2 и 4 номера (** изображение)

Question 1

2 и 4 номера (на изображение)

Question 2

Могу 2 скинуть, 4 не знаю

Question 3

давай

Question 4

Решите задачу:

$2)\; \; ctga=\sqrt2\; ,\; \; a\in (\pi ;2\pi )\\\\1+ctg^2a=\frac{1}{sin^2a}\; ,\; \; 1+(\sqrt2)^2=\frac{1}{sin^2a}\; ,\; \; 1+2=\frac{1}{sin^2a}\; ,\\\\sin^2a=\frac{1}{3}\; ,\; \; 1-cos^2a=\frac{1}{3}\; ,\; \; cos^2a=1-\frac{1}{3}\; ,\; \; cos^2a=\frac{2}{3}\; ,\\\\cosa=\pm \sqrt{\frac{2}{3}}\\\\a\in (\pi ;2\pi )\; \; \Rightarrow \\\\a)\; \; cosa\ \textless \ 0\; ,\; pri\; \; a\in (\pi ;\frac{3\pi}{2})\; ,\; cosa=-\sqrt{\frac{2}{3}}=-\frac{\sqrt6}{3}\\\\b)\; \; cosa\ \textgreater \ 0\; \; pri\; \; x\in (\frac{3\pi}{2};2\pi )\; \; \to \; \; cosa=+\sqrt{\frac{2}{3}}=\frac{\sqrt6}{3}$

$ctga=\sqrt2\ \textgreater \ 0\; \; \to \; \; a\in (\frac{3\pi}{2},2\pi )\; \; \Rightarrow \; \; cosa\ \textless \ 0\; ,\; cosa=-\frac{\sqrt6}{3}$

$\frac{\sqrt6}{cosa}=-\frac{\sqrt6}{\sqrt6/3}=-3\\\\4)\; \; (sin^4\frac{35\pi }{24}-sin^4\frac{25\pi }{24})\cdot sin\frac{25\pi }{12}\cdot sin\frac{25\pi }{6}=\\\\=[\; sin(\frac{35\pi}{24})=sin(\pi +\frac{11\pi }{24})=-sin\frac{11\pi }{24}\; ,\\\\sin\frac{25\pi }{24}=sin(\pi +\frac{\pi }{24})=-sin\frac{\pi }{24}\; ]=\\\\=(sin^4\frac{11\pi }{24}- sin^4\frac{\pi }{24})\cdot sin(2\pi +\frac{\pi }{12})\cdot sin(4\pi +\frac{\pi }{6})=\\\\=(sin\frac{11\pi }{24}-sin\frac{\pi }{24})(sin\frac{11\pi }{24}+sin\frac{11\pi }{24}+sin\frac{\pi }{24})(sin^2\frac{11\pi}{24}+sin^2\frac{\pi}{24})\cdot \\\\\cdot sin\frac{\pi}{12}\cdot sin\frac{\pi}{6}=$

$=2\, sin\frac{5\pi }{24}\cdot cos\frac{\pi }{4}\cdot 2\, sin\frac{\pi }{4}\cdot cos\frac{5\pi }{24}\cdot \frac{\sqrt3-1}{2\sqrt2}\cdot \frac{1}{2}=\\\\=2\cdot sin\frac{5\pi }{24}\cdot cos\frac{5\pi}{24}\cdot \frac{\sqrt2}{2}\cdot \frac{\sqrt2}{2}\cdot \frac{\sqrt3-1}{2\sqrt2}=\\\\=sin\frac{5\pi }{12}\cdot \frac{\sqrt3-1}{4\sqrt2}=sin(\frac{\pi}{2}-\frac{\pi}{12})\cdot \frac{\sqrt2(\sqrt3-1)}{4\sqrt2\cdot \sqrt2}=\\\\=cos\frac{\pi }{12}\cdot \frac{\sqrt6-\sqrt2}{8}=\frac{\sqrt3+1}{2\sqrt2}\cdot \frac{\sqrt6-\sqrt2}{8}=\frac{\sqrt6+\sqrt2}{4}\cdot \frac{\sqrt6-\sqrt2}{8}=$

$\frac{(\sqrt6)^2-(\sqrt2)^2}{4\cdot 8}=\frac{6-2}{4\cdot 8}=\frac{4}{4\cdot 8}=\frac{1}{8}$

Question 5

!!!!!!!!!!!!!!!!!!!!!!!!!