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Объяснение:

1)

[tex]cos\alpha =\frac{7}{9} \ \ \ \ 0^0 < \alpha < 90^0\ \ \ \ sin\frac{\alpha }{2}=?\ \ \ \ cos\frac{\alpha }{2} =?\ \ \ \ sin\alpha =?\\ sin^2\alpha +cos^2\alpha =1\\sin^2\alpha =1-cos^2\alpha =1-(\frac{7}{9})^2=1-\frac{49}{81}=\frac{81-49}{81}=\frac{32}{81}.\\ sin\alpha =б\sqrt{\frac{32}{81} }=б\frac{4\sqrt{2} }{9}\ \ \ \ 0^0 < \alpha < 90^0\ \ \ \ \Rightarrow\\ sin\alpha =\frac{4\sqrt{2} }{9} .\\[/tex]

[tex]sin^2\frac{\alpha }{2} =\frac{1-cos\alpha }{2} =\frac{1-\frac{7}{9} }{2}=\frac{\frac{2}{9} }{2}=\frac{1}{9} . \\sin\frac{\alpha }{2}=б\sqrt{\frac{1}{9} } =б\frac{1}{3} .\ \ \ \ 0^0 < \alpha < 90^0\ \ \ \ \Rightarrow\\sin\frac{\alpha }{2} =\frac{1}{3} .\\[/tex]

[tex]cos^2\frac{\alpha }{2}={\frac{1+cos\alpha }{2} } =\frac{1+\frac{7}{9} }{2} =\frac{\frac{16}{9} }{2} =\frac{8}{9} .\\cos\frac{\alpha }{2} =б\sqrt{\frac{8}{9} }=б\frac{2\sqrt{2} }{3} \ \ \ \ 0^0 < \alpha < 90^0\ \ \ \ \Rightarrow\\ cos\frac{\alpha }{2} =\frac{2\sqrt{2} }{3}.[/tex]

2)

[tex]sin\alpha =\frac{4\sqrt{2} }{9} \ \ \ \ 90^0 < \alpha < 180^0\ \ \ \ 45^0 < \frac{\alpha }{2} < 90^0\ \ \ \ sin\frac{\alpha }{2}=?\ \ \ cos\frac{\alpha }{2}=?\ \ \ \\cos\alpha =? \ \ \ \ tg\alpha =?\\ cos\alpha =-\frac{7}{9} .\\ sin\frac{\alpha }{2} =\frac{1 }{3} .\\cos\frac{\alpha }{2} =\frac{2\sqrt{2} }{3}.\\ tg\alpha =\frac{sin\alpha }{cos\alpha } =\frac{\frac{4\sqrt{2} }{9} }{-\frac{7} {9} } =-\frac{4\sqrt{2} }{7} .[/tex]