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Chứng minh các đẳng thức
a) \(\tan 3\alpha – \tan 2\alpha – \tan \alpha = \tan \alpha \tan 2\alpha \tan 3\alpha \)
b) \({{4\tan \alpha (1 – {{\tan }^2}\alpha )} \over {{{(1 + {{\tan }^2}\alpha )}^2}}} = \sin 4\alpha \)
c) \({{1 + {{\tan }^4}\alpha } \over {{{\tan }^2}\alpha + {{\cot }^2}\alpha }} = {\tan ^2}\alpha \)
d) \({{\cos \alpha \sin (\alpha – 3) – \sin \alpha \cos (\alpha – 3)} \over {\cos (3 – {\pi \over 6}) – {1 \over 2}\sin 3}} = – {{2\tan 3} \over {\sqrt 3 }}\)
Gợi ý làm bài
a) \(\tan 3\alpha – \tan 2\alpha – \tan \alpha = \tan (2\alpha + \alpha ) – \tan (2\alpha + \alpha )\)
= \({{\tan 2\alpha + \tan \alpha } \over {1 – \tan 2\alpha \tan \alpha }} – (\tan 2\alpha + tan\alpha )\)
= \((\tan 2\alpha + tan\alpha )({1 \over {1 – \tan 2\alpha \tan \alpha }} – 1)\)
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= \(\eqalign{
& {{\tan 2\alpha + \tan \alpha } \over {1 – \tan 2\alpha \tan \alpha }}(1 – 1 + \tan 2\alpha \tan \alpha ) \cr
& = \tan 3\alpha \tan 2\alpha \tan \alpha \cr} \)
b)
\(\eqalign{
& {{4\tan \alpha (1 – {{\tan }^2}\alpha )} \over {{{(1 + {{\tan }^2}\alpha )}^2}}} = {{2.2\tan \alpha } \over {1 + {{\tan }^2}\alpha }}.{{1 – {{\tan }^2}\alpha } \over {1 + {{\tan }^2}\alpha }} \cr
& = 2sin2\alpha c{\rm{os2}}\alpha {\rm{ = }}\sin 4\alpha \cr} \)
c)
\(\eqalign{
& {{1 + {{\tan }^4}\alpha } \over {{{\tan }^2}\alpha + {{\cot }^2}\alpha }} = {{1 + {{\tan }^4}\alpha } \over {{{\tan }^2}\alpha + {1 \over {{{\tan }^2}\alpha }}}} \cr
& = {{1 + {{\tan }^4}\alpha } \over {{{{{\tan }^4}\alpha + 1} \over {{{\tan }^2}\alpha }}}} = {\tan ^2}\alpha \cr} \)
d)
\(\eqalign{
& {{\cos \alpha \sin (\alpha – 3) – \sin \alpha \cos (\alpha – 3)} \over {\cos (3 – {\pi \over 6}) – {1 \over 2}\sin 3}} \cr
& = {{\sin (\alpha – 3 – \alpha )} \over {\cos 3cos{\pi \over 6} + \sin 3\sin {\pi \over 6} – {1 \over 2}\sin 3}} \cr
& = {{ – \sin 3} \over {{{\sqrt 3 } \over 2}\cos 3}} = – {{2\tan 3} \over {\sqrt 3 }} \cr} \)