Question

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Answer 1

Ответ:

$5.1.\ \ a)\ \ M\Big(\dfrac{\pi}{4}\Big)\ \ \to \ \ \ t=\dfrac{\pi}{4}=45^\circ \ \ \to \\\\sin\dfrac{\pi }{4}=\dfrac{\sqrt2}{2}\ \ \Rightarrow \ \ \ y=\dfrac{\sqrt2}{2}\\\\cos\dfrac{\pi}{4}=\dfrac{\sqrt2}{2}\ \ \Rightarrow \ \ \ x=\dfrac{\sqrt2}{2}\ \qquad \qquad \qqiuad M\Big(\dfrac{\sqrt2}{2}\ ;\ \dfrac{\sqrt2}{2}\Big)$

$b)\ \ M\Big(\dfrac{\pi}{3}\Big)\ \ \to \ \ \ t=\dfrac{\pi}{3}=60^\circ \ \ \to \\\\sin\dfrac{\pi}{3}=\dfrac{\sqrt3}{2}\ \ \Rightarrow \ \ \ y=\dfrac{\sqrt3}{2}\\\\cos\dfrac{\pi}{3}=\dfrac{1}{2}\ \ \Rightarrow \ \ \ x=\dfrac{1}{2}\ \qquad \qquad \qqiuad M\Big(\dfrac{1}{2}\ ;\ \dfrac{\sqrt3}{2}\Big)$

$c)\ \ M\Big(\dfrac{\pi}{6}\Big)\ \ \to \ \ \ t=\dfrac{\pi}{6}=30^\circ \ \ \to \\\\sin\dfrac{\pi}{6}=\dfrac{1}{2}\ \ \Rightarrow \ \ \ y=\dfrac{1}{2}\\\\cos\dfrac{\pi}{6}=\dfrac{\sqrt3}{2}\ \ \Rightarrow \ \ \ x=\dfrac{\sqrt3}{2}\ \qquad \qquad \qqiuad M\Big(\dfrac{\sqrt3}{2}\ ;\ \dfrac{1}{2}\Big)$

$d)\ \ M\Big(\dfrac{\pi}{2}\Big)\ \ \to \ \ \ t=\dfrac{\pi}{2}=90^\circ \ \ \to \\\\sin\dfrac{\pi}{2}=1\ \ \Rightarrow \ \ \ y=1\\\\cos\dfrac{\pi}{2}=0\ \ \Rightarrow \ \ \ x=0\ \qquad \qquad \qqiuad M\Big(\ 0\ ;\ 1\ \Big)$

$5.6.\ \ a)\ \ y=\dfrac{\sqrt2}{2}\ \ \ \to \ \ \ t_1=\dfrac{\pi}{4}\ ,\ \ t_2=\dfrac{3\pi }{4}\\\\b)\ \ y=\dfrac{1}{2}\ \ \ \to \ \ \ t_1=\dfrac{\pi}{6}\ ,\ t_2=\dfrac{5\pi}{6}\\\\c)\ \ y=0\ \ \ \to \ \ \ t_1=0\ ,\ \ t_2=\pi \\\\d)\ \ y=\dfrac{\sqrt3}{2}\ \ \ \to \ \ \ t_1=\dfrac{\pi}{3}\ ,\ \ t_2=\dfrac{2\pi }{3}$

$5.7.\ \ a)\ \ x=\dfrac{\sqrt3}{2}\ \ \ \to \ \ \ t_1=\dfrac{\pi}{6}\ \ ,\ \ t_2=-\dfrac{\pi}{6}\\\\ b)\ \ x=\dfrac{1}{2}\ \ \ \to \ \ \ t_1=\dfrac{\pi}{3}\ \ ,\ \ t_2=-\dfrac{\pi}{3}\\\\c)\ \ x=1\ \ \ \to \ \ \ t_1=0\ \ ,\ \ t_2=2\pi \\\\d)\ \ x=\dfrac{\sqrt2}{2}\ \ \ \to \ \ \ t_1=\dfrac{\pi}{4}\ \ ,\ \ t_2=-\dfrac{\pi}{4}$