Question

СРОЧНО АЛГЕБРА 2 ВАРИАНТ!!!

Answer 1

Ответ:

$1)\ \ \ 2sin\dfrac{\pi}{8}\cdot cos\dfrac{\pi}{8}=sin\dfrac{\pi}{4}=\dfrac{\sqrt2}{2}\\\\\\2)\ \ (cos75^\circ -sin75^\circ )^2=\underbrace{cos^275^\circ +sin^275^\circ }_{=1}-2sin75^\circ \cdot cos75^\circ =1-sin150^\circ=\\\\1-sin(180^\circ -30^\circ )=1-sin30^\circ =1-0,5=0,5\\\\\\3)\ \ \dfrac{tg\frac{\pi}{8}}{1-tg^2\frac{\pi}{8}}=\dfrac{1}{2}\cdot \dfrac{2\, tg\frac{\pi}{8}}{1-tg^2\frac{\pi}{8}}=\dfrac{1}{2}\cdot tg\dfrac{\pi}{4}=\dfrac{1}{2}\cdot 1=\dfrac{1}{2}=0,5$

$4)\ \ cos^2a=cos2a=cos^2a-(cos^2a-sin^2a)=sin^2a\\\\\\5)\ \ sina=-\dfrac{4}{5}\\\\a\in (\, \pi \ ;\ \dfrac{3\pi}{2})\ \ \to \ \ \ sina<0\ ,\ \ cosa<0\\\\cosa=-\sqrt{1-sin^2a}=-\sqrt{1-\dfrac{16}{25}}=-\sqrt{\dfrac{9}{25}}=-\dfrac{3}{5}\\\\\\sin2a=2\, sina\cdot cosa=2\cdot \dfrac{-4}{5}\cdot \dfrac{-3}{5}=\dfrac{24}{25} \\\\cos2a=cos^2a-sin^2a=\dfrac{9}{25}-\dfrac{16}{25}=-\dfrac{7}{25}\\\\\\tg2a=\dfrac{sin2a}{cos2a}=\dfrac{-4/5}{-3/5}=\dfrac{4}{3}\\\\\\ctg2a=\dfrac{1}{tg2a}=\dfrac{3}{4}$