Faaaith1
Faaaith1 Faaaith1
  • 08-11-2019
  • Mathematics
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Please help with this answer

Please help with this answer class=

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surjithayer10otahza surjithayer10otahza
  • 18-11-2019

Answer:

[tex]cos ~u=\sqrt{1-sin ^2~u} =\sqrt{1-(\frac{-3}{5})^2 } =\frac{4}{5} \\sin ~2u=2~sin~u~cos~u=2*\frac{-3}{5} *\frac{4}{5} =-\frac{24}{25} =-0.96\\cos~2~u=cos^2 u-sin^2 u=(\frac{4}{5} )^2-(\frac{-3}{5} )^2=\frac{16}{25} -\frac{9}{25} =\frac{7}{25} =0.28\\tan ~2u=\frac{sin~2u}{cos~2u} =\frac{-\frac{3}{5} }{\frac{4}{5} }=-\frac{3}{4} =-0.75[/tex]

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