Question

[Nguyễn Oanh]

cho x là góc nhọn

tính cosx,cotx nếu

a,sinx=\(\dfrac{3}{5}\)

b tanx=√3

c cosx=\(\dfrac{12}{13}\)

d cotx =1

Answer 1

[Mysterious Person]

a) ta có : \(sin^2x+cos^2x=1\Leftrightarrow\dfrac{9}{25}+cos^2x=1\Leftrightarrow cos^2x=\dfrac{16}{25}\)

\(\Rightarrow cosx=\pm\dfrac{4}{5}\)

ta có : \(tanx=\dfrac{sinx}{cosx}=\dfrac{\dfrac{3}{5}}{\pm\dfrac{4}{5}}=\pm\dfrac{3}{4}\) \(\Rightarrow cot=\dfrac{1}{tan}=\dfrac{1}{\pm\dfrac{3}{4}}=\pm\dfrac{4}{3}\)

vậy ................................................................................................

b) ta có : \(tanx=\sqrt{3}\Leftrightarrow cotx=\dfrac{1}{tanx}=\dfrac{1}{\sqrt{3}}\)

ta có : \(\dfrac{sin^2x+cos^2x}{cos^2x}=1+tan^2x\Leftrightarrow\dfrac{1}{cos^2x}=1+tan^2x\)

\(\Leftrightarrow\dfrac{1}{cos^2x}=1+\left(\sqrt{3}\right)^2=4\Rightarrow cos^2x=\dfrac{1}{4}\) \(\Leftrightarrow cos^2x=\pm\dfrac{1}{2}\)

ta có : \(sin^2x+cos^2x=1\Leftrightarrow sin^2x=1-\dfrac{1}{4}=\dfrac{3}{4}\Rightarrow sinx=\pm\dfrac{\sqrt{3}}{2}\)

vậy .............................................................................................

câu c bn làm tương tự câu a ; còn câu d bn làm tương tự câu b nha :)