Question

Тригонометрична функція
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Answer 1

$\displaystyle\bf\\\frac{3\pi }{2} < \alpha < 2\pi \ \ \ \Rightarrow \ \ \ Sin\alpha < 0\\\\\\Cos\alpha =0,8 \ \ \Rightarrow \ \ Sin\alpha =-\sqrt{1-Cos^{2}\alpha } =-\sqrt{1-0,8^{2} } =\\\\\\=-\sqrt{1-0,64}=-\sqrt{0,36} =-0,6\\\\\\\sqrt{2} Cos\Big(\alpha +\frac{\pi }{4} \Big)=\sqrt{2} \cdot(Cos\alpha \cdot Cos\frac{\pi }{4} -Sin\alpha \Cdot Sin\frac{\pi }{4} )=$

$\displaystyle\bf\\=\sqrt{2} \cdot\Big(0,8 \cdot\frac{1}{\sqrt{2} }-(-0,6) \cdot\frac{1}{\sqrt{2} }\Big)=\\\\\\=\sqrt{2} \cdot\frac{1}{2} \cdot(0,8+0,6)=1,4$