Question

If $\frac{\pi }{4}<x<\frac{\pi }{2}$, then write the value of $\sqrt{1-\sin 2x}$.

Answer 1

$$\begin{gathered} \mathrm{We}\mathrm{have}, \\ \sqrt{1-\sin 2x} \\ =\sqrt{{\sin }^{2}x+{\cos }^{2}x-2\sin x\cos x} \\ =\sqrt{{\left(\sin x-\cos x\right)}^2} \\ =\left|\sin x-\cos x\right| \\ =\sin x-\cos x \\ \left[\because \sin x>\cos x\mathrm{for}\frac{\mathrm{\pi }}{4}<x<\frac{\mathrm{\pi }}{2}\right] \\ \therefore \sqrt{1-\sin 2x}=\sin x-\cos x \end{gathered}$$