Question

Помогите сделать алгебру, пожалуйста

Answer 1

Ответ:

Используем формулы половинного аргумента:

$\boxed{\ sin^2\frac{x}{2}=\dfrac{1-cosx}{2}\ \ ,\ \ cos^2\dfrac{x}{2}=\dfrac{1+cosx}{2}\ \ ,\ \ tg\dfrac{x}{2}=\dfrac{sinx}{1+cosx}\ }$

$\displaystyle 1)\ \ cosa=-\dfrac{5}{17}\ \ \ ,\ \ \ \ tg\dfrac{a}{2}=\dfrac{sina}{1+cosa}\\\\a\in \Big(\dfrac{\pi}{2}\ ;\ \pi \Big)\ \ \Rightarrow \ \ sina > 0\\\\sina=\sqrt{1-cos^2a}=\sqrt{1-\dfrac{25}{289}}=\dfrac{\sqrt{264}}{17}=\dfrac{2\sqrt{66}}{17}\\\\tg\frac{a}{2}=\frac{\dfrac{2\sqrt{66}}{17}}{1-\dfrac{5}{17}}=\frac{2\sqrt{66}}{12}=\frac{\sqrt{66}}{6}\approx 1,35$

$\displaystyle 2)\ \ cos\beta =-\frac{38}{43}\\\\\beta \in \Big(\pi \, ;\, \frac{3\pi}{2}\, \Big)\ \ \to \ \ \ \frac{\beta }{2}\in \Big(\frac{\pi}{2}\, ;\, \frac{3\pi}{4}\, \Big)\ \ \ \Rightarrow \ \ \ sin\frac{\beta }{2} > 0\ \ (2\ chetvert\flat )\\\\\\sin\frac{\beta }{2}=+\sqrt{\frac{1-cos\beta }{2}}=\sqrt{\frac{1+\frac{38}{43}}{2}}=\sqrt{\frac{81}{86}}=\frac{9}{\sqrt{86}}\approx 0,97$

$\displaystyle cos^2\beta +sin\frac{\beta }{2}+sin^2\beta =(\underbrace{sin^2\beta +cos^2\beta }_{1})+sin\frac{\beta }{2}=1+sin\frac{\beta }{2}=1+\frac{9}{\sqrt{86}}=\\\\=\frac{9+\sqrt{86}}{\sqrt{86}}\approx 1,97$

$\displaystyle 3)\ \ cosx=\frac{4}{11}\\\\\\x\in \Big(\frac{3\pi}{2}\, ;\, 2\pi \, \Big)\ \ \to \ \ \ \frac{x}{2}\in \Big(\frac{3\pi}{4}\, ;\, \pi \ \Big)\ \ \Rightarrow \ \ sin\frac{x}{2} > 0\ ,\ cos\frac{x}{2} < 0\\\\\\sin\frac{x}{2}=+\sqrt{\frac{1-cosx}{2}}=\sqrt{\frac{1-\frac{4}{11}}{2}}=\sqrt{\frac{7}{22} }\approx 0,564\approx 0,6\\\\\\cos\frac{x}{2}=-\sqrt{\frac{1+cosx}{2}}=-\sqrt{\frac{1+\frac{4}{11}}{2}}=-\sqrt{\frac{15}{22} }\approx -0,826\approx -0,8$

$sin\dfrac{x}{2}+cos\dfrac{x}{2}+2=0,6-0,8+2=1,8$