Toán Tính $tan$$\frac{x}{2}$ biết ${sinx + cosx =}$ $\frac{1}{5}$ 10/09/2021 By Athena Tính $tan$$\frac{x}{2}$ biết ${sinx + cosx =}$ $\frac{1}{5}$
$\sin x+\cos x=\dfrac{1}{5}$ $\Leftrightarrow \sin^2x+\cos^2x+2\sin x\cos x=\dfrac{1}{25}$ $\Leftrightarrow \sin x\cos x=\dfrac{ \dfrac{1}{25}-1}{2}=\dfrac{-12}{25}$ $\sin x$, $\cos x$ 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 $\sin x=\dfrac{-3}{5}; \cos x=\dfrac{4}{5}$ $\cos^2\dfrac{x}{2}=\dfrac{1+\cos x}{2}=\dfrac{9}{10}$ $\dfrac{1}{\cos^2\dfrac{x}{2}}=1+\tan^2\dfrac{x}{2}$ $\to \tan\dfrac{x}{2}=\pm\dfrac{1}{3}$ Trả lời
$\begin{array}{l}\left\{ \begin{array}{l}\sin x + \cos x = \dfrac{1}{5}\\{\sin ^2}x + {\cos ^2}x = 1\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}\sin x = \dfrac{1}{5} – \cos x\\{\left( {\dfrac{1}{5} – \cos x} \right)^2} + {\cos ^2}x = 1\end{array} \right. \Rightarrow \left[ \begin{array}{l}\cos x = \dfrac{4}{5} \Rightarrow \sin x = \dfrac{{ – 3}}{5}\\\cos x = \dfrac{{ – 3}}{5} \Rightarrow \sin x = \dfrac{4}{5}\end{array} \right.\\\left[ \begin{array}{l}\tan x = – \dfrac{3}{4}\\\tan x = – \dfrac{4}{3}\end{array} \right. \Rightarrow \left[ \begin{array}{l}\dfrac{{ – 3}}{4} = \dfrac{{2\tan \dfrac{x}{2}}}{{1 – {{\tan }^2}\dfrac{x}{2}}} \Rightarrow \dfrac{3}{4}{\tan ^2}\dfrac{x}{2} – 2\tan \dfrac{x}{2} – \dfrac{3}{4} = 0 \Rightarrow \left[ \begin{array}{l}\tan \dfrac{x}{2} = 3\\\tan \dfrac{x}{2} = \dfrac{{ – 1}}{3}\end{array} \right.\\\dfrac{{ – 4}}{3} = \dfrac{{2\tan \dfrac{x}{2}}}{{1 – {{\tan }^2}\dfrac{x}{2}}} \Rightarrow \dfrac{4}{3}{\tan ^2}\dfrac{x}{2} – 2\tan \dfrac{x}{2} – \dfrac{4}{3} = 0 \Rightarrow \left[ \begin{array}{l}\tan \dfrac{x}{2} = 2\\\tan \dfrac{x}{2} = \dfrac{{ – 1}}{2}\end{array} \right.\end{array} \right.\end{array}$ Trả lời
$\sin x+\cos x=\dfrac{1}{5}$
$\Leftrightarrow \sin^2x+\cos^2x+2\sin x\cos x=\dfrac{1}{25}$
$\Leftrightarrow \sin x\cos x=\dfrac{ \dfrac{1}{25}-1}{2}=\dfrac{-12}{25}$
$\sin x$, $\cos x$ 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 $\sin x=\dfrac{-3}{5}; \cos x=\dfrac{4}{5}$
$\cos^2\dfrac{x}{2}=\dfrac{1+\cos x}{2}=\dfrac{9}{10}$
$\dfrac{1}{\cos^2\dfrac{x}{2}}=1+\tan^2\dfrac{x}{2}$
$\to \tan\dfrac{x}{2}=\pm\dfrac{1}{3}$
$\begin{array}{l}
\left\{ \begin{array}{l}
\sin x + \cos x = \dfrac{1}{5}\\
{\sin ^2}x + {\cos ^2}x = 1
\end{array} \right. \Leftrightarrow \left\{ \begin{array}{l}
\sin x = \dfrac{1}{5} – \cos x\\
{\left( {\dfrac{1}{5} – \cos x} \right)^2} + {\cos ^2}x = 1
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\cos x = \dfrac{4}{5} \Rightarrow \sin x = \dfrac{{ – 3}}{5}\\
\cos x = \dfrac{{ – 3}}{5} \Rightarrow \sin x = \dfrac{4}{5}
\end{array} \right.\\
\left[ \begin{array}{l}
\tan x = – \dfrac{3}{4}\\
\tan x = – \dfrac{4}{3}
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
\dfrac{{ – 3}}{4} = \dfrac{{2\tan \dfrac{x}{2}}}{{1 – {{\tan }^2}\dfrac{x}{2}}} \Rightarrow \dfrac{3}{4}{\tan ^2}\dfrac{x}{2} – 2\tan \dfrac{x}{2} – \dfrac{3}{4} = 0 \Rightarrow \left[ \begin{array}{l}
\tan \dfrac{x}{2} = 3\\
\tan \dfrac{x}{2} = \dfrac{{ – 1}}{3}
\end{array} \right.\\
\dfrac{{ – 4}}{3} = \dfrac{{2\tan \dfrac{x}{2}}}{{1 – {{\tan }^2}\dfrac{x}{2}}} \Rightarrow \dfrac{4}{3}{\tan ^2}\dfrac{x}{2} – 2\tan \dfrac{x}{2} – \dfrac{4}{3} = 0 \Rightarrow \left[ \begin{array}{l}
\tan \dfrac{x}{2} = 2\\
\tan \dfrac{x}{2} = \dfrac{{ – 1}}{2}
\end{array} \right.
\end{array} \right.
\end{array}$