Trong \[\Delta ABC\], nếu \[\frac{{\sin B}}{{\sin C}} = 2\cos A\] thì \[\Delta ABC\] là tam giác có tính chất nào sau đây?

Vì \(a,b\) là các góc nhọn nên \(\cos a > 0,\cos b > 0\).

Ta có \(\cos a = \sqrt {1 - {{\sin }^2}a}  = \frac{{15}}{{17}} \Rightarrow \tan a = \frac{{\sin a}}{{\cos a}} = \frac{8}{{15}}\);

          \(\cos b = \sqrt {\frac{1}{{1 + {{\tan }^2}b}}}  = \frac{{12}}{{13}} \Rightarrow \sin b = \cos b\tan b = \frac{5}{{13}}{\rm{. }}\)

Khi đó, \(\sin \left( {a - b} \right) = \sin a\cos b - \cos a\sin b = \frac{8}{{17}} \cdot \frac{{12}}{{13}} - \frac{{15}}{{17}} \cdot \frac{5}{{13}} = \frac{{21}}{{221}}\).

          \(\cos \left( {a + b} \right) = \cos a\cos b - \sin a\sin b = \frac{{15}}{{17}} \cdot \frac{{12}}{{13}} - \frac{8}{{17}} \cdot \frac{5}{{13}} = \frac{{140}}{{221}}\)

          \(\tan \left( {a + b} \right) = \frac{{\tan a + \tan b}}{{1 - \tan a\tan b}} = \frac{{\frac{8}{{15}} + \frac{5}{{12}}}}{{1 - \frac{8}{{15}} \cdot \frac{5}{{12}}}} = \frac{{171}}{{140}}.\)

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

\(\begin{array}{l}\frac{\pi }{2} < x < \pi  \Rightarrow \frac{\pi }{4} < \frac{x}{2} < \frac{\pi }{2} \Rightarrow \sin \frac{x}{2} > 0.\\\sin \frac{x}{2} = \sqrt {\frac{{1 - \cos x}}{2}}  = \sqrt {\frac{{1 - \frac{1}{5}}}{2}}  = \frac{{\sqrt {10} }}{5}\end{array}\).

\(\begin{array}{l}\frac{\pi }{2} < x < \pi  \Rightarrow \frac{\pi }{4} < \frac{x}{2} < \frac{\pi }{2} \Rightarrow \cos \frac{x}{2} > 0.\\\cos \frac{x}{2} = \sqrt {\frac{{1 + \cos x}}{2}}  = \sqrt {\frac{{1 + \frac{1}{5}}}{2}}  = \frac{{\sqrt {15} }}{5}.\\\tan \frac{x}{2} = \frac{{\sin \frac{x}{2}}}{{\cos \frac{x}{2}}} = \frac{{\frac{{\sqrt {10} }}{5}}}{{\frac{{\sqrt {15} }}{5}}} = \frac{{\sqrt 6 }}{3}.\\\cot \frac{x}{2} = \frac{{\cos \frac{x}{2}}}{{\sin \frac{x}{2}}} = \frac{{\frac{{\sqrt {15} }}{5}}}{{\frac{{\sqrt {10} }}{5}}} = \frac{{\sqrt 6 }}{2}.\end{array}\)

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