Rút gọn biểu thức \(A = \frac{{\cos x + \cos 2x + \cos 3x}}{{\sin x + \sin 2x + \sin 3x}}\) ta được

Vì \(\cos A = \frac{4}{5} \Rightarrow \sin A = \frac{3}{5}\); \(\cos B = \frac{5}{{13}} \Rightarrow \sin B = \frac{{12}}{{13}}\) vì \(0 < A,B < \pi \).

Có \(\widehat A + \widehat B + \widehat C = \pi \)\( \Rightarrow \widehat C = \pi - \left( {\widehat A + \widehat B} \right)\).

Do đó \(A = 130\cos C - 1\)\( = 130\cos \left[ {\pi - \left( {A + B} \right)} \right] - 1\)\( = - 130\cos \left( {A + B} \right) - 1\)

\( = - 130\left[ {\cos A.\cos B - \sin A.\sin B} \right] - 1\)\( = - 130\left[ {\frac{4}{5}.\frac{5}{{13}} - \frac{3}{5}.\frac{{12}}{{13}}} \right] - 1 = 31\).

Trả lời: 31.

a) \(\sin \alpha = \frac{2}{3}\) với \(0 < \alpha < \frac{\pi }{2}\) suy ra \(\cos \alpha = \frac{{\sqrt 5 }}{3}\) \( \Rightarrow \sin 2\alpha = 2\sin \alpha .\cos \alpha = 2.\frac{2}{3}.\frac{{\sqrt 5 }}{3} = \frac{{4\sqrt 5 }}{9}\).

b) \(\cos \left( {\alpha + \frac{{3\pi }}{2}} \right) = \cos \left( {\alpha + 2\pi - \frac{\pi }{2}} \right) = \cos \left( {\alpha - \frac{\pi }{2}} \right) = \cos \left( {\frac{\pi }{2} - \alpha } \right) = \sin \alpha = \frac{2}{3}\).

c) \(\sqrt 2 \cos \left( {\alpha + \frac{\pi }{4}} \right) = \cos \alpha - \sin \alpha = \frac{{\sqrt 5 - 2}}{3}\).

d) Có \(\sin \alpha = \frac{2}{3};\cos \alpha = \frac{{\sqrt 5 }}{3}\) \( \Rightarrow \tan \alpha = \frac{2}{{\sqrt 5 }},\cot \alpha = \frac{{\sqrt 5 }}{2}\).

Do đó \(D = \frac{{\cot \alpha + \tan \alpha }}{{\cot \alpha - \tan \alpha }} = \frac{{\frac{{\sqrt 5 }}{2} + \frac{2}{{\sqrt 5 }}}}{{\frac{{\sqrt 5 }}{2} - \frac{2}{{\sqrt 5 }}}} = 9\).

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