1. ## Maximum value

Find the maximum value for A :
$A = xy + x\sqrt {1 - y^2 } + y\sqrt {1 - x^2 } - \sqrt {\left( {1 - x^2 } \right)\left( {1 - y^2 } \right)}
$

2. Originally Posted by dhiab
Find the maximum value for A :
$A = xy + x\sqrt {1 - y^2 } + y\sqrt {1 - x^2 } - \sqrt {\left( {1 - x^2 } \right)\left( {1 - y^2 } \right)}
$
Let $x=\sin\theta,y=\sin\phi$,then $A=\sin\theta\sin\phi+\sin\theta\cos\phi+\cos\theta \sin\phi-\cos\theta\cos\phi=$ $\sin (x+y)+\cos (x+y)=\sqrt{2}\sin (x+y+\frac{\pi}{4})\leq \sqrt{2}$

3. Originally Posted by ynj
Let $x=\sin\theta,y=\sin\phi$,then $A=\sin\theta\sin\phi+\sin\theta\cos\phi+\cos\theta \sin\phi-\cos\theta\cos\phi=$ $\sin (x+y)+\cos (x+y)=\sqrt{2}\sin (x+y+\frac{\pi}{4})\leq \sqrt{2}$
Hello :
Remark
$\sqrt {1 - \sin ^2 \theta } = \sqrt {\cos ^2 \theta } = \left| {\cos \theta } \right|
$

I'think you have another cases

4. Originally Posted by dhiab
Hello :
Remark
$\sqrt {1 - \sin ^2 \theta } = \sqrt {\cos ^2 \theta } = \left| {\cos \theta } \right|
$

I'think you have another cases
yeah, but you can assume $x,y>0$, since $A(x,y)\leq A(|x|,|y|)$!!!