Айниятро исбот намоед:
\(\sin{9\alpha}+\sin{10\alpha}+\sin{11\alpha}+\sin{12\alpha}=4\cos{\frac{\alpha}{2}}\cos{\alpha}\sin{\frac{21\alpha}{2}}\)
\(\sin{9\alpha}+\sin{11\alpha}=2\sin{\frac{9+11\alpha}{2}}\cos{\frac{9-11\alpha}{2}}=2\sin{10\alpha}\cos{\alpha}\)
\(\sin{10\alpha}+\sin{12\alpha}=2\sin{\frac{10+12\alpha}{2}}\cos{\frac{10-12\alpha}{2}}=2\sin{12\alpha}\cos{\alpha}\)
\(\sin{9\alpha}+\sin{10\alpha}+\sin{11\alpha}+\sin{12\alpha}=2\sin{10\alpha}\cos{\alpha}+2\sin{11\alpha}\cos{\alpha}=\)
\(=2\cos{\alpha}(\sin{10\alpha}+\sin{11\alpha})\)
\(\sin{10\alpha}+\sin{11\alpha}=2\sin{\frac{10+11\alpha}{2}}\cos{\frac{10-11\alpha}{2}}=2\sin{\frac{21\alpha}{2}}\cos{\frac{\alpha}{2}}\)
\(\sin{9\alpha}+\sin{10\alpha}+\sin{11\alpha}+\sin{12\alpha}=4\cos{\frac{\alpha}{2}}\cos{\alpha}\sin{\frac{21\alpha}{2}}\)
Айният исбот шуд.