Register Now

Login

Lost Password

Lost your password? Please enter your email address. You will receive a link and will create a new password via email.

222-9+11+12:2*14+14 = ? ( )

Toán Lớp 9: Cho `sin(\alpha)-cos(\alpha)=1/5`. Tính `cot(\alpha)`

Toán Lớp 9: Cho sin(\alpha)-cos(\alpha)=1/5. Tính cot(\alpha)

Comments ( 2 )

  1. Giải đáp:
     \(\left[ \begin{array}{l}\cot\alpha=\dfrac{\cos\alpha}{\sin\alpha}=\dfrac43\\cot\alpha=\dfrac{\cos\alpha}{\sin\alpha}=\dfrac34\end{array} \right.\)
    Lời giải và giải thích chi tiết:
    sin\alpha-cos\alpha=1/5
    <=>sin\alpha=cos\alpha+1/5
    Mặt khác:sin^2\alpha+cos^2\alpha=1
    <=>(cos\alpha+1/5)^2+cos^2\alpha=1
    <=>cos^2\alpha+2/5cos\alpha+1/25+cos^2\alpha=1
    <=>2cos^2\alpha+2/5cos\alpha-24/25=0
    <=>cos^2\alpha+cos\alpha/5-12/25=0
    <=>25cos^2\alpha+5cos\alpha-12=0
    <=> \(\left[ \begin{array}{l}\cos\alpha=\dfrac35\\\cos\alpha=-\dfrac45\end{array} \right.\) 
    <=> \(\left[ \begin{array}{l}\sin\alpha=\cos\alpha+\dfrac15=\dfrac45\\\sin\alpha=\cos\alpha+\dfrac15=-\dfrac35\end{array} \right.\) 
    <=> \(\left[ \begin{array}{l}\cot\alpha=\dfrac{\cos\alpha}{\sin\alpha}=\dfrac43\\cot\alpha=\dfrac{\cos\alpha}{\sin\alpha}=\dfrac34\end{array} \right.\) 

  2. Giải đáp:
    $\left[ \begin{array}{l}\cot\alpha=\dfrac{4}{3}\\\cot\alpha=\dfrac{3}{4}\end{array} \right.$
    Lời giải và giải thích chi tiết:
    $\sin\alpha-\cos\alpha=\dfrac{1}{5}$
    $⇒\left(\sin\alpha-\cos\alpha\right)^2=\dfrac{1}{25}$
    $⇒\sin^2\alpha-2\sin\alpha.\cos\alpha+\cos^2\alpha=\dfrac{1}{25}$
    $⇒1-2\sin\alpha.\cos\alpha=\dfrac{1}{25}$
    $⇒2\sin\alpha.\cos\alpha=\dfrac{24}{25}$
    $⇒\sin\alpha.\cos\alpha=\dfrac{12}{25}$
    $⇒\dfrac{\sin\alpha.\cos\alpha}{\sin^2\alpha}=\dfrac{12}{25\sin^2\alpha}$
    $⇒\dfrac{\cos\alpha}{\sin\alpha}=\dfrac{12}{25}.\dfrac{\sin^2\alpha+\cos^2\alpha}{\sin^2\alpha}$
    $⇒\cot\alpha=\dfrac{12}{25}.\left(\dfrac{\sin^2\alpha}{\sin^2\alpha}+\dfrac{\cos^2\alpha}{\sin^2\alpha}\right)$
    $⇒25\cot\alpha=12+12\cot^2\alpha$
    $⇒12\cot^2\alpha-25\cot\alpha+12=0$
    $⇒\left[ \begin{array}{l}\cot\alpha=\dfrac{4}{3}\\\cot\alpha=\dfrac{3}{4}\end{array} \right.$
    Vậy $\left[ \begin{array}{l}\cot\alpha=\dfrac{4}{3}\\\cot\alpha=\dfrac{3}{4}\end{array} \right.$.

Leave a reply

222-9+11+12:2*14+14 = ? ( )

About Lan Anh