Question

Cho cotg α =\(\dfrac{a^2-b^2}{2ab}\) trong đó α là góc nhọn, a > b > 0. Tính cos α

Answer 1

\(1+cot^2a=\dfrac{1}{sin^2a}\)

\(\Leftrightarrow\dfrac{1}{sin^2a}=1+\dfrac{\left(a^2-b^2\right)^2}{4a^2b^2}=\dfrac{4a^2b^2+a^4-2a^2b^2+b^4}{4a^2b^2}\)

\(\Leftrightarrow sin^2a=\dfrac{4a^2b^2}{a^4+2a^2b^2+b^4}=\left(\dfrac{2ab}{\left(a^2+b^2\right)}\right)^2\)

=>\(cos^2a=\dfrac{a^4+2a^2b^2+b^4-4a^2b^2}{\left(a^2+b^2\right)^2}\)

\(\Leftrightarrow cos^2a=\dfrac{\left(a^2-b^2\right)^2}{\left(a^2+b^2\right)^2}\)

hay \(cosa=\dfrac{\left(a^2-b^2\right)}{a^2+b^2}\)