Tính \(S = {\cos ^2}5^\circ + {\cos ^2}10^\circ + {\cos ^2}15^\circ + ... + {\cos ^2}80^\circ + {\cos ^2}85^\circ \).

a) Do \(90^\circ < x < 180^\circ \) nên \(\cos x < 0\).

b) Ta có \({\cos ^2}x = 1 - {\sin ^2}x = 1 - \frac{{25}}{{169}} = \frac{{144}}{{169}}\).

Do đó \(P = 2{\sin ^2}x - {\cos ^2}x = 2.\left( {\frac{{25}}{{169}}} \right) - \frac{{144}}{{169}} = - \frac{{94}}{{169}}\).

c) Do \(\cos x < 0\) và \({\cos ^2}x = \frac{{144}}{{169}}\) nên \(\cos x = - \frac{{12}}{{13}}\).

Suy ra \(\tan x = \frac{{\sin x}}{{\cos x}} = - \frac{5}{{12}}\).

d) Ta có \(A = \frac{{{{\sin }^2}x}}{{1 + {{\cos }^2}x}} = \frac{{{{\sin }^2}x}}{{1 + 1 - {{\sin }^2}x}} = \frac{{{{\left( {\frac{5}{{13}}} \right)}^2}}}{{2 - {{\left( {\frac{5}{{13}}} \right)}^2}}} = \frac{{25}}{{313}}\).

Đáp án: a) Sai;   b) Đúng; c) Sai; d) Đúng.

a) Ta có \(\frac{{\sin \alpha - \cos \alpha }}{{2\sin \alpha + 3\cos \alpha }} = \frac{{\frac{{\sin \alpha }}{{\cos \alpha }} - 1}}{{2\frac{{\sin \alpha }}{{\cos \alpha }} + 3}}\)\( = \frac{{\tan \alpha - 1}}{{2.\tan \alpha + 3}}\)\( = \frac{{ - 2 - 1}}{{2.\left( { - 2} \right) + 3}} = 3\).

b) Vì \(90^\circ < \alpha < 180^\circ \) nên \(\cos \alpha < 0\).

c) Có \({\cos ^2}\alpha = \frac{1}{{1 + {{\tan }^2}\alpha }} = \frac{1}{{1 + {{\left( { - 2} \right)}^2}}} = \frac{1}{5}\).

d) Có \(\sin \left( {180^\circ - \alpha } \right) = \sin \alpha \).

Có \({\sin ^2}\alpha = 1 - {\cos ^2}\alpha = 1 - \frac{1}{5} = \frac{4}{5} \Rightarrow \sin \alpha = \frac{{2\sqrt 5 }}{5}\) vì \(90^\circ < \alpha < 180^\circ \).

Đáp án: a) Đúng;   b) Sai; c) Đúng; d) Sai.