Advertisements (Quảng cáo)
Rút gọn
a) \({{{{\sin }^2}2\alpha + 4{{\sin }^4}\alpha – 4{{\sin }^2}\alpha c{\rm{o}}{{\rm{s}}^2}\alpha } \over {4 – {{\sin }^2}2\alpha – 4{{\sin }^2}\alpha }}\)
b) \(3 – 4\cos 2a + \cos 4a\)
c) \(\cos 4a – \sin 4a\cot 2a\)
d) \({{{\mathop{\rm cota}\nolimits} + \tan a} \over {1 + \tan 2a\tan a}}\)
Gợi ý làm bài
a)
\(\eqalign{
& {{{{\sin }^2}2\alpha + 4{{\sin }^2}4\alpha – 4{{\sin }^2}\alpha {{\cos }^2}\alpha } \over {4 – {{\sin }^2}2\alpha – 4{{\sin }^2}\alpha }} \cr
& = {{{{\sin }^2}2\alpha + 4{{\sin }^4}\alpha – {{\sin }^2}2\alpha } \over {4{{\cos }^2}a – 4{{\sin }^2}2\alpha {{\cos }^2}\alpha }} \cr} \)
\( = {{4{{\sin }^2}\alpha } \over {4co{s^2}\alpha (1 – {{\sin }^2}\alpha )}} = {\tan ^4}\alpha \)
b)
Advertisements (Quảng cáo)
\(\eqalign{
& 3 – 4\cos 2a + \cos 4a \cr
& = 3 – 4(1 – 2{\sin ^2}a) + (1 – 2{\sin ^2}2a) \cr} \)
\(\eqalign{
& = 8{\sin ^2}a – 8{\sin ^2}a{\cos ^2}a \cr
& = 8{\sin ^2}a(1 – {\cos ^2}a) \cr} \)
\( = 8{\sin ^4}a\)
c)
\(\eqalign{
& \cos 4a – \sin 4a\cot 2a \cr
& = 2{\cos ^2}2a – 1 – 2\sin 2a\cos 2a{{\cos 2a} \over {\sin 2a}} = – 1 \cr} \)
d) \({{\cot a + \tan a} \over {1 + \tan 2a\tan a}} = {{{{\cos a} \over {\sin a}} + {{\sin a} \over {\cos a}}} \over {1 + {{\sin 2a\sin a} \over {\cos 2a\cos a}}}}\)
\( = {1 \over {\sin a\cos a}}.{{\cos acos2a} \over {\cos 2a\cos a + \sin 2a\sin a}}\)
\( = {2 \over {\sin 2a}}.{{\cos acos2a} \over {\cos (2a – a)}} = 2\cot 2a\)