{a_n}=\cfrac{1}{a_n-1} - \cfrac{1}{a_{n+1}-1}$裂项完成
步骤：因式分解；移系数；取倒数；裂项；互移。

$\cfrac{1}{a_n}=\cfrac{1}{a_n-1} - \cfrac{1}{a_{n+1}-1}$
取倒前的那一步：$a_{n+1}-1=a_n(a_n-1)\Rightarrow \cfrac{1}{a_n}=\cfrac{a_n-1}{a_{n+1}-1} $
$A_n=1-\cfrac{1}{a_{n+1}}\quad B_n=\cfrac{a_1-1}{a_{n+1}-1} $

步骤：因式分解；移系数；取倒数；裂项；互移。
$ a_{n+1}-2=\cfrac{1}{2} a_n(a_n-2)\Rightarrow \cfrac{1}{a_{n+1}-2} =\cfrac{2}{ a_n(a_n-2)} $
$ \cfrac{1}{a_{n+1}-2} =\cfrac{a_n-(a_n-2)}{a_n(a_n-2)} =\cfrac{1}{a_n-2}-\cfrac{1}{a_n} $
$\Rightarrow \cfrac{1}{a_n} =\cfrac{1}{a_n-2}-\cfrac{1}{a_{n+1}-2} $
$a_1=\cfrac{5}{2} \quad \cfrac{1}{a_1}+ \cfrac{1}{a_2}+...+\cfrac{1}{a_{2020}} \quad $
$=(\cfrac{1}{a_1 -2}-\cfrac{1}{a_2-2})+(\cfrac{1}{a_2-2} -\cfrac{1}{a_3-2})+\cdots +(\cfrac{1}{a_{2020}-2}-\cfrac{1}{a_{2021}-2} )$
$\Rightarrow =\cfrac{1}{\cfrac{5}{2} -2}-\cfrac{1}{a_{2021}-2} =2-\cfrac{1}{a_{2021}-2}$
$a_{n+1}-a_n=\cfrac{1}{2}(a_n-2)^2\gt0 , a_n\nearrow 易证a_{2023}\gt 3$

${\color{Red} PART \quad4　\quad三角裂项 }$
${\color{Red}一、模型总结：来自差角公式：}$
${\color{Red} \quad①\quad a_n=\tan n\cdot \tan (n+1)} $
$\tan 1=\tan [(n+1)-n]=\cfrac{\tan (n+1)-\tan n}{1+\tan n\cdot \tan (n+1)} \Rightarrow $
$1+\tan n\cdot \tan (n+1)=\cfrac{\tan (n+1)-\tan n}{\tan 1} \Rightarrow 1+\tan n\cdot \tan (n+1) =\cfrac{1}{\tan 1} [\tan (n+1)-\tan n]$
${\color{Red} \tan n\cdot \tan (n+1) =\cfrac{1}{\tan 1} [\tan (n+1)-\tan n]-1} $
${\color{Red} S_n =\cfrac{1}{\tan 1} [\tan (n+1)-\tan 1]-n} $
${\color{Green} \quad ②\quad a_n=\cfrac{1}{\sin n\sin (n+1)} } $
${\color{Green} \because \sin 1}=\sin[(n+1)-n]=\sin (n+1)\cos n-\cos (n+1)\sin n$
右边分子分母$\times \sin 1 \quad \therefore $
$a_n=\cfrac{\sin 1}{\sin 1\sin n\sin (n+1)}=\cfrac{1}{\sin 1}\cdot \cfrac{\sin (n+1)\cos n-\cos (n+1)\sin n}{\sin n\sin (n+1)} $
${\color{Green} a_n= \cfrac{1}{\sin n\sin (n+1)} } =\cfrac{1}{\sin 1}\cdot [\cfrac{1}{\tan n} -\cfrac{1}{\tan (n+1)} ]$
${\color{Red}\quad ③\quad a_n=\cfrac{1}{\cos n\cos (n+1)} } $
同上，分子分母$\times \sin1 \quad a_n=\cfrac{1}{\sin1} \cdot\cfrac{\sin1}{\cos n\cos (n+1)} =$
$a_n=\cfrac{1}{\sin1} \cdot\cfrac{\sin (n+1)\cos n-\cos (n+1)\sin n}{\cos n\cos (n+1)} =\cfrac{1}{\sin1} \cdot[\tan (n+1)-\tan n]$
${\color{Red}\quad a_n=\cfrac{1}{\sin1} \cdot[\tan (n+1)-\tan n]} $

$7a_4=63\Rightarrow a_4=9,a_5=9,d=a_5-a_4=3,a_1=0,a_n=3n-3$
$b_n=\cfrac{\sin 3}{\cos(3n-3)\cos (3n)} =\cfrac{\sin [3n-(3n-3)]}{\cos(3n-3)\cos (3n)} =\cfrac{\sin 3n\cos(3n-3)-\cos 3n \sin (3n-3)}{\cos(3n-3)\cos (3n)}=\tan 3n -\tan (3n-3)$
$Ｓ_n=b_1+b_2+\dots +b_n =\tan 3 -{\color{Red} \tan 0} +\tan 6-\tan 3+\dots +{\color{Red} \tan 3n} -\tan (3n-3)\quad$位置对称！！
$S_n=\tan 3n$

抽象数列：
${\color{Green} 1、a_{n+m}=a_n+a_m{\color{Orange} \Rightarrow a_n=na_1=nd} }; $
${\color{Red} 2、a_{n+m}=a_na_m\Rightarrow a_n=a_1^n=q^n} $
${\color{Blue} 3、a_{m+n}+a_{m-n}=2a_m+2a_n\Rightarrow a_n=n^2} $
${\color{Tan} 4、a_{m+n}+a_{m-n}=2a_ma_n\Rightarrow a_n=\cos (\omega n)} $

$q^9=27\therefore q^3=3, \cfrac{a_1a_5+a_3a_6}{a^2_6+2a_3^2}$
$=\cfrac{(q^3)^2+q^3q^6}{(q^6)^2+2(q^3)^3} =\cfrac{q^6(1+q^3)}{q^6(q^6+2q^3)} =\cfrac{4}{15}$

令$m=1,d=a_1=2,S_{k+2}-S_k=a_{k+1}+a_{k+2}=26,k=6$

令$q=1,a_1=q=2$

$\{a_n\}=n^2,求S_n\quad$
$n^3-(n-1)^3=n^3-(n^3-3n^2+3n-1)=3n^2-3n+1\quad $
$(n-1)^3-(n-2)^3=3(n-1)^2-3(n-1)+1\quad $
$(n-2)^3-(n-3)^3=3(n-2)^2-3(n-2)+1\quad $
$(n-3)^3-(n-4)^3=3(n-3)^2-3(n-3)+1\quad $
$\dots \dots \dots $
$2^3-1^3=3\cdot2^2-3\cdot 2+1\quad $
$1^3-0^3=3\cdot1^2-3\cdot 1+1\quad $
左右两边相加，$n^3=3S_n-3\cfrac{n(n+1)}{2}+n$
$2n^3=6S_n-3n(n+1)+2n=6S_n-3n^2-n\Rightarrow 6S_n=2n^3+3n^2+n$
$6S_n=n(2n^2+3n+1)=n(n+1)(2n+1)\Rightarrow S_n=n(2n^2+3n+1)=\cfrac{1}{6}\cdot n(n+1)(2n+1)$
$S_7=7\times 8\times 15{\div} 6=140$

令$a_n=\cos (\omega n)\Rightarrow a_1=\cos \omega =\cfrac{1}{2} $
$\therefore \omega =\cfrac{\pi}{3} ,T=\cfrac{2\pi }{\omega }=6$
$2023=337\times 6+1,a_1=a_{2023}=\cfrac{1}{2} $

正切分式定理:
${\color{Red} \cfrac{\tan A}{\tan B}+ \cfrac{\tan A}{\tan C} =\cfrac{2a^2}{b^2+c^2-a^2} \qquad} $
${\color{Green} \cfrac{\tan B}{\tan A}+ \cfrac{\tan B}{\tan C} =\cfrac{2b^2}{a^2+c^2-b^2} \qquad} $
${\color{Violet} \cfrac{\tan C}{\tan A}+ \cfrac{\tan C}{\tan B} =\cfrac{2c^2}{a^2+b^2-c^2} \qquad } \cfrac{\tan C}{\tan A}+ \cfrac{\tan C}{\tan B} =\cfrac{2c^2}{a^2+b^2-c^2} \qquad $
$证明：\cfrac{\tan A}{\tan B}=\cfrac{\sin A\cos B}{\cos A \sin B}=\cfrac{a\times \cfrac{a^2+c^2-b^2}{2ac} }{b\times \cfrac{b^2+c^2-a^2}{2bc}}=\cfrac{a^2+c^2-b^2}{b^2+c^2-a^2} $
$\cfrac{\tan A}{\tan C}= \cfrac{\tan A}{\tan C}=\cfrac{\sin A \cos C}{\sin C \cos A}=\cfrac{a\times \cfrac{a^2+b^2-c^2}{2ab} }{c\times \cfrac{b^2+c^2-a^2}{2bc}}=\cfrac{a^2+b^2-c^2}{b^2+c^2-a^2} $
${\color{Red} 两式相加得证 } $

${\color{Red} 此题要用到正切恒等式及中线长定理 } $

${\color{Red} 此题用正切分式定理的证明过程或特殊值法，即分母相等 } $

常用的几个抽象函数模型：
${\color{Red} ①f(x+y)=f(x)+f(y)\Leftrightarrow f(x)=kx;\qquad} $
${\color{Green} ②f(x+y)=f(x)+f(y)+c\Leftrightarrow f(x)=kx+b;\qquad}$
${\color{Red} ③f(xy)=f(x)+f(y)\Leftrightarrow f(x)=\log_{a}{\left | x \right | } ;\qquad} $
${\color{Green} ④f(x+y)=f(x)f(y)\Leftrightarrow y=a^x} $
${\color{Red}⑤ f(x+y)+f(x-y)=2f(x)f(y)} {\color{Green} \Leftrightarrow y=\cos \omega x} $
${\color{Red}⑥ f(x+y)+f(x-y)=f(x)f(y)} {\color{Green} \Leftrightarrow y=A\cos \omega x} $
${\color{Green} ㈦⑦f(x+y）=f(x)+f(y)+2xy\Leftrightarrow f(x)=x^2} $
${\color{Peach} ⑧f(x+y）=f(x)f(\cfrac{\pi}{2}-y)+f(y)f(\cfrac{\pi}{2}-x) \Longrightarrow f(x)=\sin x} $
${\color{Purple} ⑨f^2(x)-f^2(y)=f(x+y)f(x-y)\Leftrightarrow \sin^2x-\sin^2y=\sin (x+y)\sin (x-y)}$正弦平方差公式
请宝贝们，证明一下正弦平方差定理，在右边往左边证。

设$f(x)=A\cos \omega x\Rightarrow f(x+y)=A(\cos\omega x\cos \omega y-\sin \omega x\sin \omega y);\quad $
$f(x-y)=A(\cos\omega x\cos \omega y+\sin \omega x\sin \omega y)\Rightarrow 左边=f(x+y)+f(x-y)=2A\cos\omega x\cos \omega y$
$右边=f(x)f(y)=A^2\cos\omega x\cos \omega y\mapsto 2A=A^2,A=2$
$f(1)=2\cos \omega =1\Rightarrow \omega=\cfrac{\pi}{3} \Rightarrow f(x)=2\cos \cfrac{\pi}{3} x
$
$2(\cos \cfrac{\pi}{3} +\cos \cfrac{2\pi}{3}+\cos \cfrac{3\pi}{3}+\cos \cfrac{4\pi}{3})=-3$

构造：$f(x)=x^2\log \left | x \right | \qquad $
$f(xy)=(xy)^2\log \left | (xy) \right | =x^2y^2(\log \left | x \right |+\left |y \right |)=y^2f(x)+x^2f(y)$

同第一题$f(x)=2\cos \omega$

左边$＝\log xy+m=\log x+\log y+m;\quad $
右边$=f(x)+f(y)-1=\log x+m+\log y+m-1; \quad $
$m=2m-1\Rightarrow m=1$
$f(4)=\log_{a}{4} +1=2\Rightarrow a=4;\therefore f(x)=\log_{4}{x} +1$
$\therefore f(\cfrac{1}{2} )=\log_{4}{\cfrac{1}{2}} +1=\log_{2^2}{2^{-1}} +1=-\cfrac{1}{2}+1=\cfrac{1}{2}$

设$f(x)=\cos \omega x\because f(4)=\cos 4\omega =-1\Rightarrow \omega =\cfrac{\pi}{4} $
$\therefore f(x)=\cos \cfrac{\pi}{4} x$

$\sin x$

设$f(x)=ax^2+bx+c,故左边＝f(x)+f(y)=ax^2+bx+c+ay^2+by+c;\quad $
右边＝$f(x+y)-xy-1=a(x+y)^2+b(x+y)+c-xy-1=ax^2+ay^2+2axy+b(x+y)+c-xy-1；$
左边＝右边；$\begin{cases} c-1=2c\\2a-1=0\end{cases}$
$a=\cfrac{1}{2},c=-1,f(x) =\cfrac{1}{2}x^2+bx-1,f(1)=1,\Rightarrow b=\cfrac{3}{2}$
$f(x) =\cfrac{1}{2}x^2+\cfrac{3}{2}x-1$

设$f(x)=\sin \omega x,f(1)=\sin \omega =1 \Rightarrow \omega =\cfrac{\pi}{2}$
$f(2x+1)为偶函数，故f(2x+1)=f(-2x+1),即f(x)关于x=1对称；$
$f(0)=0,sin为奇函数，关于(2,0)对称，T=\cfrac{2\pi}{\omega}=4$

在三角形$\bigtriangleup ABC中，有\sin ^2A-\sin ^2B=\sin (A+B)\sin (A-B)$
$\sin (A+B)\sin (A-B)=(\sin A\cos B+\cos A\sin B)(\sin A\cos B-\cos A\sin B)$
$=\sin^2 A\cos^2 B-\cos^2 A\sin^2 B=\sin^2 A(1-\sin^2 B)-(1-\sin^2 A)\sin^2 B$
=$\sin ^2A-\sin ^2B$

AD

$\cfrac{3t-t}{1+3t^2} =\cfrac{1}{\sqrt{3} }$

$\cfrac{2\sqrt{6}}{9 } $

$A=\cfrac{\pi}{6}$

$第1题 、b^2-a^2=ac{\color{Red} \Rightarrow } \sin ^2B-\sin ^2A=\sin A \sin C\Rightarrow \sin (B+A) \sin (B-A)=\sin A\sin C$
${\color{Red} \Rightarrow } \sin (B-A)=\sin A {\color{Red} \Rightarrow }B-A=A,B=2A或B-A+A=\pi$
$锐角三角形{\color{Red} \Rightarrow } \begin{cases} 0\lt A\lt \cfrac{\pi}{2}\\ 0\lt 2A\lt \cfrac{\pi}{2}\\ 0\lt \pi -3A\lt \cfrac{\pi}{2}\end{cases}{\color{Red} \Rightarrow} A\in (\cfrac{\pi}{6}, \cfrac{\pi}{4})$
$c选\Rightarrow \cfrac{1}{\cfrac{1}{\tan A} -\cfrac{1}{\tan B} }=\cfrac{\sin A\sin 2A}{-\sin A}=-\sin 2A=\cfrac{\sin A\sin B}{\sin(A-B)} =-\sin 2A$
$D{\color{Green} \Rightarrow 2\sin C=\sin A+\sin B} \Rightarrow 2\sin (A+B)=\sin A+\sin B\Rightarrow 2\sin 3A=\sin A+\sin 2A$
$2[\sin A\cos 2A+\cos A\sin 2A]=\sin A(1+2\cos A)\Rightarrow 2[\sin A(2\cos ^2A-1)+2\sin A\cos ^2A)]=$
$2\sin A(4\cos ^2-1)=\sin A(1+2\cos A)\Rightarrow 8\cos ^2-2=1+2\cos A\Rightarrow 8t^2-2t-3=0,t=\cfrac{3}{4}$
$第二题：数量积，共起点或共终点，夹角就是三角形的顶角。\tan (B-C)展开是含Bc角的，可见是求BC两角的关系。$
$ca\cos B-ba\cos C=-\cfrac{1}{2}c^2\Rightarrow ac\cdot \cfrac{a^2+c^2-b^2}{2ac}-ab\cdot\cfrac{a^2+b^2-c^2}{2ab}=-\cfrac{1}{2}c^2$
$\Rightarrow c^2-b^2=-\cfrac{1}{2} a^2 \Rightarrow \sin ^2C -\sin ^2B= -\cfrac{1}{2} \sin ^2A $
$\sin (C+B)\sin(C-B) =-\cfrac{1}{2} \sin ^2A\Rightarrow \sin (C-B)=-\cfrac{1}{2} \sin A$
$\Rightarrow \sin (C-B)=-\cfrac{1}{2} \sin (B+C)\Rightarrow \cfrac{3}{2} \sin C\cos B=\cfrac{1}{2}\cos C\sin B$
$3\tan C=\tan B$
$第三题：\frac{b}{a} -1=b^2-a^2\Rightarrow \frac{b-a}{a}= b^2-a^2\Rightarrow 1=a^2+ab$
$c=1\Rightarrow c^2=1\Rightarrow a^2=ab+b^2,此时与第一题相同了。$
$\sin ^2C-\sin ^2 A=\sin A\sin B\Rightarrow \sin (C+A)\sin (C-A)=\sin A\sin B\Rightarrow C=2A
B=\pi -3A$
$\sin 3A-\sin A=\sin (A+2A)-\sin A=3\sin A-4\sin^3 A-\sin A=2\sin A-4\sin^3 A$
$令\sin A=t,没有锐角的限制。\pi -3A\gt 0 \Rightarrow A\in (0,\frac{\pi }{6} ),2t-4t^3(t\in (0,\frac{\sqrt{3}}{2} )$
$第四题：2b^2-2a^2=c^2\Rightarrow 2\sin^2B -2\sin^2 A=\sin ^2C\Rightarrow 2\sin (B-A)=\sin C$
$2\sin (B-A)=\sin c=\sin (B+A)\Rightarrow \sin B\cos B=3\cos B\sin A\Rightarrow \tan B=3\tan A$
$设\tan A=t ，\tan （B-A)=\cfrac{3t-t}{1+3t^2} =\cfrac{2t}{1+3t^2}=\cfrac{2}{\frac{1}{t} +3t}$
$\Rightarrow \cfrac{1}{t} =3t\Rightarrow t=\frac{\sqrt{3} }{3} $