Bài tập 23 trang 201 SGK Toán 10 NC
Bài tập 23 trang 201 SGK Toán 10 NC
Chứng minh các biểu thức sau không phụ thuộc vào α
a) \(\sqrt {{{\sin }^4}\alpha + 4\left( {1 - {{\sin }^2}\alpha } \right)} + \sqrt {{{\cos }^4}\alpha + 4{{\sin }^2}\alpha } \)
b) \(2\left( {{{\sin }^6}\alpha + {{\cos }^6}\alpha } \right) - 3\left( {{{\cos }^4}\alpha + {{\sin }^4}\alpha } \right)\)
c) \(\frac{2}{{\tan \alpha - 1}} + \frac{{\cot \alpha + 1}}{{\cot \alpha - 1}}\left( {\tan \alpha \ne 1}\right) \)
a)
\(\begin{array}{*{20}{l}}
{\sqrt {{{\sin }^4}\alpha + 4\left( {1 - {{\sin }^2}\alpha } \right)} + \sqrt {{{\cos }^4}\alpha + 4{{\sin }^2}\alpha } }\\
{ = \sqrt {{{\left( {2 - {{\sin }^2}\alpha } \right)}^2}} + \sqrt {{{\left( {2 - {{\cos }^2}\alpha } \right)}^2}} }\\
{ = 2 - {{\sin }^2}\alpha + 2 - {{\cos }^2}\alpha \left( {{{\sin }^2}\alpha ,{{\cos }^2}\alpha \le 1} \right)}\\
{ = 4 - \left( {{{\sin }^2}\alpha + {{\cos }^2}\alpha } \right) = 4 - 1 = 3}
\end{array}\)
b)
\(\begin{array}{*{20}{l}}
{2\left( {{{\sin }^6}\alpha + {{\cos }^6}\alpha } \right) - 3\left( {{{\cos }^4}\alpha + {{\sin }^4}\alpha } \right)}\\
\begin{array}{l}
= 2 - 6{\sin ^2}\alpha .{\cos ^2}\alpha \\
\,\,\,\,\, - 3\left( {1 - 2{{\sin }^2}\alpha .{{\cos }^2}\alpha } \right)
\end{array}\\
{ = 2 - 3 = - 1}
\end{array}\)
c)
\(\begin{array}{*{20}{l}}
{\frac{2}{{\tan \alpha - 1}} + \frac{{\cot \alpha + 1}}{{\cot \alpha - 1}}}\\
{ = \frac{2}{{\frac{1}{{\cot \alpha }} - 1}} + \frac{{\cot \alpha + 1}}{{\cot \alpha - 1}}}\\
\begin{array}{l}
= \frac{{2\cot \alpha }}{{1 - \cot \alpha }} + \frac{{\cot \alpha + 1}}{{\cot \alpha - 1}}\\
= \frac{{\cot \alpha - 1}}{{1 - \cot \alpha }} = - 1
\end{array}
\end{array}\)
-- Mod Toán 10
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