Question

A=\(\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\cdot cosa}}}\) với 0<a<bi/2.rút gọn biểu thức A ta được A=cos\(\frac{a}{n}\)hãy cho biết n thuộc khoảng nào

Answer 1

\(A=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}cosa}}}\)

\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\left(2cos^2\frac{a}{2}-1\right)}}}\)

\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+cos^2\frac{a}{2}-\frac{1}{2}}}}\)

\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}cos\frac{a}{2}}}\)

\(=\sqrt{\frac{1}{2}+\frac{1}{2}\sqrt{\frac{1}{2}+\frac{1}{2}\left(2cos^2\frac{a}{4}-1\right)}}\)

\(=\sqrt{\frac{1}{2}+\frac{1}{2}cos\frac{a}{4}}=\sqrt{\frac{1}{2}+\frac{1}{2}\left(cos^2\frac{a}{8}-1\right)}\)

\(=cos\frac{a}{8}\Rightarrow n=8\)