Advertisements (Quảng cáo)
a) Chứng minh \(\cos \dfrac{{2\pi }}{9}\cos \dfrac{{4\pi }}{9}\cos \dfrac{{8\pi }}{9} = – \dfrac{1}{8}\) bằng cách nhân cả hai vế với \(\sin \dfrac{{2\pi }}{9}.\)
b) Chứng minh rằng\(\cos \dfrac{{2\pi }}{9} + \cos \dfrac{{8\pi }}{9} = 2\cos \dfrac{{5\pi }}{9}\cos \dfrac{\pi }{3} = \cos \dfrac{{5\pi }}{9},\)
Từ đó suy ra \(\cos \dfrac{{2\pi }}{9} + \cos \dfrac{{4\pi }}{9} + \cos \dfrac{{8\pi }}{9} = 0\) .
c) Từ b) suy ra rằng \({\cos ^2}\dfrac{{2\pi }}{9} + {\cos ^2}\dfrac{{4\pi }}{9} + {\cos ^2}\dfrac{{8\pi }}{9} = \dfrac{3}{2}\).
d) Từ b và c) suy ra rằng:
\(\cos \dfrac{{2\pi }}{9}\cos \dfrac{{4\pi }}{9} + \cos \dfrac{{4\pi }}{9}\cos \dfrac{{8\pi }}{9} + \cos \dfrac{{8\pi }}{9}\cos \dfrac{{2\pi }}{9} = – \dfrac{3}{4}\) .
e) Từ a), b) và d) suy ra rằng
\(\left( {X – \cos \dfrac{{2\pi }}{9}} \right)\left( {X – \cos \dfrac{{4\pi }}{9}} \right)\left( {X – \cos \dfrac{{8\pi }}{9}} \right) = {X^3} – \dfrac{3}{4}X + \dfrac{1}{8},\)
từ đó ta có \(\left( {1 – \cos \dfrac{{2\pi }}{9}} \right)\left( {1 – \cos \dfrac{{4\pi }}{9}} \right)\left( {1 – \cos \dfrac{{8\pi }}{9}} \right) = \dfrac{3}{8}.\)
Suy ra
• \(\sin \dfrac{\pi }{9}\sin \dfrac{{2\pi }}{9}\sin \dfrac{{4\pi }}{9} = \dfrac{{\sqrt 3 }}{8}.\)
• \(\sin \dfrac{{5\pi }}{9}\sin \dfrac{{7\pi }}{9}\sin \dfrac{{8\pi }}{9} = \dfrac{{\sqrt 3 }}{8}.\)
f) Từ e) suy ra rằng
\(\sin \dfrac{\pi }{9}\sin \dfrac{{2\pi }}{9}\sin \dfrac{{3\pi }}{9}\sin \dfrac{{4\pi }}{9}\sin \dfrac{{5\pi }}{9}\sin \dfrac{{6\pi }}{9}\sin \dfrac{{7\pi }}{9}\sin \dfrac{{8\pi }}{9} = \dfrac{9}{{256}}.\)
(Chú ý. Người ta chứng minh được rằng không thể dùng thước và compa để dựng đa giác đều chín cạnh nội tiếp trong một đường tròn cho trước.)
a) Ta có:
\(\begin{array}{l}\sin \dfrac{{2\pi }}{9}\cos \dfrac{{2\pi }}{9}\cos \dfrac{{4\pi }}{9}\cos \dfrac{{8\pi }}{9}\\ = \dfrac{1}{2}\sin \dfrac{{4\pi }}{9}\cos \dfrac{{4\pi }}{9}\cos \dfrac{{8\pi }}{9}\\ = \dfrac{1}{4}\sin \dfrac{{8\pi }}{9}\cos \dfrac{{8\pi }}{9}\\ = \dfrac{1}{8}\sin \dfrac{{16\pi }}{9}\\ = \dfrac{1}{8}\sin \left( {2\pi – \dfrac{{2\pi }}{9}} \right)\\ = – \dfrac{1}{8}\sin \dfrac{{2\pi }}{9}\end{array}\)
Từ đó: \(\cos \dfrac{{2\pi }}{9}\cos \dfrac{{4\pi }}{9}\cos \dfrac{{8\pi }}{9} = – \dfrac{1}{8}.\)
b) Ta có
Advertisements (Quảng cáo)
\(\begin{array}{l}\cos \dfrac{{2\pi }}{9} + \cos \dfrac{{8\pi }}{9} = 2\cos \dfrac{{5\pi }}{9}\cos \dfrac{\pi }{3}\\ = \cos \dfrac{{5\pi }}{9} = \cos \left( {\pi – \dfrac{{4\pi }}{9}} \right)\\ = – \cos \dfrac{{4\pi }}{9}\end{array}\)
từ đó \(\cos \dfrac{{2\pi }}{9} + \cos \dfrac{{4\pi }}{9} + \cos \dfrac{{8\pi }}{9} = 0.\)
c) Do
\(\begin{array}{l}\cos \dfrac{{2\pi }}{9} = 2{\cos ^2}\dfrac{\pi }{9} – 1 = 2{\cos ^2}\dfrac{{8\pi }}{9} – 1,\\cos\dfrac{{4\pi }}{9} = 2{\cos ^2}\dfrac{{2\pi }}{9} – 1\\\cos \dfrac{{8\pi }}{9} = 2{\cos ^2}\dfrac{{4\pi }}{9} – 1,\end{array}\)
nên từ b) suy ra
\({\cos ^2}\dfrac{{2\pi }}{9} + {\cos ^2}\dfrac{{4\pi }}{9} + {\cos ^2}\dfrac{{8\pi }}{9} = \dfrac{3}{2}.\)
d) Với mọi số A, B, C ta có:
\(AB + BC + CA = \dfrac{1}{2}\left[ {{{\left( {A + B + C} \right)}^2} – {A^2} – {B^2} – {C^2}} \right]\) nên
\(\begin{array}{l}\cos \dfrac{{2\pi }}{9}\cos \dfrac{{4\pi }}{9} + \cos \dfrac{{4\pi }}{9}\cos \dfrac{{8\pi }}{9} + \cos \dfrac{{8\pi }}{9}\cos \dfrac{{2\pi }}{9}\\ = \dfrac{1}{2}\left[ {{{\left( {\cos \dfrac{{2\pi }}{9} + \cos \dfrac{{4\pi }}{9} + \cos \dfrac{{8\pi }}{9}} \right)}^2} – \left( {{{\cos }^2}\dfrac{{2\pi }}{9} + {{\cos }^2}\dfrac{{4\pi }}{9} + {{\cos }^2}\dfrac{{8\pi }}{9}} \right)} \right]\\ = – \dfrac{1}{2}.\dfrac{3}{2} = – \dfrac{3}{4}.\end{array}\)
e) Ta có
\(\begin{array}{l}\left( {X – \cos \dfrac{{2\pi }}{9}} \right)\left( {X – \cos \dfrac{{4\pi }}{9}} \right)\left( {X – \cos \dfrac{{8\pi }}{9}} \right)\\ = {X^3} – \left( {\cos \dfrac{{2\pi }}{9} + \cos \dfrac{{4\pi }}{9} + \cos \dfrac{{8\pi }}{9}} \right){X^2}\\ + \left( {\cos \dfrac{{2\pi }}{9}\cos \dfrac{{4\pi }}{9} + \cos \dfrac{{4\pi }}{9}\cos \dfrac{{8\pi }}{9} + \cos \dfrac{{8\pi }}{9}\cos \dfrac{{2\pi }}{9}} \right)X\\ – \cos \dfrac{{2\pi }}{9}\cos \dfrac{{4\pi }}{9}\cos \dfrac{{8\pi }}{9}\\ = {X^3} – \dfrac{3}{4}X + \dfrac{1}{8}.\end{array}\)
Từ đó \(\left( {1 – \cos \dfrac{{2\pi }}{9}} \right)\left( {1 – \cos \dfrac{{4\pi }}{9}} \right)\left( {1 – \cos \dfrac{{8\pi }}{9}} \right) = \dfrac{3}{8}\), tức là
\(2{\sin ^2}\dfrac{\pi }{9}.2{\sin ^2}\dfrac{{2\pi }}{9}.2{\sin ^2}\dfrac{{4\pi }}{9} = \dfrac{3}{8}\),
suy ra
\(\sin \dfrac{\pi }{9}.\sin \dfrac{{2\pi }}{9}.\sin \dfrac{{4\pi }}{9} = \dfrac{{\sqrt 3 }}{8}\)
Đẳng thức này lại cho ta \(\sin \dfrac{{5\pi }}{9}\sin \dfrac{{7\pi }}{9}\sin \dfrac{{8\pi }}{9} = \dfrac{{\sqrt 3 }}{8}.\)
f) Từ e) ta suy ra:
\(\begin{array}{l}\sin \dfrac{\pi }{9}\sin \dfrac{{2\pi }}{9}\sin \dfrac{{3\pi }}{9}\sin \dfrac{{4\pi }}{9}\sin \dfrac{{5\pi }}{9}\sin \dfrac{{6\pi }}{9}\sin \dfrac{{7\pi }}{9}\sin \dfrac{{8\pi }}{9}\\ = \dfrac{{\sqrt 3 }}{8}.\dfrac{{\sqrt 3 }}{8}\sin \dfrac{\pi }{3}\sin \dfrac{{2\pi }}{3} = \dfrac{9}{{256}}.\end{array}\)