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  4. If A= sin 45°+ cos 45° and B= sin 44°+ cos 44°, then

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Q. If $A=\sin 45^{\circ}+\cos 45^{\circ}$ and $B=\sin 44^{\circ}+\cos 44^{\circ}$, then

Trigonometric Functions

Solution:

$A=\frac{1}{\sqrt{2}} +\frac{1}{\sqrt{2}}=\sqrt{2} \sin 90^{\circ}$
$B =\sqrt{2}\left[\frac{1}{\sqrt{2}} \sin 44^{\circ}+\frac{1}{\sqrt{2}} \cos 44^{\circ}\right]$
$=\sqrt{2} \sin \left(45^{\circ}+44^{\circ}\right)$
$=\sqrt{2} \sin 89^{\circ}< \sqrt{2} \sin 90^{\circ}=\sqrt{2}$
$\therefore A > B$