прошу, хотя бы первые два, нужно доказать тождество

прошу, хотя бы первые два, нужно доказать тождество
- Предмет:
  Алгебра
- Автор:
  kailey
- 5 лет назад

Ответы 1

12.
$\frac{sin\alpha+sin\frac{\alpha}{2}}{1+cos\alpha+cos\frac{\alpha}{2}}=\frac{sin(2\cdot\frac{\alpha}{2})+sin\frac{\alpha}{2}}{1+cos(2\cdot\frac{\alpha}{2)}+cos\frac{\alpha}{2}}=$
$\frac{2sin\frac{\alpha}{2} cos\frac{\alpha}{2}+sin\frac{\alpha}{2}}{1+cos^2\frac{\alpha}{2}-sin^2\frac{\alpha}{2}+cos\frac{\alpha}{2}}=\frac{sin\frac{\alpha}{2}(2 cos\frac{\alpha}{2}+1)}{1-sin^2\frac{\alpha}{2}+cos^2\frac{\alpha}{2}+cos\frac{\alpha}{2}}=$
$\frac{sin\frac{\alpha}{2}(2 cos\frac{\alpha}{2}+1)}{cos^2\frac{\alpha}{2}+cos^2\frac{\alpha}{2}+cos\frac{\alpha}{2}}=\frac{sin\frac{\alpha}{2}(2 cos\frac{\alpha}{2}+1)}{2cos^2\frac{\alpha}{2}+cos\frac{\alpha}{2}}=$
$\frac{sin\frac{\alpha}{2}(2 cos\frac{\alpha}{2}+1)}{cos\frac{\alpha}{2}(2cos\frac{\alpha}{2}+1)}=\frac{sin\frac{\alpha}{2}}{cos\frac{\alpha}{2}}=tg\frac{\alpha}{2}$
13.
$\frac{1+cos\frac{\alpha}{2}-sin\frac{\alpha}{2}}{1-cos\frac{\alpha}{2}-sin\frac{\alpha}{2}}=\frac{1+cos(2\cdot\frac{\alpha}{4})-sin(2\cdot\frac{\alpha}{4})}{1-cos(2\cdot\frac{\alpha}{4})-sin(2\cdot\frac{\alpha}{4})}=$
$\frac{1+cos^2\frac{\alpha}{4}-sin^2\frac{\alpha}{4}-2sin\frac{\alpha}{4}cos\frac{\pi}{4}}{1-cos^2\frac{\alpha}{4}+sin^2\frac{\pi}{4}-2sin\frac{\alpha}{4} cos\frac{\pi}{4}}=\frac{1-sin^2\frac{\alpha}{4}+cos^2\frac{\alpha}{4}-2sin\frac{\alpha}{4}cos\frac{\pi}{4}}{sin^2\frac{\alpha}{4}+sin^2\frac{\pi}{4}-2sin\frac{\alpha}{4} cos\frac{\pi}{4}}=$
$\frac{cos^2\frac{\alpha}{4}+cos^2\frac{\alpha}{4}-2sin\frac{\alpha}{4}cos\frac{\pi}{4}}{2sin^2\frac{\pi}{4}-2sin\frac{\alpha}{4} cos\frac{\pi}{4}}=\frac{2cos^2\frac{\alpha}{4}-2sin\frac{\alpha}{4}cos\frac{\pi}{4}}{2sin^2\frac{\pi}{4}-2sin\frac{\alpha}{4} cos\frac{\pi}{4}}=$
$\frac{-2cos\frac{\alpha}{4}(-cos\frac{\alpha}{4}+sin\frac{\alpha}{4})}{2sin\frac{\pi}{4}(sin\frac{\pi}{4}-cos\frac{\pi}{4})}=\frac{-cos\frac{\alpha}{4}}{sin\frac{\pi}{4}}=-ctg\frac{\pi}{4}$
14.
$\frac{4sin^4\frac{\alpha}{4}}{1-cos^2\frac{\alpha}{2}}=\frac{4sin^4\frac{\alpha}{4}}{sin^2\frac{\alpha}{2}}=$
$\frac{4sin^4\frac{\alpha}{4}}{sin^2(2\cdot\frac{\alpha}{4})}=\frac{4sin^4\frac{\alpha}{4}}{(sin(2\cdot\frac{\alpha}{4}))^2}=$
$\frac{4sin^4\frac{\alpha}{4}}{(2sin\frac{\alpha}{4}cos\frac{\alpha}{4})^2}=\frac{4sin^4\frac{\alpha}{4}}{4sin^2\frac{\alpha}{4}cos^2\frac{\alpha}{4}}=$
$\frac{sin^2\frac{\alpha}{4}}{cos^2\frac{\alpha}{4}}=tg^2\frac{\alpha}{4}$
15.
$\frac{2sin\alpha-sin2\alpha}{2sin\alpha+sin2\alpha}=\frac{2sin\alpha-2sin\alpha cos\alpha}{2sin\alpha+2sin\alpha cos\alpha}=$
$\frac{2sin\alpha(1-cos\alpha)}{2sin\alpha(1+cos\alpha)}=\frac{1-cos\alpha}{1+cos\alpha}=$
$\frac{1-cos(2\cdot\frac{\alpha}{2})}{1+cos(2\cdot\frac{\alpha}{2})}=\frac{1-cos^2\frac{\alpha}{2}+sin^2\frac{\alpha}{2}}{1+cos^2\frac{\alpha}{2}-sin^2\frac{\alpha}{2}}=$
$\frac{sin^2\frac{\alpha}{2}+sin^2\frac{\alpha}{2}}{1-sin^2\frac{\alpha}{2}+cos^2\frac{\alpha}{2}}=\frac{2sin^2\frac{\alpha}{2}}{cos^2\frac{\alpha}{2}+cos^2\frac{\alpha}{2}}=$
$\frac{2sin^2\frac{\alpha}{2}}{2cos^2\frac{\alpha}{2}}=tg^2\frac{\alpha}{2}$
- Автор:
  charles80
- 5 лет назад
- 0
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