Cho \({\rm{cos}}x = - \dfrac{3}{5}\) với \(\dfrac{\pi }{2} < x < \pi \). Tính
Cho \({\rm{cos}}x = - \dfrac{3}{5}\) với \(\dfrac{\pi }{2} < x < \pi \). Tính \({\rm{sin}}x,{\rm{cos}}2x\)
Ta có: \({\rm{co}}{{\rm{s}}^2}x + {\rm{si}}{{\rm{n}}^2}x = 1\)
\( \Leftrightarrow {\rm{si}}{{\rm{n}}^2}x = 1 - {\rm{co}}{{\rm{s}}^2}x \Leftrightarrow {\rm{si}}{{\rm{n}}^2}x = 1 - {\left( { - \dfrac{3}{5}} \right)^2} = \dfrac{{16}}{{25}} \Leftrightarrow \left[ {\begin{array}{*{20}{l}}{{\rm{sin}}x = \dfrac{4}{5}}\\{{\rm{sin}}x = - \dfrac{4}{5}}\end{array}} \right.\)
Do \(\dfrac{\pi }{2} < x < \pi \Rightarrow 0 < {\rm{sin}}x < 1\).
Suy ra: \({\rm{sin}}x = \dfrac{4}{5}\)
Khi đó: \({\rm{cos}}2x = {\rm{co}}{{\rm{s}}^2}x - {\rm{si}}{{\rm{n}}^2}x = {\left( { - \dfrac{3}{5}} \right)^2} - {\left( {\dfrac{4}{5}} \right)^2} = - \dfrac{7}{{25}}\)
Vậy \({\rm{sin}}x = \dfrac{4}{5},{\rm{cos}}2x = - \dfrac{7}{{25}}\)
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