Bài tập 11 trang 11 SBT Toán 11 Tập 1 Cánh diều - CD

a) Ta có:

• \(\cos \left( {\frac{{7\pi }}{8}} \right) = \cos \left( {\pi - \frac{\pi }{8}} \right) = - \cos \frac{\pi }{8}\)
• \(\cos \left( {\frac{{5\pi }}{8}} \right) = \cos \left( {\pi - \frac{{3\pi }}{8}} \right) = - \cos \frac{{3\pi }}{8}\)

\( \Rightarrow A = {\cos ^2}\frac{\pi }{8} + {\cos ^2}\frac{{3\pi }}{8} + {\cos ^2}\frac{{5\pi }}{8} + {\cos ^2}\frac{{7\pi }}{8}\)

\( = {\cos ^2}\frac{\pi }{8} + {\cos ^2}\frac{{3\pi }}{8} + {\cos ^2}\frac{{3\pi }}{8} + {\cos ^2}\frac{\pi }{8} = 2\left( {{{\cos }^2}\frac{\pi }{8} + {{\cos }^2}\frac{{3\pi }}{8}} \right)\)

Mặt khác, vì \(\cos \frac{{3\pi }}{8} = \sin \left( {\frac{\pi }{2} - \frac{{3\pi }}{8}} \right) = \sin \frac{\pi }{8}\)

Từ đó \(A = 2\left( {{{\cos }^2}\frac{\pi }{8} + {{\sin }^2}\frac{\pi }{8}} \right) = 2\).

b) Ta có: \(\sin \frac{{9\pi }}{5} = \sin \left( { - \frac{\pi }{5} + 2\pi } \right) = \sin \left( { - \frac{\pi }{5}} \right) = - \sin \frac{\pi }{5}\)

\(\Rightarrow \sin \frac{{9\pi }}{5} + \sin \frac{\pi }{5} = 0\)

Tương tự ta có \(\sin \frac{{8\pi }}{5} + \sin \frac{{2\pi }}{5} = 0\), \(\sin \frac{{7\pi }}{5} + \sin \frac{{3\pi }}{5} = 0\), \(\sin \frac{{6\pi }}{5} + \sin \frac{{4\pi }}{5} = 0\).

Như vậy \(B = 0 + 0 + 0 + 0 + \sin \frac{{5\pi }}{5} = \sin \pi = 0\).

c) Ta có: \(\tan {89^o} = \cot \left( {{{90}^o} - {{89}^o}} \right) = \cot {1^o}\).

              \(\tan {88^o} = \cot \left( {{{90}^o} - {{88}^o}} \right) = \cot {2^o}\).

              ...

              \(\tan {46^o} = \cot \left( {{{90}^o} - {{46}^o}} \right) = \cot {44^o}\).

Do đó \(C = \left( {\tan {1^o}.\tan {{89}^o}} \right)\left( {\tan {2^o}.\tan {{88}^o}} \right)...\left( {\tan {{44}^o}.\tan {{46}^o}} \right)\tan {45^o}\).

-- Mod Toán 11 HỌC247