Question

[Bùi Minh Phúc]

cho α nhọn biết cosα-sinα=1/5. Tính cotα

Answer 1

[Lê Song Phương]

Bạn nên nhớ \(\sin^2\alpha+\cos^2\alpha=1\)

Lại thêm \(\cos\alpha-\sin\alpha=\dfrac{1}{5}\Leftrightarrow\sin\alpha=\cos\alpha-\dfrac{1}{5}\) nên ta có \(\left(\cos\alpha-\dfrac{1}{5}\right)^2+\cos^2\alpha=1\) \(\Leftrightarrow\cos^2\alpha-\dfrac{2}{5}\cos\alpha+\dfrac{1}{25}+\cos^2\alpha-1=0\)

\(\Leftrightarrow2\cos^2\alpha-\dfrac{2}{5}\cos\alpha-\dfrac{24}{25}=0\)

\(\Leftrightarrow\cos^2\alpha-\dfrac{1}{5}\cos\alpha-\dfrac{12}{25}=0\)

\(\Leftrightarrow25\cos^2\alpha-5\cos\alpha-12=0\)

Đặt \(\cos\alpha=p\left(0< p< 1\right)\) thì ta có \(25p^2-5p-12=0\)

Ta có \(\Delta=\left(-5\right)^2-4.25\left(-12\right)=1225>0\), vậy:

\(x_1=\dfrac{-\left(-5\right)+\sqrt{1225}}{2.25}=\dfrac{4}{5}\left(nhận\right)\)

\(x_2=\dfrac{-\left(-5\right)-\sqrt{1225}}{2.25}=-\dfrac{3}{5}\left(loại\right)\)

Vậy ta có \(\cos\alpha=\dfrac{4}{5}\). Ta lại có \(\sin\alpha=\cos\alpha-\dfrac{1}{5}=\dfrac{4}{5}-\dfrac{1}{5}=\dfrac{3}{5}\)

Mà \(\cot\alpha=\dfrac{\cos\alpha}{\sin\alpha}=\dfrac{\dfrac{4}{5}}{\dfrac{3}{5}}=\dfrac{4}{3}\). Vậy \(\cot\alpha=\dfrac{4}{3}\)