§2. Giá trị lượng giác của một cung

\(\dfrac{tanx}{sinx}-\dfrac{sinx}{cotx}=cosx\)

\(\Leftrightarrow\dfrac{\dfrac{sinx}{cosx}}{sinx}-\dfrac{sinx}{\dfrac{cosx}{sinx}}=cosx\)

\(\Leftrightarrow\dfrac{1}{cosx}-\dfrac{sin^2x}{cosx}=cosx\)

\(\Leftrightarrow\dfrac{cos^2x}{cosx}=cosx\)

\(\Rightarrowđpcm\)

\(tan^2x-sin^2x=tan^2x.sin^2x\)

\(\Leftrightarrow\dfrac{sin^2x}{cos^2x}-sin^2x=\dfrac{sin^2x}{cos^2x}.sin^2x\)

\(\Leftrightarrow\dfrac{sin^2x\left(1-cos^2x\right)}{cos^2x}=\dfrac{sin^4x}{cos^2x}\)

\(\Leftrightarrow\dfrac{sin^2x.sin^2x}{cos^2x}=\dfrac{sin^4x}{cos^2x}\)

\(\Rightarrowđpcm\)

\(cosC=\dfrac{a^2+b^2-c^2}{2ab}=\dfrac{5}{7}\)

\(\Rightarrow sinC=\sqrt{1-cos^2C}=\dfrac{2\sqrt{6}}{7}\)

Áp dụng công thức trung tuyến:

\(AM=m_a=\dfrac{\sqrt{2\left(b^2+c^2\right)-a^2}}{2}=2\sqrt{7}\)

\(\Rightarrow R=\dfrac{AM}{2sinC}=\dfrac{7\sqrt{42}}{12}\)

\(tana+cota=tana+\frac{1}{tana}=\frac{tan^2a+1}{tana}>0\)

\(\Rightarrow tana>0\) mà \(tana=\frac{sina}{cosa}\Rightarrow sina>0\) (do cos và tan đều dương)

\(\Rightarrow sina=\sqrt{1-cos^2a}=0,6\)

\(\tan A=\frac{\sin A}{\cos A}=3\Rightarrow\sin A=3\cos A\Leftrightarrow\sin^3A=27\cos^3A\)

\(B=\frac{\sin A-\cos A}{\sin^3A+3\cos^3A+2\sin A}=\frac{3\cos A-\cos A}{27\cos^3A+27\cos^3A+6\cos A}=\frac{1}{27\cos^2A+3}\) (*)

Ta có công thức:\(1+\tan^2A=\frac{1}{\cos^2A}\Leftrightarrow\cos^2A=\frac{1}{1+\tan^2A}=\frac{1}{1+3^2}=\frac{1}{10}\)

thay vào (*) ta được: \(B=\frac{1}{27\cos^2A+3}=\frac{1}{27\times\frac{1}{10}+3}=\frac{10}{57}\)

Tính theo kiểu lớp 9 :)

\(cos\alpha=\frac{2}{3}\Rightarrow sin\alpha=\sqrt{1-cos^2\alpha}=\sqrt{1-\left(\frac{2}{3}\right)^2}=\frac{\sqrt{5}}{3}\Rightarrow\left\{{}\begin{matrix}tan\alpha=\frac{\frac{\sqrt{5}}{3}}{\frac{2}{3}}=\frac{\sqrt{5}}{2}\\cotg\alpha=\frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}}=\frac{2\sqrt{5}}{5}\end{matrix}\right.\)

\(A=\frac{tan\alpha+3cotg\alpha}{tan\alpha+cotg\alpha}=\frac{\frac{\sqrt{5}}{2}+3.\frac{2\sqrt{5}}{5}}{\frac{\sqrt{5}}{2}+\frac{2\sqrt{5}}{5}}=\frac{17}{9}\)