Question

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

Answer 1

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

$=2+cosccdot (cos(a-b)-cosc)=[; c=180^circ -(a+b) ]=\\=2+cosccdot(cos(a-b)-cos(180^circ -(a+b)))=\\=[; cos(180^circ - alpha )=-cos alpha ; ]=\\=2+cosccdot (cos(a-b)+cos(a+b))=\\=2+cosccdot 2cos frac{a-b+a+b}{2} cdot cos frac{a-b-(a+b)}{2} =\\=2+cosccdot 2cosacdot cos(-b)=\\=2+2cosacdot cosbcdot coscqquad Rightarrow \\sin^2a+sin^2b+sin^2c-2cosacdot cosbcdot cosc=2$

Answer 2

Cпасибо