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$A=\sin^22A+\sin^22B+\sin^22C=\dfrac{1-\cos4A}{2}+\dfrac{1-\cos4B}{2}+\dfrac{1-\cos4C}{2}=\dfrac{3}{2}-\dfrac{1}{2}(\cos4A+\cos4B+\cos4C)$


$\cos4A+\cos4B+\cos4C=2.cos(2A+2B).cos(2A-2B)+2.\cos^22C-1$

$=2.\cos2C.\cos(2A-2B)+2.\cos^22C-1=2.\cos2C.[\cos(2A-2B)+\cos(2A+2B)]-1$

$=4.\cos2A.\cos2B.\cos2C-1$


\Rightarrow $A=2-2..\cos2A.\cos2B.\cos2C$
 
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