\(Bài.1:\\ sin\alpha=\dfrac{3}{5}\Rightarrow cos\alpha=\sqrt{1-\left(sin\alpha\right)^2}=\sqrt{1-\left(\dfrac{3}{5}\right)^2}=\dfrac{4}{5}\\ \Rightarrow tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\dfrac{3}{5}}{\dfrac{4}{5}}=\dfrac{3}{4}\\ Bài.2:\\ cos\alpha=\dfrac{1}{4}\Rightarrow sin\alpha=\sqrt{1-\left(\dfrac{1}{4}\right)^2}=\dfrac{\sqrt{15}}{4}\\ \Rightarrow tan\alpha=\dfrac{sin\alpha}{cos\alpha}=\dfrac{\dfrac{\sqrt{15}}{4}}{\dfrac{1}{4}}=\sqrt{15}\)
Ví dụ 1 :
Ta có :
\(sinB=0,8\)
mà \(sinB=sin\left(90^o-C\right)=cosC\)
\(\Rightarrow cosC=0,8\)
\(sin^2C+cos^2C=1\)
\(\Leftrightarrow sin^2C=1-cos^2C=1-0,64=0,36\)
\(\Leftrightarrow sinC=0,36\)
\(tanC=\dfrac{sinC}{cosC}=\dfrac{0,36}{0,8}=0,45\)
\(\Rightarrow cotC=\dfrac{1}{tanC}=\dfrac{1}{0,45}=\dfrac{100}{45}=\dfrac{20}{9}\)
Ví dụ 3 :
a) \(tan\alpha=\dfrac{1}{3}\)
\(1+tan^2\alpha=\dfrac{1}{cos^2\alpha}\)
\(\Leftrightarrow cos^2\alpha=\dfrac{1}{1+tan^2\alpha}=\dfrac{1}{1+\dfrac{1}{9}}=\dfrac{9}{10}\)
\(\Leftrightarrow cos\alpha=\dfrac{3}{\sqrt[]{10}}=\dfrac{3\sqrt[]{10}}{10}\)
\(sin\alpha=tan\alpha.cos\alpha=\dfrac{1}{3}.\dfrac{3\sqrt[]{10}}{10}=\dfrac{\sqrt[]{10}}{10}\)
giúp em với ạ. Em cảm ơn ạ
