Ai giúp em câu a, b, c với ạ

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Ai giúp em câu a, b, c với ạ
ai-giup-em-cau-a-b-c-voi-a

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Ladonna 5 years 2021-05-19T02:10:20+00:00 1 Answers 27 views 0

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  1. Doris
    0
    2021-05-19T02:12:05+00:00
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    a) $\dfrac{{\sin 2x + \sin 10x + \sin 12x}}{{\sin 4x + \sin 8x + \sin 12x}} = \dfrac{{2\sin 7x.\cos 5x + 2\sin 5x.\cos 5x}}{{2\sin 8x.\cos 4x + 2\sin 4x.\cos 4x}} = \dfrac{{2\cos 5x\left( {\sin 7x + \sin 5x} \right)}}{{2\cos 4x.\left( {\sin 8x + \sin 4x} \right)}} = \dfrac{{\cos 5x.2.\sin 6x.\cos x}}{{\cos 4x.2.\sin 6x.\cos 2x}} = \dfrac{{\cos 5x.\cos x}}{{\cos 4x.\cos 2x}}$

    b) $\begin{array}{l}
    \tan 3a = \tan (2a + a) = \dfrac{{\tan 2a + \tan a}}{{1 – \tan 2a.\tan a}}\\
     \Rightarrow \tan 3a\left( {1 – \tan 2a.\tan a} \right) = \tan 2a + \tan a\\
     \Rightarrow \tan 3a – \tan 2a – \tan a = \tan 3a.\tan 2a.\tan a
    \end{array}$

    c)$\tan 4x – \dfrac{1}{{\cos 4x}} = \dfrac{{\sin 4x}}{{\cos 4x}} – \dfrac{1}{{\cos 4x}} = \dfrac{{\sin 4x – 1}}{{\cos 4x}} = \dfrac{{2\sin 2x\cos 2x – 1}}{{{{\cos }^2}2x – {{\sin }^2}2x}} = \dfrac{{2\sin 2x.\cos 2x – {{\sin }^2}2x – {{\cos }^2}2x}}{{\left( {\cos 2x – \sin 2x} \right)\left( {\cos 2x + \sin 2x} \right)}} = \dfrac{{ – {{\left( {\cos 2x – \sin 2x} \right)}^2}}}{{\left( {\cos 2x – \sin 2x} \right)\left( {\cos 2x + \sin 2x} \right)}} = \dfrac{{\sin 2x – \cos 2x}}{{\cos 2x + \sin 2x}}$

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