Question

a, Cho 0<x<\(\dfrac{\Pi}{4}\) .Chứng minh : sinx<cosx

b, Cho \(\dfrac{\Pi}{4}< x< \dfrac{\Pi}{2}\) .Chứng minh : sinx> cosx

 

Answer 1

Ta có \(sinx-cosx=\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)\)

a, Do \(0< x< \dfrac{\pi}{4}\Rightarrow-\dfrac{\pi}{4}< x-\dfrac{\pi}{4}< 0\)

⇒ \(\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)\) < 0

⇒ sinx - cosx < 0

=> sinx < cosx

b, Do \(\dfrac{\pi}{4}< x< \dfrac{\pi}{2}\Rightarrow0< x-\dfrac{\pi}{4}< \dfrac{\pi}{4}\)

⇒ \(\sqrt{2}sin\left(x-\dfrac{\pi}{4}\right)\) > 0

⇒ sinx - cosx > 0

=> sinx > cosx