$\sin (a+b)\cdot \sin (a-b)=\sin^2 a-\sin^2 b\quad 正弦平方差公式，伪平方差公式$
$右边＝(\sin a+\sin b)\cdot (\sin a-\sin b)={\color{Green} 2\sin \cfrac{a+b}{2}}{\color{Red} \cos \cfrac{a-b}{2}\cdot 2{\color{Green} \cos \cfrac{a+b}{2}} \sin \cfrac{a-b}{2}}$
$\sin (a+b)\cdot \sin (a-b)=左边$
若从左边往右边证也是可以的。积化和差
$左边=(\sin a\cos b+\cos a\sin b)(\sin a\cos b-\cos a\sin b)=\sin^2 a\cos^2 b-\cos^2 a\sin^2 b$
$\cfrac{1}{4}[\sin(a+b)+\sin(a-b)]^2-\cfrac{1}{4}[\sin(a+b)-\sin(a-b)]^2=\sin(a+b)\sin(a-b)$
$正弦三倍角公式，可以这样记.令a=2\alpha,b=\alpha,代入\sin 3\alpha \sin \alpha =\sin^2 2\alpha -\sin ^2 \alpha$
$=4\sin^2\alpha \cos^2\alpha -\sin^2 \alpha=\sin^2\alpha(4\sin^2\alpha -1)$
$\Rightarrow \sin 3\alpha =\sin \alpha (4-4\cos^2\alpha -1)=3\sin \alpha -4\sin^2 \alpha$
$证明:\sin^2A+\sin^2B+\sin^2C=2+2\cos A\cos B\cos C$
$右边=2+[\cos (A+B)+\cos (A-B)]\cos C=2-\cos^2C+\cos (A-B)\cos C$
$=2-(1-\sin^2C)-\cos (A-B)\cos (A+B)=2-1+\sin^2C-\cfrac{1}{2}(\cos 2A+\cos 2B)$
$=1+\sin^2C-\cfrac{1}{2}(1-2\sin^2A+1-2\sin^2B)=\sin^2A+\sin^2B+\sin^2C$

$证明:\cos^2A+\cos^2B+\cos^2C=1-2\cos A\cos B\cos C$
$右边=1-[\cos (A+B)+\cos (A-B)]\cos C=1+\cos^2C-\cos (A-B)\cos C$
$=1+\cos^2C+\cos (A-B)\cos(A+B)=1+\cos^2C+\cfrac{1}{2}(\cos2A+\cos 2B)$
$=1+\cos^2C+\cfrac{1}{2}(2\cos^2A-2\cos^2B-1)=\cos^2A+\cos^2B+\cos^2C$
$题目:A,B,C为三角形内角,如果\sin^2A+\sin^2B+\sin^2C=2,请问是什么三角形?$
$题目:A,B,C为三角形内角,如果\cos^2A+\cos^2B+\cos^2C=1,\cos A\cos B \sin C=\cfrac{\sqrt{3}}{4},$
$则C=(\quad)$