Bài 56 trang 218 SGK Đại số 10 Nâng cao
Tính: ...
Tính:
Tính
a)
(sinalpha ,{ m{ }}cos2alpha ,{ m{ }}sin2alpha ,,cos {alpha over 2},sin {alpha over 2}) biết
(cos alpha = {4 over 5} ) và (- {pi over 2} < alpha < 0 )
b) ( an ({pi over 4} - alpha )) biết
(left{ matrix{
cos alpha = - {9 over {11}} hfill cr
pi < alpha < {{3pi } over 2} hfill cr}
ight.)
c) ({sin ^4}alpha - {cos ^4}alpha ) biết (cos2alpha = {3 over 5})
d)
(cos (alpha - eta )) biết (left{ matrix{
sin alpha - sin eta = {1 over 3} hfill cr
cos alpha - cos eta = {1 over 2} hfill cr}
ight.)
e) (sin {pi over {16}}sin {{3pi } over {16}}sin {{5pi } over {16}}sin {{7pi } over {16}})
Đáp án
a) Ta có:
(eqalign{
& - {pi over 2} < alpha < 0 Rightarrow sin alpha < 0cr& Rightarrow sin alpha = - sqrt {1 - {{cos }^2}alpha } = - {3 over 5} cr
& sin 2alpha = 2sin alpha cos alpha = - {{24} over {25}} cr
& cos 2alpha = 2{cos ^2}alpha - 1 = {7 over {25}} cr
& cos {alpha over 2} = sqrt {{{1 + cos alpha } over 2}} = {{3sqrt {10} } over {10}};cr&sin {alpha over 2} =- sqrt {{{1 - cos alpha } over 2}} = - {{sqrt {10} } over {10}} cr} )
b) Vì (pi < alpha < {{3pi } over 2} Rightarrow an alpha > 0)
Do đó:
(eqalign{
& an alpha = sqrt {{1 over {{{cos }^2}}} - 1} = {{2sqrt {10} } over 9} cr
& an ({pi over 4} - alpha ) = {{1 - an alpha } over {1 + an alpha }} = {{121 - 36sqrt {10} } over {41}} cr} )
c) Ta có:
(eqalign{
& {sin ^4}alpha - {cos ^4}alpha = ({sin ^2}alpha - {cos ^2}alpha )({sin ^2}alpha + {cos ^2}alpha ) cr
& = {sin ^2}alpha - {cos ^2}alpha = - cos 2alpha = - {3 over 5} cr} )
d) Ta có:
(eqalign{
& {(sin alpha - sin eta )^2} = {({1 over 3})^2}cr& Rightarrow {sin ^2}alpha + {sin ^2}eta - 2sin alpha sin eta = {1 over 9},,,,,,(1) cr
& {(cosalpha - cos eta )^2} = {({1 over 2})^2}cr& Rightarrow {cos ^2}alpha + {cos ^2}eta - 2cos alpha cos eta = {1 over 4},,,(2) cr} )
Cộng từng vế của (1) và (2), ta được:
(1 + 1 - 2(cosalpha cos eta + sin alpha sin eta ) = {1 over 9} + {1 over 4} = {{13} over {36}})
Từ đó: (cos (alpha - eta ) = {{59} over {72}})
e) Ta có:
(eqalign{
& sin {pi over {16}}sin {{3pi } over {16}}sin {{5pi } over {16}}sin {{7pi } over {16}}cr& = sin {pi over {16}}sin {{3pi } over {16}}sin ({pi over 2} - {{3pi } over 6})sin ({pi over 2} - {pi over {16}}) cr
& = sin {pi over {16}}sin {{3pi } over {16}}cos {{3pi } over {16}}cos {pi over {16}}cr& = ({1 over 2}sin {pi over 8})({1 over 2}sin {{3pi } over 8}) cr
& = {1 over 4}sin {pi over 8}sin ({pi over 2} - {pi over 8}) cr&= {1 over 4}sin{pi over 8}cos {pi over 8} = {1 over 8}sin {pi over 4} = {{sqrt 2 } over {16}} cr} )
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