Айниятро исбот намоед:
\(\cos{\alpha}+\sin{\alpha}+\cos{3\alpha}+\sin{3\alpha}=2\sqrt{2}\cos{\alpha}\sin{(\frac{\pi}{4}+2\alpha)}\)
A\( = \cos{\alpha}+\sin{\alpha}+\cos{3\alpha}+\sin{3\alpha}\)
\(\cos{\alpha}+\cos{3\alpha}=2\cos{\frac{\alpha+3\alpha}{2}}\cos{\frac{\alpha-3\alpha}{2}=}\)
\(=2\cos{2\alpha}\cos{\alpha}\)
\(\sin{\alpha}+\sin{3\alpha}=2\sin{\frac{\alpha+3\alpha}{2}}\cos{\frac{\alpha-3\alpha}{2}=}\)
\(=2\sin{2\alpha}\cos{\alpha}\)
A\( = 2\cos{2\alpha}\cos{\alpha}+2\sin{2\alpha}\cos{\alpha}=2\cos{\alpha}(\cos{2\alpha}+\sin{2\alpha})\)
\(\cos{2\alpha}+\sin{2\alpha}=\sqrt{2}(\frac{\sqrt{2}}{2}\cos{2\alpha}+\frac{\sqrt{2}}{2}\sin{2\alpha})\)
\(\frac{\sqrt{2}}{2}\cos{2\alpha}+\frac{\sqrt{2}}{2}\sin{2\alpha}=\sin{\frac{\pi}{4}}\cos{2\alpha}+\cos{\frac{\pi}{4}}\sin{2\alpha}=\)
\(=\sin{(\frac{\pi}{4}+2\alpha)}\)
\(\cos{2\alpha}+\sin{2\alpha}=\sqrt{2}\sin{(\frac{\pi}{4}+2\alpha)}\)
A\(=2\cos{\alpha}\cdot\sqrt{2}\sin{(\frac{\pi}{4}+2\alpha)}=\)
\(=2\sqrt{2}\cos{\alpha}\sin{(\frac{\pi}{4}+2\alpha)}\)
\(\cos{\alpha}+\sin{\alpha}+\cos{3\alpha}+\sin{3\alpha}=2\sqrt{2}\cos{\alpha}\sin{(\frac{\pi}{4}+2\alpha)}\)
Айният исбот шуд.