Chứng minh rằng các biểu thức sau là những hằng số không phụ thuộc \(\alpha ,\beta \).
LG a
\(\sin 6\alpha \cot 3\alpha - c{\rm{os6}}\alpha \)
Lời giải chi tiết:
\(\sin 6\alpha \cot 3\alpha - c{\rm{os6}}\alpha \) \(= 2\sin 3\alpha \cos 3\alpha .\dfrac{{\cos 3\alpha }}{{\sin 3\alpha }} \) \( - (2{\cos ^2}3\alpha - 1)\)
=\(2{\cos ^2}3\alpha - 2{\cos ^2}3\alpha + 1 = 1\)
LG b
\({{\rm{[}}\tan ({90^0} - \alpha ) - \cot ({90^0} + \alpha ){\rm{]}}^2}\) \( -{{\rm{[}}c{\rm{ot(18}}{{\rm{0}}^0} + \alpha ) + \cot ({270^0} + \alpha ){\rm{]}}^2}\);
Lời giải chi tiết:
\({{\rm{[}}\tan ({90^0} - \alpha ) - \cot ({90^0} + \alpha ){\rm{]}}^2} \) \( - {{\rm{[}}c{\rm{ot(18}}{{\rm{0}}^0} + \alpha ) + \cot ({270^0} + \alpha ){\rm{]}}^2}\)
=\({(\cot \alpha + \tan \alpha )^2} - {(\cot \alpha - \tan \alpha )^2}\)
=\({\cot ^2}\alpha + 2 + {\tan ^2}\alpha \) \( - {\cot ^2}\alpha + 2 - {\tan ^2}\alpha = 4\)
LG c
\((\tan \alpha - \tan \beta )cot(\alpha - \beta ) - \tan \alpha \tan \beta \)
Lời giải chi tiết:
\((\tan \alpha - \tan \beta )cot(\alpha - \beta ) - \tan \alpha \tan \beta \) \( = \dfrac{{\tan \alpha - \tan \beta }}{{\tan (\alpha - \beta )}} - \tan \alpha \tan \beta \)
= \(1 + \tan \alpha \tan \beta - \tan \alpha \tan \beta = 1\)
LG d
\((\cot \dfrac{\alpha }{3} - \tan \dfrac{\alpha }{3})\tan \dfrac{{2\alpha }}{3}\).
Lời giải chi tiết:
\((\cot \dfrac{\alpha }{3} - \tan \dfrac{\alpha }{3})\tan \dfrac{{2\alpha }}{3} \) \( = (\dfrac{{\cos \dfrac{\alpha }{3}}}{{\sin \dfrac{\alpha }{3}}} - \dfrac{{\sin \dfrac{\alpha }{3}}}{{\cos \dfrac{\alpha }{3}}})\dfrac{{\sin \dfrac{{2\alpha }}{3}}}{{\cos \dfrac{{2\alpha }}{3}}}\)
=\(\dfrac{{{{\cos }^2}\dfrac{\alpha }{3} - {{\sin }^2}\dfrac{\alpha }{3}}}{{\sin \dfrac{\alpha }{3}\cos \dfrac{\alpha }{3}}}.\dfrac{{\sin \dfrac{{2\alpha }}{3}}}{{\cos \dfrac{{2\alpha }}{3}}}\) \( = \dfrac{{\cos \dfrac{{2\alpha }}{3}}}{{\dfrac{1}{2}\sin \dfrac{{2\alpha }}{3}}}.\dfrac{{\sin \dfrac{{2\alpha }}{3}}}{{\cos \dfrac{{2\alpha }}{3}}} = 2\)
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