Косинусы двух углов треугольника равны 1/3 и 2/3. Найдите синус третьего угла

[latex]cosa = \frac{1}{3} [/latex] [latex]cosb = \frac{2}{3} [/latex] [latex]sinc = ?[/latex] [latex]sin^2x+cos^2x=1[/latex] [latex]sinx= \sqrt{(1-cos^2x)} [/latex] [latex]sina= \sqrt{1- \frac{1}{9} } = \sqrt{ \frac{8}{9} } = \frac{ \sqrt{8} }{3} = \frac{2 \sqrt{2} }{3} [/latex] [latex]sinb= \sqrt{(1- \frac{4}{9} )}= \sqrt{ \frac{5}{9} } = \frac{ \sqrt{5} }{3} [/latex] [latex]cos(x+y)=cosx*cosy-sinx*siny[/latex] [latex]cos(a+b)= \frac{1}{3}* \frac{2}{3} -\frac{2 \sqrt{2} }{3}*\frac{ \sqrt{5} }{3}= \frac{2-2 \sqrt{10} }{9} [/latex] [latex]cos(x-y)=cosx*cosy+sinx*siny[/latex] [latex]cos(180-c)=cos(180)*cosc+sin(180)*sinc=-1*cosc+0*sinc=-cosc[/latex] [latex]a+b+c=180[/latex] [latex]cos(a+b) = cos(180-c)[/latex] [latex]cosc = - \frac{2-2 \sqrt{10} }{9} [/latex] [latex]sinc=\sqrt{1- (\frac{2-2 \sqrt{10} }{9})^{2} }=\sqrt{1- \frac{(2-2 \sqrt{10} )^{2}}{81} }= \sqrt{ \frac{81-(2-2 \sqrt{10})^{2} }{81} } =[/latex] [latex]=\frac{\sqrt{ 81-(2-2 \sqrt{10})^{2} }}{9}= \frac{ \sqrt{81-(4-8 \sqrt{10}+4*10) } }{9} = \frac{ \sqrt{37+8 \sqrt{10} } }{9}=[/latex] [latex]=\frac{ \sqrt{32+8 \sqrt{10} + 5 } }{9}=\frac{ \sqrt{16*2+8 \sqrt{2}\sqrt{5} + 5 } }{9}=\frac{ \sqrt{(4\sqrt{2})^{2}+2*4 \sqrt{2}\sqrt{5} + \sqrt{5}^{2} } }{9}=[/latex] [latex]=\frac{ (\sqrt{(4\sqrt{2}+ \sqrt{5})^{2} } }{9}=\frac{ 4\sqrt{2}+ \sqrt{5} }{9}[/latex] Ответ: [latex]sinc=\frac{ 4\sqrt{2}+ \sqrt{5} }{9}[/latex]