Question

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

• A. $A>B$
• B. $A<B$
• C. $A=B$
• D. none of these

Answer 1

Answer: (A)

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