Chứng minh rằng các biểu thức sau là những hằng số không phụ thuộc α, β
a) \(\sin 6\alpha \cot 3\alpha - \cos 6\alpha\);
b) \({\left[ {\tan \left( {{{90}^0} - \alpha } \right) - \cot \left( {{{90}^0} + \alpha } \right)} \right]^2} - {\left[ {\cot \left( {{{180}^0} + \alpha } \right) + \cot \left( {{{270}^0} + \alpha } \right)} \right]^2}\);
c) \(\left( {\tan \alpha - \tan \beta } \right)\cot \left( {\alpha - \beta } \right) - \tan \alpha \tan \beta\);
d) \(\left( {\cot \frac{\alpha }{3} - \tan \frac{\alpha }{3}} \right).\tan \frac{{2\alpha }}{3}\)
a)
\(\begin{array}{l}
\sin 6\alpha \cot 3\alpha - \cos 6\alpha \\
= 2\sin 3\alpha \cos 3\alpha .\frac{{\cos 3\alpha }}{{\sin 3\alpha }} - \left( {2{{\cos }^2}3\alpha - 1} \right)\\
= 2{\cos ^2}3\alpha - 2{\cos ^2}3\alpha + 1 = 1
\end{array}\)
b)
\(\begin{array}{l}
{\left[ {\tan \left( {{{90}^0} - \alpha } \right) - \cot \left( {{{90}^0} + \alpha } \right)} \right]^2} - {\left[ {\cot \left( {{{180}^0} + \alpha } \right) + \cot \left( {{{270}^0} + \alpha } \right)} \right]^2}\\
= {\left( {\cot \alpha + \tan \alpha } \right)^2} - {\left( {\cot \alpha - \tan \alpha } \right)^2}\\
= {\cot ^2}\alpha + 2 + {\tan ^2}\alpha - {\cot ^2}\alpha + 2 - {\tan ^2}\alpha = 4
\end{array}\)
c)
\(\begin{array}{l}
\left( {\tan \alpha - \tan \beta } \right)\cot \left( {\alpha - \beta } \right) - \tan \alpha \tan \beta \\
= \frac{{\tan \alpha - \tan \beta }}{{\tan \left( {\alpha - \beta } \right)}} - \tan \alpha \tan \beta \\
= 1 + \tan \alpha \tan \beta - \tan \alpha \tan \beta = 1
\end{array}\)
d)
\(\begin{array}{l}
\left( {\cot \frac{\alpha }{3} - \tan \frac{\alpha }{3}} \right).\tan \frac{{2\alpha }}{3}\\
= \left( {\frac{{\cos \frac{\alpha }{3}}}{{\sin \frac{\alpha }{3}}} - \frac{{\sin \frac{\alpha }{3}}}{{\cos \frac{\alpha }{3}}}} \right).\frac{{\sin \frac{{2\alpha }}{3}}}{{\cos \frac{{2\alpha }}{3}}}\\
= \frac{{{{\cos }^2}\frac{\alpha }{3} - {{\sin }^2}\frac{\alpha }{3}}}{{\sin \frac{\alpha }{3}\cos \frac{\alpha }{3}}}.\frac{{\sin \frac{{2\alpha }}{3}}}{{\cos \frac{{2\alpha }}{3}}}\\
= \frac{{\cos \frac{{2\alpha }}{3}}}{{\frac{1}{2}\sin \frac{{2\alpha }}{3}}}.\frac{{\sin \frac{{2\alpha }}{3}}}{{\cos \frac{{2\alpha }}{3}}} = 2
\end{array}\)
-- Mod Toán 10