正弦平方差公式

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} $