Радикали

[Spider Iovkov]

Да се представи във вид на произведение разликата [tex]A=\sqrt{1+sinx}-\sqrt{1-sinx}[/tex], ако [tex]-\frac{7\pi}{2}\le x\le-\frac{5\pi}{2}[/tex] и квадратните корени са аритметични.

[Реклама]

[martosss]

[Spider Iovkov]

[tex]A=\sqrt{1+sinx}-\sqrt{1-sinx}=\sqrt{1+cos(\frac{\pi}{2}-x)}-\sqrt{1-cos(\frac{\pi}{2}-x)}=\sqrt{2cos^2 (\frac{\pi}{4}-\frac{x}{2})}-\sqrt{2sin^2 (\frac{\pi}{4}-\frac{x}{2})}=\sqrt{2}|cos(\frac{\pi}{4}-\frac{x}{2})|-\sqrt{2}|sin(\frac{\pi}{4}-\frac{x}{2})|=\sqrt{2}(|cos(\frac{\pi}{4}-\frac{x}{2})|-|sin(\frac{\pi}{4}-\frac{x}{2})|)[/tex].

Тъй като [tex]\frac{5\pi}{4}\le -\frac{x}{2}\le\frac{7\pi}{4}[/tex], то [tex]\frac{5\pi}{4}+\frac{\pi}{4}\le \frac{\pi}{4}-\frac{x}{2}\le \frac{7\pi}{4}+\frac{\pi}{4}[/tex], тоест [tex]\frac{3\pi}{2}\le \frac{\pi}{4}-\frac{x}{2}\le2\pi[/tex]
[tex]\Rightarrow |cos(\frac{\pi}{4}-\frac{x}{2})|=cos(\frac{\pi}{4}-\frac{x}{2}), |sin(\frac{\pi}{4}-\frac{x}{2})|=-sin(\frac{\pi}{4}-\frac{x}{2})[/tex].

[tex]\Rightarrow A=\sqrt{2}[cos(\frac{\pi}{4}-\frac{x}{2})+sin(\frac{\pi}{4}-\frac{x}{2})]=\sqrt{2}[sin(\frac{\pi}{4}+\frac{x}{2})+sin(\frac{\pi}{4}-\frac{x}{2})]=2\sqrt{2}sin{\frac{\pi}{4}}cos{\frac{x}{2}}=2cos{\frac{x}{2}}[/tex]