Cho tan a/2 = 1/căn bậc hai 2. Tính sin a, cos a, tan a
Cho \(\tan \frac{a}{2} = \frac{1}{{\sqrt 2 }}\). Tính sin a, cos a, tan a.
Cho \(\tan \frac{a}{2} = \frac{1}{{\sqrt 2 }}\). Tính sin a, cos a, tan a.
Ta có \(\sin a = 2\sin \frac{a}{2}\cos \frac{a}{2} = \frac{{2\sin \frac{a}{2}\cos \frac{a}{2}}}{{{{\sin }^2}\frac{a}{2} + {{\cos }^2}\frac{a}{2}}}\) (do \({\sin ^2}\frac{a}{2} + {\cos ^2}\frac{a}{2} = 1\))
\( = \frac{{2\tan \frac{a}{2}}}{{{{\tan }^2}\frac{a}{2} + 1}} = \frac{{2.\frac{1}{{\sqrt 2 }}}}{{{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2} + 1}} = \frac{{2\sqrt 2 }}{3}\).
\(\cos a = {\cos ^2}\frac{a}{2} - {\sin ^2}\frac{a}{2} = \frac{{{{\cos }^2}\frac{a}{2} - {{\sin }^2}\frac{a}{2}}}{{{{\sin }^2}\frac{a}{2} + {{\cos }^2}\frac{a}{2}}}\)\( = \frac{{1 - {{\tan }^2}\frac{a}{2}}}{{{{\tan }^2}\frac{a}{2} + 1}} = \frac{{1 - {{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2}}}{{{{\left( {\frac{1}{{\sqrt 2 }}} \right)}^2} + 1}} = \frac{1}{3}\).
\(\tan a = \frac{{\sin a}}{{\cos a}} = \frac{{\frac{{2\sqrt 2 }}{3}}}{{\frac{1}{3}}} = 2\sqrt 2 \).