Question

ПОМОГИТЕ СРОЧНО, ПОЖАЛУЙСТА 7, 8, 9

Answer 1

Ответ:

$7)\ \ ctga=-0,2\\\\1+ctg^2a=\dfrac{1}{sin^2a}\ \ \ \to \ \ \ sin^2a=\dfrac{1}{1+ctg^2a}=\dfrac{1}{1+0,04}=\dfrac{25}{26}\\\\\\\dfrac{3\pi}{2}<a<2\pi \ \ \ \Rightarrow \ \ \ sina<0\ \ \ i\ \ \ sina=-\dfrac{5}{\sqrt{26}}\\\\\\tga=\dfrac{1}{ctga}=\dfrac{1}{-0,2}=-5\\\\\\cosa=ctga\cdot sina=-0,2\cdot \dfrac{-5}{\sqrt{26}}=\dfrac{1}{\sqrt{26}}$

$8)\ \ sina=a^2+a-1\\\\|sina|\leq 1\ \ \ \Rightarrow \ \ \ |\, a^2+a-1|\leq 1\ \ ,\ \ \left\{\begin{array}{l}a^2+a-1\leq 1\\a^2+a-1\geq -1\end{array}\right\ \ \\\\\\\left\{\begin{array}{l}a^2+a-2\leq 0\\a^2+a\geq 0\end{array}\right\ \ \left\{\begin{array}{l}(a+2)(a-1)\leq 0\\a(a+1)\geq 0\end{array}\right\ \ \left\{\begin{array}{l}a\in [-2\, ;\ 1\ ]\\a\in (-\infty ;-1\ ]\cup [\ 0\ ;+\infty \, )\end{array}\right$

$a\in [-2\ ;-1\ ]\cup [\ 0\ ;\ 1\ ]$

$\displaystyle 9)\ \ ctga\cdot \Big(\frac{2\cdot sin\frac{\pi}{6}}{cosa}-cosa\Big)=ctga\cdot\Big(\frac{2\cdot \frac{1}{2}}{cosa}-cosa\Big)=\frac{cosa}{sina}\cdot \frac{1-cos^2a}{cosa}=\\\\\\=\frac{cosa}{sina}\cdot \frac{sin^2a}{cosa}=sina\\\\\\sina=sina$