Cho tam giác ABC có \(\widehat A - \widehat B = 120^\circ \) và \(\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} = \frac{1}{{32}}\). Giá trị của cos2C (làm tròn kết quả đến hàng phần trăm) bằng bao nhiêu?

\(\sin \frac{A}{2}\sin \frac{B}{2}\sin \frac{C}{2} = \frac{1}{2}\left[ {\cos \left( {\frac{A}{2} - \frac{B}{2}} \right) - \cos \left( {\frac{A}{2} + \frac{B}{2}} \right)} \right]\sin \frac{C}{2}\)

\( = \frac{1}{2}\left[ {\cos 60^\circ  - \cos \left( {\frac{{180^\circ  - C}}{2}} \right)} \right]\sin \frac{C}{2}\)\( = \frac{1}{2}\left[ {\frac{1}{2} - \sin \frac{C}{2}} \right]\sin \frac{C}{2}\)\( = \frac{1}{2}\left[ {\frac{1}{2}\sin \frac{C}{2} - {{\sin }^2}\frac{C}{2}} \right] = \frac{1}{{32}}\)

\( \Rightarrow \frac{1}{2}\sin \frac{C}{2} - {\sin ^2}\frac{C}{2} = \frac{1}{{16}}\)\( \Leftrightarrow \sin \frac{C}{2} = \frac{1}{4}\).

Suy ra \({\cos ^2}\frac{C}{2} = 1 - {\sin ^2}\frac{C}{2} = 1 - \frac{1}{{16}} = \frac{{15}}{{16}}\).

Có \(\cos 2C = 1 - 2{\sin ^2}C = 1 - 2.{\left( {2\sin \frac{C}{2}.\cos \frac{C}{2}} \right)^2}\)\( = 1 - 8.{\sin ^2}\frac{C}{2}.{\cos ^2}\frac{C}{2}\)\( = 1 - 8.\frac{1}{{16}}.\frac{{15}}{{16}} = \frac{{17}}{{32}} \approx 0,53\).

Trả lời: 0,53.