Алгебра: 3×cosx/4×cosx/2×sinx/4 =1-ctgx/1-ctg²x

3×cosx/4×cosx/2×sinx/4 =1-ctgx/1-ctg²x

Ответ:

maiorov27

06.10.2020 12:39

$3\cdot cos \frac{x}{4}\cdot cos \frac{x}{2} \cdot sin \frac{x}{4}= \frac{1-ctgx}{1-ctg^2x}$

$\star \; \; 3\cdot cos \frac{x}{4} \cdot cos \frac{x}{2} \cdot sin \frac{x}{4}= 3\cdot ( \underbrace {sin\frac{x}{4} \cdot cos \frac{x}{4}}_{1/2\cdot sin(2\cdot x/4)})\cdot cos \frac{x}{2}=\\\\=\frac{3}{2}\cdot sin(\frac{2x}{4})\cdot cos \frac{x}{2}= \frac{3}{2}\cdot sin \frac{x}{2}\cdot cos \frac{x}{2}= \frac{3}{4}\cdot \frac{1}{2} \cdot sin\frac{2x}{2} =\frac{3}{8}\cdot sinx\; \ ;\star \\\\\\\star \; \; \frac{1-ctgx}{1-ctg^2x} = \frac{1-ctgx}{(1-ctgx)(1+ctgx)} = \frac{1}{1+ctgx}= \frac{1}{1+\frac{cosx}{sinx}} =\frac{sinx}{sinx+cosx}\; \; \star$

$\frac{3}{8}\cdot sinx= \frac{sinx}{sinx+cosx}\\\\ \frac{3}{8}\cdot sinx(sinx+cosx)=sinx\\\\ 3sinx(sinx+cosx)-8sinx=0\\\\sinx\cdot (3sinx+cosx-8)=0\\\\a)\; \; sinx=0\; ,\; \; \; x=\pi n,\; n\in Z\\\\b)\; \; 3sinx+cosx=8\; |:\sqrt{10}\\\\\frac{3}{\sqrt{10}}\cdot sinx+\frac{1}{\sqrt{10}}\cdot cosx=\frac{8}{\sqrt{10}}$

$( \frac{3}{\sqrt{10}})^2+( \frac{1}{\sqrt{10}})^2=1\; \; \to \; \; \; \frac{3}{\sqrt{10}}=sin\varphi \; ,\; \; \frac{1}{\sqrt{10}} =cos\varphi \\\\tg\varphi =3\; \; \to \; \; \varphi =arctg3\\\\sin\varphi \cdot sinx+cos\varphi \cdot cosx=\frac{8}{\sqrt{10}}$

$cos(x-\varphi )=\frac{8}{\sqrt{10}}\; ,\; \; \; \frac{8}{\sqrt{10}}\approx \frac{8}{3,16}\approx 2,53\ \textgreater \ 1\; \; \to \\\\net\; \; teshenij\; ,\; t.k.\; \; |cos \alpha | \leq 1\\\\Otvet:\; \; x=\pi n,\; n\in Z\; .$