Если cos15°-sin15°=,то найдите A.там это 4cos15°

Решите задачу:

$\cos15^{\circ}-\sin15^{\circ}=\frac{\alpha}{4\cos15^{\circ}} \\ \\ \alpha =4\cos15^{\circ}(\cos15^{\circ}-\sin15^{\circ})=2\cdot(2\cos^2 15^{\circ}-2\sin15^{\circ}\cos15^{\circ})=\\\\=2\cdot(1+\cos(2\cdot15^{\circ})-\sin(2\cdot15^{\circ}))=2\cdot(1+\cos30^{\circ}-\sin30^{\circ})=\\\\=2\cdot(1+\frac{\sqrt{3}}{2}-\frac{1}{2})=2\cdot(\frac{1}{2}+\frac{\sqrt{3}}{2})=1+\sqrt{3}$