Tính $tan$$\frac{x}{2}$ biết ${sinx + cosx =}$ $\frac{1}{5}$

Question

Tính $tan$$\frac{x}{2}$ biết ${sinx + cosx =}$ $\frac{1}{5}$

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Neala 5 years 2021-05-19T12:49:24+00:00 2 Answers 24 views 0

Answers ( )

  1. thienan
    0
    2021-05-19T12:50:28+00:00
    Reply

    $\sin x+\cos x=\dfrac{1}{5}$

    $\Leftrightarrow \sin^2x+\cos^2x+2\sin x\cos x=\dfrac{1}{25}$

    $\Leftrightarrow \sin x\cos x=\dfrac{ \dfrac{1}{25}-1}{2}=\dfrac{-12}{25}$

    $\sin x$, $\cos x$ là nghiệm phương trình:

    $t^2-\dfrac{1}{5}t-\dfrac{12}{25}$

    $\to t_1=\dfrac{4}{5}; t_2=\dfrac{-3}{5}$

    – Nếu $\sin x=\dfrac{4}{5}; \cos x=\dfrac{-3}{5}$:

    $\cos^2\dfrac{x}{2}=\dfrac{1+\cos x}{2}= \dfrac{1}{5}$

    $\dfrac{1}{\cos^2\dfrac{x}{2} }=1+\tan^2\dfrac{x}{2}$

    $\to \tan\dfrac{x}{2}=\pm 2$ 

    – Nếu $\sin x=\dfrac{-3}{5}; \cos x=\dfrac{4}{5}$

    $\cos^2\dfrac{x}{2}=\dfrac{1+\cos x}{2}=\dfrac{9}{10}$

    $\dfrac{1}{\cos^2\dfrac{x}{2}}=1+\tan^2\dfrac{x}{2}$

    $\to \tan\dfrac{x}{2}=\pm\dfrac{1}{3}$

  2. Verity
    0
    2021-05-19T12:50:45+00:00
    Reply

    $\begin{array}{l}
    \left\{ \begin{array}{l}
    \sin x + \cos x = \dfrac{1}{5}\\
    {\sin ^2}x + {\cos ^2}x = 1
    \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
    \sin x = \dfrac{1}{5} – \cos x\\
    {\left( {\dfrac{1}{5} – \cos x} \right)^2} + {\cos ^2}x = 1
    \end{array} \right. \Rightarrow \left[ \begin{array}{l}
    \cos x = \dfrac{4}{5} \Rightarrow \sin x = \dfrac{{ – 3}}{5}\\
    \cos x = \dfrac{{ – 3}}{5} \Rightarrow \sin x = \dfrac{4}{5}
    \end{array} \right.\\
    \left[ \begin{array}{l}
    \tan x =  – \dfrac{3}{4}\\
    \tan x =  – \dfrac{4}{3}
    \end{array} \right. \Rightarrow \left[ \begin{array}{l}
    \dfrac{{ – 3}}{4} = \dfrac{{2\tan \dfrac{x}{2}}}{{1 – {{\tan }^2}\dfrac{x}{2}}} \Rightarrow \dfrac{3}{4}{\tan ^2}\dfrac{x}{2} – 2\tan \dfrac{x}{2} – \dfrac{3}{4} = 0 \Rightarrow \left[ \begin{array}{l}
    \tan \dfrac{x}{2} = 3\\
    \tan \dfrac{x}{2} = \dfrac{{ – 1}}{3}
    \end{array} \right.\\
    \dfrac{{ – 4}}{3} = \dfrac{{2\tan \dfrac{x}{2}}}{{1 – {{\tan }^2}\dfrac{x}{2}}} \Rightarrow \dfrac{4}{3}{\tan ^2}\dfrac{x}{2} – 2\tan \dfrac{x}{2} – \dfrac{4}{3} = 0 \Rightarrow \left[ \begin{array}{l}
    \tan \dfrac{x}{2} = 2\\
    \tan \dfrac{x}{2} = \dfrac{{ – 1}}{2}
    \end{array} \right.
    \end{array} \right.
    \end{array}$

     

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