Chứng minh rằng
a) \({{1 - \cos 2a + \sin 2a} \over {1 + \cos 2a + \sin 2a}} = \tan a\)
b) \({{\cot a + {\mathop{\rm tana}\nolimits} } \over {1 + \tan 2\tan a}} = 2\cot 2a\)
c) \({{\sqrt 2 - {\mathop{\rm sina}\nolimits} - \cos a} \over {\sin a - \cos a}} = - \tan \left( {{a \over 2} - {\pi \over 8}} \right)\)
d) \(\cos 2a - \cos 3a - \cos 4a + \cos 5a = - 4\sin {a \over 2}\sin a\cos {{7a} \over 2}\)
Gợi ý làm bài
a)
\(\eqalign{
& {{1 - \cos 2a + \sin 2a} \over {1 + \cos 2a + \sin 2a}} \cr
& = {{2{{\sin }^2}a + 2\sin a\cos a} \over {1 + 2{{\cos }^2}a - 1 + 2\sin a\cos a}} \cr} \)
\( = {{2\sin a(\sin a + {\mathop{\rm cosa}\nolimits} )} \over {2\cos a(\sin a + \cos a)}} = \tan a\)
b) \({{\cot a + \tan a} \over {1 + \tan 2a\tan a}} = {{{1 \over {\tan a}} + \tan a} \over {1 + {{2tana} \over {1 - {{\tan }^2}a}}}}\)
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\( = {{1 + {{\tan }^2}a} \over {\tan a}}:{{1 - {{\tan }^2}a + 2{{\tan }^2}a} \over {1 - {{\tan }^2}a}}\)
\( = {{1 - {{\tan }^2}a} \over {\tan a}} = 2\cot 2a\)
c) \({{\sqrt 2 - \sin a - \cos a} \over {\sin a - \cos a}} = {{\sqrt 2 - \sqrt 2 sin(a + {\pi \over 4})} \over {\sqrt 2 sin(a - {\pi \over 4})}}\)
\( = {{1 - \sin (a + {\pi \over 4})} \over {sin(a - {\pi \over 4})}} = {{sin{\pi \over 2} - sin(a + {\pi \over 4})} \over {sin(a - {\pi \over 4})}}\)
\(\eqalign{
& = {{\cos \left( {{a \over 2} + {{3\pi } \over 8}} \right)sin\left( {{\pi \over 8} - {a \over 2}} \right)} \over {2sin\left( {{a \over 2} - {\pi \over 8}} \right)\cos \left( {{a \over 2} - {\pi \over 8}} \right)}} \cr
& = {{sin\left( { - {a \over 2} + {\pi \over 8}} \right)sin\left( {{\pi \over 8} - {a \over 2}} \right)} \over {sin\left( {{a \over 2} - {\pi \over 8}} \right)sin\left( {{a \over 2} - {\pi \over 8}} \right)}} \cr} \)
\( = {{ - sin\left( {{a \over 2} - {\pi \over 8}} \right)} \over {\cos \left( {{a \over 2} - {\pi \over 8}} \right)}} = - \tan \left( {{a \over 2} - {\pi \over 8}} \right)\)
d)
\(\eqalign{
& \cos 2a - \cos 3a - \cos 4a + \cos 5a \cr
& = (\cos 2a - \cos 4a) + (\cos 5a - \cos 3a) \cr} \)
\(\eqalign{
& = - 2\sin 3a\sin ( - a) - 2\sin 4a\sin a \cr
& = 2\sin a(\sin 3a - \sin 4a) \cr} \)
\(\eqalign{
& = 4\sin a\cos {{7a} \over 2}\sin \left( { - {a \over 2}} \right) \cr
& = - 4\sin {a \over 2}\sin a\cos {{7a} \over 2} \cr} \)