3.468 Given: $\sin \alpha-\cos \alpha=$. Find $A=\frac{\sin 4 \alpha+\sin 10 \alpha-\sin 6 \alpha}{\cos 2 \alpha+1-2 \sin ^{2} 4 \alpha}$.
Solution. $A=\frac{(\sin 4 \alpha+\sin 10 \alpha)-\sin 6 \alpha}{\cos 2 \alpha+\cos 8 \alpha}=\frac{2 \sin 7 \alpha \cdot \cos 3 \alpha-2 \sin 3 \alpha \cdot \cos 3 \alpha}{2 \cos 3 \alpha \cdot \cos 5 \alpha}=$

$=\frac{\sin 7 \alpha-\sin 3 \alpha}{\cos 5 \alpha}=2 \sin 2 \alpha . \quad(\sin \alpha-\cos \alpha)^{2}=m^{2} \Rightarrow 1-m^{2}=\sin 2 \alpha \Rightarrow$

$\Rightarrow A=2\left(1-m^{2}\right)$.

Answer: $A=2\left(1-m^{2}\right)$.