Bài 23 trang 218 Sách bài tập (SBT) Toán Đại số 10
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a) ({{1 - cos 2a + sin 2a} over {1 + cos 2a + sin 2a}} = an a)
b) ({{cot a + {mathop{ m tana} olimits} } over {1 + an 2 an a}} = 2cot 2a)
c) ({{sqrt 2 - {mathop{ m sina} olimits} - cos a} over {sin a - cos a}} = - an left( {{a over 2} - {pi over 8}} ight))
d) (cos 2a - cos 3a - cos 4a + cos 5a = - 4sin {a over 2}sin acos {{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 + 2sin acos a} over {1 + 2{{cos }^2}a - 1 + 2sin acos a}} cr} )
( = {{2sin a(sin a + {mathop{ m cosa} olimits} )} over {2cos a(sin a + cos a)}} = an a)
b) ({{cot a + an a} over {1 + an 2a an a}} = {{{1 over { an a}} + an a} over {1 + {{2tana} over {1 - {{ an }^2}a}}}})
( = {{1 + {{ an }^2}a} over { an a}}:{{1 - {{ an }^2}a + 2{{ an }^2}a} over {1 - {{ an }^2}a}})
( = {{1 - {{ an }^2}a} over { an a}} = 2cot 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} + {{3pi } over 8}}
ight)sinleft( {{pi over 8} - {a over 2}}
ight)} over {2sinleft( {{a over 2} - {pi over 8}}
ight)cos left( {{a over 2} - {pi over 8}}
ight)}} cr
& = {{sinleft( { - {a over 2} + {pi over 8}}
ight)sinleft( {{pi over 8} - {a over 2}}
ight)} over {sinleft( {{a over 2} - {pi over 8}}
ight)sinleft( {{a over 2} - {pi over 8}}
ight)}} cr} )
( = {{ - sinleft( {{a over 2} - {pi over 8}} ight)} over {cos left( {{a over 2} - {pi over 8}} ight)}} = - an left( {{a over 2} - {pi over 8}} ight))
d)
(eqalign{
& cos 2a - cos 3a - cos 4a + cos 5a cr
& = (cos 2a - cos 4a) + (cos 5a - cos 3a) cr} )
(eqalign{
& = - 2sin 3asin ( - a) - 2sin 4asin a cr
& = 2sin a(sin 3a - sin 4a) cr} )
(eqalign{
& = 4sin acos {{7a} over 2}sin left( { - {a over 2}}
ight) cr
& = - 4sin {a over 2}sin acos {{7a} over 2} cr} )
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