Bài 4 Trang 155 Sgk Đại Số 10
(displaystyle 2sin 2alpha - sin 4alpha over 2sin 2alpha + sin 4alpha )
Phương pháp giải:
Áp dụng các công thức:
(eginarrayl+ )cos2alpha = 1 - 2sin ^2alpha = 2cos ^2alpha - 1.\+ ) analpha = dfracsin alpha cos alpha .\+ ) analpha .cotalpha = 1.endarray)
Lời giải đưa ra tiết:
(eqalign & 2sin 2alpha - sin 4alpha over 2sin 2alpha + sin 4alpha cr& = frac2sin 2alpha - sin left( 2.2alpha ight)2sin 2alpha + sin left( 2.2alpha ight)cr&= 2sin 2alpha - 2sin 2alpha .cos2alpha over 2sin 2alpha + 2sin 2alpha .cos2alpha cr & = frac2sin 2alpha left( 1 - cos 2alpha ight)2sin 2alpha left( 1 + cos 2alpha ight)cr &= 1 - cos 2alpha over 1 + cos 2alpha = frac1 - left( 1 - 2sin ^2alpha ight)1 + left( 2cos ^2alpha - 1 ight)cr&= 2sin ^2alpha over 2cos ^2alpha = fracsin ^2alpha cos ^2alpha cr&=left( fracsin alpha cos alpha ight)^2= an^2alpha.cr )
LG b
( an alpha (1 + cos ^2alpha over sin alpha - sin alpha ))
Lời giải chi tiết:
(eqalign& an alpha left(1 + cos ^2alpha over sin alpha - sin alpha ight ) cr&= sin alpha over cos alpha left(1 + cos ^2alpha - sin ^2alpha over sin alpha ight) cr & = fracsin alpha cos alpha .fracsin ^2alpha + cos ^2alpha + cos ^2alpha - sin ^2alpha sin alpha cr &= sin alpha over cos alpha .2cos ^2alpha over sin alpha = 2cos alpha. cr )
LG c
(displaystyle sin (pi over 4 - alpha ) + cos (pi over 4 - alpha ) over sin (pi over 4 - alpha ) - cos (pi over 4 - alpha ))
Lời giải đưa ra tiết:
(displaystyle sin (pi over 4 - alpha ) + cos (pi over 4 - alpha ) over sin (pi over 4 - alpha ) - cos (pi over 4 - alpha ))
(eginarrayl = dfraccos left( fracpi 4 - alpha ight)left< fracsin left( fracpi 4 - alpha ight)cos left( fracpi 4 - alpha ight) + 1 ight>cos left( fracpi 4 - alpha ight)left< fracsin left( fracpi 4 - alpha ight)cos left( fracpi 4 - alpha ight) - 1 ight>\ = dfraccos left( fracpi 4 - alpha ight)left< an left( fracpi 4 - alpha ight) + 1 ight>cos left( fracpi 4 - alpha ight)left< an left( fracpi 4 - alpha ight) - 1 ight>\ = dfrac an left( fracpi 4 - alpha ight) + 1 an left( fracpi 4 - alpha ight) - 1\ = left< an left( fracpi 4 - alpha ight) + 1 ight>:left< an left( fracpi 4 - alpha ight) - 1 ight>\ = left( frac an fracpi 4 - an alpha 1 + an fracpi 4. an alpha + 1 ight):left( frac an fracpi 4 - an alpha 1 + an fracpi 4. an alpha - 1 ight)\ = left( frac1 - an alpha 1 + an alpha + 1 ight):left( frac1 - an alpha 1 + an alpha - 1 ight)\ = frac1 - an alpha + 1 + an alpha 1 + an alpha :frac1 - an alpha - 1 - an alpha 1 + an alpha \ = frac21 + an alpha :frac - 2 an alpha 1 + an alpha \ = frac21 + an alpha .frac1 + an alpha - 2 an alpha \ = - frac1 an alpha = - cot alpha endarray)
Cách khác:
LG d
(displaystyle sin 5alpha - sin 3alpha over 2cos 4alpha )
Lời giải bỏ ra tiết:
(displaystyle sin 5alpha - sin 3alpha over 2cos 4alpha ) (displaystyle = 2cos 5alpha + 3alpha over 2sin 5alpha - 3alpha over 2 over 2cos 4alpha ) (displaystyle = frac2cos 4alpha sin alpha 2cos 4alpha )
