Question

Cho tan x = -2\(\sqrt{2}\) . Tính giá trị lượng giác còn lại

Answer 1

\(tanx=-2\sqrt{2}\)

=>\(cotx=\dfrac{1}{-2\sqrt{2}}=\dfrac{-\sqrt{2}}{4}\)

\(tanx=-2\sqrt{2}\)

=>\(\dfrac{sinx}{cosx}=-2\sqrt{2}\)

=>sin x và cosx khác dấu 

\(1+tan^2x=\dfrac{1}{cos^2x}\)

=>\(\dfrac{1}{cos^2x}=1+8=9\)

=>\(cos^2x=\dfrac{1}{9}\)

=>\(\left[{}\begin{matrix}cosx=\dfrac{1}{3}\\cosx=-\dfrac{1}{3}\end{matrix}\right.\)

TH1: \(cosx=\dfrac{1}{3}\)

\(tanx=-2\sqrt{2}\)

=>\(\dfrac{sinx}{cosx}=-2\sqrt{2}\)

=>\(sinx=-2\sqrt{2}\cdot cosx=-2\sqrt{2}\cdot\dfrac{1}{3}=-\dfrac{2\sqrt{2}}{3}\)

TH2: cosx=-1/3

=>\(sinx=-2\sqrt{2}\cdot cosx=-2\sqrt{2}\cdot\dfrac{-1}{3}=\dfrac{2\sqrt{2}}{3}\)