Question

CMR:\(tan\left(\frac{\pi}{4}-x\right)=\frac{1-sin2x}{Cos2x}\)

Answer 1

Ta có 2 công thức: \(\left\{{}\begin{matrix}sinx+cosx=\sqrt{2}cos\left(x-\frac{\pi}{4}\right)\\sinx-cosx=\sqrt{2}sin\left(x-\frac{\pi}{4}\right)\end{matrix}\right.\)

\(\Rightarrow tan\left(\frac{\pi}{4}-x\right)=-tan\left(x-\frac{\pi}{4}\right)=-\frac{sin\left(x-\frac{\pi}{4}\right)}{cos\left(x-\frac{\pi}{4}\right)}=-\frac{sinx-cosx}{sinx+cosx}\)

\(=\frac{cosx-sinx}{cosx+sinx}=\frac{\left(cosx-sinx\right)^2}{cos^2x-sin^2x}=\frac{1-2sinx.cosx}{cos2x}=\frac{1-sin2x}{cos2x}\)