Доказать тождество (sina)^2+(sinb)^2+(sinc)^2-2cosacosbcosc=2, a+b+c=180^0

Решите задачу:

$sin^2a+sin^2b+sin^2c-2cosa\cdot cosb\cdot cosc=2\; ,\; \; a+b+c=180^\circ \\\\sin^2a+sin^2b+sin^2c=2+2cosa\cdot cosb\cdot cosc\\\\sin^2a+sin^2b+sin^2c= \frac{1-cos2a}{2} + \frac{1-cos2b}{2} + \frac{1-cos2c}{2} =\\\\= \frac{3}{2} -\frac{1}{2} (cos2a+cos2b+cos2c)=\frac{1}{2}(3-(cos2a+cos2b)-cos2c)=\\\\=\frac{1}{2}(3-2cos(a+b)\cdt cos(a-b)- (2cos^2c-1))=\\\\=\frac{1}{2}(4-2cos(180^\circ -c)\cdot cos(a-b)-2cos^2c)=\\\\=\frac{1}{2}(4+2cosc\cdot cos(a-b)-2cos^2c)=2+cosc\cdot cos(a-b)-cos^2c=$

$=2+cosc\cdot (cos(a-b)-cosc)=[\; c=180^\circ -(a+b)\ ]=\\\\=2+cosc\cdot(cos(a-b)-cos(180^\circ -(a+b)))=\\\\=[\; cos(180^\circ - \alpha )=-cos \alpha \; ]=\\\\=2+cosc\cdot (cos(a-b)+cos(a+b))=\\\\=2+cosc\cdot 2cos \frac{a-b+a+b}{2} \cdot cos \frac{a-b-(a+b)}{2} =\\\\=2+cosc\cdot 2cosa\cdot cos(-b)=\\\\=2+2cosa\cdot cosb\cdot cosc\qquad \Rightarrow \\\\sin^2a+sin^2b+sin^2c-2cosa\cdot cosb\cdot cosc=2$