if 2 cos A/2=(root 1+sinA)+(root 1-sinA)then A/2 lies between?

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Please find below the solution to the asked query:

$$\begin{gathered} 2\cos \frac{\mathrm{A}}{2}=\sqrt{1+\mathrm{sinA}}+\sqrt{1-\mathrm{sinA}} \\ \Rightarrow 2\cos \frac{\mathrm{A}}{2}=\sqrt{{\cos }^2\frac{\mathrm{A}}{2}+{\sin }^2\frac{\mathrm{A}}{2}+2\sin \frac{\mathrm{A}}{2}\cos \frac{\mathrm{A}}{2}}+\sqrt{{\cos }^2\frac{\mathrm{A}}{2}+{\sin }^2\frac{\mathrm{A}}{2}-2\sin \frac{\mathrm{A}}{2}\cos \frac{\mathrm{A}}{2}} \\ \Rightarrow 2\cos \frac{\mathrm{A}}{2}=\sqrt{{\left(\cos \frac{\mathrm{A}}{2}+\sin \frac{\mathrm{A}}{2}\right)}^2}+\sqrt{{\left(\cos \frac{\mathrm{A}}{2}-\sin \frac{\mathrm{A}}{2}\right)}^2} \\ \Rightarrow 2\cos \frac{\mathrm{A}}{2}=\left|\cos \frac{\mathrm{A}}{2}+\sin \frac{\mathrm{A}}{2}\right|+\left|\cos \frac{\mathrm{A}}{2}-\sin \frac{\mathrm{A}}{2}\right|...\left(\mathrm{i}\right) \\ \mathrm{We} \mathrm{know} \mathrm{that} \\ \left|\mathrm{x}\right|=\left\{\begin{array}{ll}\mathrm{x}& \mathrm{x}\ge 0\\ -\mathrm{x}& \mathrm{x}\le 0\end{array}\right. \\ \mathrm{For} \left(\mathrm{i}\right) \mathrm{to} \mathrm{be} \mathrm{true}, \mathrm{mod} \mathrm{must} \mathrm{open} \mathrm{with} \mathrm{positive} \mathrm{sign} \mathrm{i}.\mathrm{e}. \\ \cos \frac{\mathrm{A}}{2}+\sin \frac{\mathrm{A}}{2}\ge 0 \mathrm{and} \cos \frac{\mathrm{A}}{2}-\sin \frac{\mathrm{A}}{2}\ge 0 \\ \Rightarrow \frac{1}{\sqrt{2}}\cos \frac{\mathrm{A}}{2}+\frac{1}{\sqrt{2}}\sin \frac{\mathrm{A}}{2}\ge 0 \mathrm{and} \frac{1}{\sqrt{2}}\cos \frac{\mathrm{A}}{2}-\frac{1}{\sqrt{2}}\sin \frac{\mathrm{A}}{2}\ge 0 \\ \Rightarrow \sin \left(\frac{\mathrm{\pi }}{4}\right)\cos \frac{\mathrm{A}}{2}+\cos \left(\frac{\mathrm{\pi }}{4}\right)\sin \frac{\mathrm{A}}{2}\ge 0 \mathrm{and} \sin \left(\frac{\mathrm{\pi }}{4}\right)\cos \frac{\mathrm{A}}{2}-\cos \left(\frac{\mathrm{\pi }}{4}\right)\sin \frac{\mathrm{A}}{2}\ge 0 \\ \mathrm{Multiply} \mathrm{second} \mathrm{inequality} \mathrm{by} -1 \\ \Rightarrow \sin \left(\frac{\mathrm{\pi }}{4}\right)\cos \frac{\mathrm{A}}{2}+\cos \left(\frac{\mathrm{\pi }}{4}\right)\sin \frac{\mathrm{A}}{2}\ge 0 \mathrm{and} \cos \left(\frac{\mathrm{\pi }}{4}\right)\sin \frac{\mathrm{A}}{2}-\sin \left(\frac{\mathrm{\pi }}{4}\right)\cos \frac{\mathrm{A}}{2}\le 0 \\ \Rightarrow \sin \left(\frac{\mathrm{A}}{2}+\frac{\mathrm{\pi }}{4}\right)\ge 0 \mathrm{and} \sin \left(\frac{\mathrm{A}}{2}-\frac{\mathrm{\pi }}{4}\right)\le 0 \\ \Rightarrow \sin \left(\frac{\mathrm{A}}{2}+\frac{\mathrm{\pi }}{4}\right)\ge 0 \mathrm{and} \sin \left(-\frac{\mathrm{A}}{2}+\frac{\mathrm{\pi }}{4}\right)\ge 0 \\ \mathrm{Now} \mathrm{we} \mathrm{make} \mathrm{two} \mathrm{graphs} \mathrm{and} \mathrm{see} \mathrm{where} \mathrm{they} \mathrm{are} \mathrm{positive} \mathrm{together}. \end{gathered}$$


$$\begin{gathered} \Rightarrow x\in \left[0,\frac{\mathrm{\pi }}{2}\right]\cup \left[\frac{7\mathrm{\pi }}{2},4\mathrm{\pi }\right] \\ \sin \frac{\mathrm{x}}{2} \mathrm{has} \mathrm{period} \mathrm{of} 4\mathrm{\pi } \\ \mathbf{\Rightarrow }\mathbf{x}\mathbf{\in }\left[\mathbf{4}\mathbf{k\pi }\mathbf{,}\mathbf{4}\mathbf{k\pi }\mathbf{+}\frac{\mathbf{\pi }}{\mathbf{2}}\right]\mathbf{\cup }\left[\mathbf{4}\mathbf{k\pi }\mathbf{+}\frac{\mathbf{7}\mathbf{\pi }}{\mathbf{2}}\mathbf{,}\mathbf{4}\mathbf{k\pi }\mathbf{+}\mathbf{4}\mathbf{\pi }\right] \end{gathered}$$


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