Câu 4 trang 155 SGK Đại số 10

Câu 4 trang 155 SGK Đại số 10

Rút gọn biểu thức

Bài 4. Rút gọn biểu thức

a) ({{2sin 2alpha  - sin 4alpha } over {2sin 2alpha  + sin 4alpha }})

b) ( an alpha ({{1 + {{cos }^2}alpha } over {sin alpha }} - sin alpha ))

c) ({{sin ({pi  over 4} - alpha ) + cos ({pi  over 4} - alpha )} over {sin ({pi  over 4} - alpha ) - cos ({pi  over 4} - alpha )}})

d) ({{sin 5alpha  - sin 3alpha } over {2cos 4alpha }})

Trả lời:

a)

(eqalign{
& {{2sin 2alpha - sin 4alpha } over {2sin 2alpha + sin 4alpha }} = {{2sin 2alpha - 2sin 2alpha .cos2alpha } over {2sin 2alpha + 2sin 2alpha .cos2alpha }} cr
& = {{1 - cos 2alpha } over {1 + cos 2alpha }} = {{2{{sin }^2}alpha } over {2{{cos }^2}alpha }} = an^2alphacr} )

b)

(eqalign{
& an alpha left({{1 + {{cos }^2}alpha } over {sin alpha }} - sin alpha ight ) = {{sin alpha } over {cos alpha }}left({{1 + {{cos }^2}alpha - {{sin }^2}alpha } over {sin alpha }} ight) cr
& = {{sin alpha } over {cos alpha }}.{{2{{cos }^2}alpha } over {sin alpha }} = 2cos alpha cr} )

c)

(eqalign{
& = {{ an left({pi over 4} - alpha ight) + 1} over { anleft({pi over 4} - alpha ight) - 1}} = left({{ an {pi over 4} - an alpha } over {1 + an {pi over 4}. an alpha }} + 1 ight):left({{ an {pi over 4} - an alpha } over {1 + an {pi over 4}. an alpha }} - 1 ight) cr
& = left({{1 - an alpha + 1 + an alpha } over {1 + an alpha }}  ight):left({{1 - an alpha - 1 - an alpha } over {1 + an alpha }}  ight) cr
& = {{ - 1} over { an alpha }} = - cot alpha cr} ) 

d) 

({{sin 5alpha  - sin 3alpha } over {2cos 4alpha }} = {{2cos {{5alpha  + 3alpha } over 2}sin {{5alpha  - 3alpha } over 2}} over {2cos 4alpha }} = sin alpha )

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