Предмет: Алгебра
Автор: Ananasik16122005
Вопрос задан 6 лет назад

Помогити
Упростить
1. sin2α − sin2α cos2α
2. sin2α + cos2α + tg2α
3.tgα/ctgα+ 1

Ответы

Ответ дал: Applegate

Ответ:

$1) \displaystyle sin^3\alpha cos\alpha;\\2)\frac{sin2\alpha cos2\alpha +cos^22\alpha+sin2\alpha}{cos2\alpha};\\3) \frac{1}{cos^2\alpha}$

Объяснение:

$\displaystyle1) sin2\alpha-sin2\alpha cos2\alpha=2sin\alpha cos\alpha-2sin\alpha cos\alpha cos2\alpha=2sin\alpha cos\alpha (1-cos2\alpha)=sin\alpha cos\alpha*\frac{1-cos2\alpha}{2} =sin\alpha cos\alpha*sin^2\alpha=sin^3\alpha cos\alpha;$

$\displaystyle 2) sin2\alpha+cos2\alpha+tg2\alpha=sin2\alpha+cos2\alpha+\frac{2tg\alpha}{1-tg^2\alpha} =sin2\alpha +cos2\alpha+\frac{2*\frac{sin\alpha}{cos\alpha} }{1-\frac{sin^2\alpha}{cos^2\alpha} } =sin2\alpha+cos2\alpha+\frac{\frac{2sin\alpha}{cos\alpha} }{\frac{cos^2\alpha-sin^2\alpha}{cos^2\alpha} } =sin2\alpha+cos2\alpha+\frac{2sin\alpha}{cos\alpha} *\frac{cos^2\alpha}{cos^2\alpha-sin^2\alpha} =sin2\alpha+cos2\alpha+\frac{2sin\alpha cos\alpha}{cos^2\alpha-sin^2\alpha} =$

$\displaystyle =sin2\alpha^{(cos2\alpha}+cos2\alpha^{(cos2\alpha}+\frac{sin2\alpha}{cos2\alpha} =\frac{sin2\alpha cos2\alpha +cos^22\alpha+sin2\alpha}{cos2\alpha}$

$\displaystyle 3) \frac{tg\alpha}{ctg\alpha} +1=\frac{\frac{sin\alpha}{cos\alpha} }{\frac{cos\alpha}{sin\alpha} } +1=\frac{sin\alpha}{cos\alpha} *\frac{sin\alpha}{cos\alpha} +1=\frac{sin^2\alpha}{cos^2\alpha} +1^{(cos^2\alpha}=\frac{sin^2\alpha+cos^2\alpha}{cos^2\alpha} =\frac{1}{cos^2\alpha}$

_____________________

справочные материалы:

$\displaystyle sin^2\alpha+cos^2\alpha=1;\\tg2\alpha=\frac{2tg\alpha}{1-tg^2\alpha} ,~~~\alpha\neq \frac{\pi }{2} +\pi k,~~k\in Z;\\cos2\alpha=cos^2\alpha-sin^2\alpha;\\sin2\alpha=2sin\alpha cos\alpha;\\tg\alpha=\frac{sin\alpha}{cos\alpha} ;\\ctg\alpha=\frac{cos\alpha}{sin\alpha} ;\\sin^2\alpha=\frac{1-cos2\alpha}{2}$