cho cos a = 4/5 và 3pi/2 30/06/2021 Bởi aihong cho cos a = 4/5 và 3pi/2 { "@context": "https://schema.org", "@type": "QAPage", "mainEntity": { "@type": "Question", "name": " cho cos a = 4/5 và 3pi/2
$\begin{array}{l} \dfrac{{3\pi }}{2} < a < 2\pi \to \cos a > 0,\sin a < 0\\ \sin a = – \sqrt {1 – {{\cos }^2}a} = – \sqrt {1 – {{\left( {\dfrac{4}{5}} \right)}^2}} = – \sqrt {1 – \dfrac{{16}}{{25}}} = – \dfrac{3}{5}\\ \cos 2a = 2{\cos ^2}a – 1 = 2.{\left( {\dfrac{4}{5}} \right)^2} – 1 = \dfrac{{32}}{{25}} – 1 = \dfrac{7}{{25}}\\ \dfrac{1}{{\sin 2a}} – \cot 2a = \dfrac{1}{{\sin 2a}} – \dfrac{{\cos 2a}}{{\sin 2a}}\\ = \dfrac{{1 – \cos 2a}}{{\sin 2a}} = \dfrac{{1 – \left( {1 – 2{{\sin }^2}a} \right)}}{{2\sin a.\cos a}}\\ = \dfrac{{2{{\sin }^2}a}}{{2\sin a\cos a}} = \dfrac{{\sin a}}{{\cos a}} = \tan a \end{array}$ Bình luận
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$\begin{array}{l} \dfrac{{3\pi }}{2} < a < 2\pi \to \cos a > 0,\sin a < 0\\ \sin a = – \sqrt {1 – {{\cos }^2}a} = – \sqrt {1 – {{\left( {\dfrac{4}{5}} \right)}^2}} = – \sqrt {1 – \dfrac{{16}}{{25}}} = – \dfrac{3}{5}\\ \cos 2a = 2{\cos ^2}a – 1 = 2.{\left( {\dfrac{4}{5}} \right)^2} – 1 = \dfrac{{32}}{{25}} – 1 = \dfrac{7}{{25}}\\ \dfrac{1}{{\sin 2a}} – \cot 2a = \dfrac{1}{{\sin 2a}} – \dfrac{{\cos 2a}}{{\sin 2a}}\\ = \dfrac{{1 – \cos 2a}}{{\sin 2a}} = \dfrac{{1 – \left( {1 – 2{{\sin }^2}a} \right)}}{{2\sin a.\cos a}}\\ = \dfrac{{2{{\sin }^2}a}}{{2\sin a\cos a}} = \dfrac{{\sin a}}{{\cos a}} = \tan a \end{array}$