Câu 9 trang 161 SGK Đại số 10
Câu 9 trang 161 SGK Đại số 10 Tính: ...
Câu 9 trang 161 SGK Đại số 10
Tính:
Bài 9. Tính
a) (4(cos{24^0} + cos {48^0} - cos {84^0} - cos {12^0}))
b) (96sqrt 3 sin {pi over {48}}cos {pi over {48}}cos {pi over {24}}cos {pi over {12}}cos {pi over 6})
c) ( an {9^0} - an {63^0} + an {81^0} - an {27^0})
Trả lời:
a)
(eqalign{
& cos{24^0} + cos {48^0} = cos ({36^0} - {12^0}) + cos ({36^0} + {12^0}) cr
& = 2cos {36^0}cos {12^0} cr
& cos {84^0} + cos {12^0} = 2cos {36^0}cos {48^0} cr
& 4(cos {24^0} + cos {48^0} - cos {84^0} - cos {12^0}) = 8cos {36^0}(cos {12^0} - cos {48^0}) cr
& = 8cos {36^0}.2sin {30^0}.sin {18^0} = 8cos {36^0}sin {18^0} cr
& = 8cos {36^0}.sqrt {{{1 - cos {{36}^0}} over 2}} cr} )
Đặt (36^0= x) ta có:
(eqalign{
& sin3x{
m{ }} = {
m{ }}sin{
m{ }}left( {{{180}^0} - 3x}
ight) = sin2x cr
& Leftrightarrow 3sin x - 4{sin ^3}x = 2sin xcos x cr
& Leftrightarrow 3 - 4(1 - {cos ^2}x) = 2{mathop{
m cosx}
olimits} cr
& Leftrightarrow 4co{s^2}x - 2cos x - 1 = 0 cr
& Rightarrow {mathop{
m cosx}
olimits} = cos {36^0} = {{1 + sqrt 5 } over 4} cr} )
Vậy :
(4(cos{24^0} + cos {48^0} - cos {84^0} - cos {12^0}) = 2(1 + sqrt 5 )sqrt {{{3 - sqrt 5 } over 8}} = 2)
b)
(eqalign{
& 96sqrt 3 sin {pi over {48}}cos {pi over {48}}cos {pi over {24}}cos {pi over {12}}cos {pi over 6} cr
& = 48sqrt 3 sin {pi over {24}}cos {pi over {24}}cos {pi over {12}}cos {pi over 6} cr
& = 24sqrt 3 sin {pi over {12}}cos {pi over {12}}cos {pi over 6} cr
& = 12sqrt 3 sin {pi over 6}cos {pi over 6} = 6sqrt 3 sin {pi over 3} = 9 cr} )
c)
(eqalign{
& an {9^0} - an {63^0} + an {81^0} - an {27^0} cr
& = {{cos {{81}^0}} over {sin {{81}^0}}} + {{sin {{81}^0}} over {cos {{81}^0}}} - ({{cos {{27}^0}} over {sin {{27}^0}}} + {{sin {{27}^0}} over {cos {{27}^0}}}) cr
& = {1 over {sin {{81}^0}.cos{{81}^0}}} - {1 over {sin {{27}^0}.cos{{27}^0}}} cr
& = {2 over {sin {{18}^0}}} - {2 over {sin {{54}^0}}} = {2 over {cos {{72}^0}}} - {2 over {cos {{36}^0}}} cr
& = {2 over {2{{cos }^2}{{36}^0} - 1}} - {2 over {cos {{36}^0}}} cr} )
Thay (cos {36^0} = {{1 + sqrt 5 } over 4}) ta được: ( an {9^0} - an {63^0} + an {81^0} - an {27^0} = 4)
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