Қалалық Жәутіков олимпиадасы 9 сынып, 2004 жыл

Теңдікті дәлелдеңдер: $\cos \dfrac{\pi }{20}\cdot \cos \dfrac{3\pi }{20}\cdot \cos \dfrac{7\pi }{20}\cdot \cos \dfrac{9\pi }{20}+\cos \dfrac{\pi }{15}\cdot \cos \dfrac{2\pi }{15}\cdot \cos \dfrac{4\pi }{15}\cdot \cos \dfrac{8\pi }{15}=0.$

Комментарий/решение

Көбейтіндіні қосындыға түрлендіру формуласы бойынша: $\cos\left ( \frac{\pi }{20} \right )\cos\left ( \frac{9\pi }{20} \right )=\frac{1}{2}\left ( \cos\left ( \frac{\pi }{2} \right )+\cos\left ( \frac{2\pi}{5} \right ) \right )=\frac{1}{2}\cos\left ( \frac{2\pi }{5}\right ).$

$\cos\left ( \frac{3\pi }{20} \right )\cos\left ( \frac{7\pi }{20} \right )=\frac{1}{2}\left ( \cos\left ( \frac{\pi }{2} \right )+\cos\left ( \frac{\pi}{5} \right ) \right )=\frac{1}{2}\cos\left ( \frac{\pi }{5}\right ).$

Сонда $\frac{1}{2}\cos\left ( \frac{\pi }{5}\right )\cdot \frac{1}{2}\cos\left ( \frac{2\pi }{5} \right )=\frac{1}{4}\cos\left ( \frac{\pi }{5}\right )\cdot \cos\left ( \frac{2\pi }{5}\right ).$

$A=\cos 36^{\circ}, B=\cos 72^{\circ}$ деп белгілейік.

$A=\cos 36^{\circ}=\cos \left ( 90^{\circ}-54^{\circ} \right )=\sin 54^{\circ}, B=\cos 72^{\circ}=\cos \left ( 90^{\circ}-18^{\circ} \right )=\sin 18^{\circ} .$

$A-B=\cos 36^{\circ}-\cos72^{\circ}=2\sin 54^{\circ}\sin 18^{\circ}=2AB.$

Көбейтіндісін қарастырайық: $AB=\sin 54^{\circ}\sin 18^{\circ}=\frac{\sin 18^{\circ}\sin 36^{\circ}\sin 54^{\circ}\sin 72^{\circ}}{\sin 36^{\circ}\sin 72^{\circ}}=\frac{\sin 18^{\circ}\sin 36^{\circ}\cos 36^{\circ}\cos 18^{\circ}}{\sin 36^{\circ}\sin 72^{\circ}}.$

$\sin 18^{\circ}\cos 18^{\circ}=\frac{\sin 36^{\circ}}{2}, \cos 36^{\circ}\sin 36^{\circ}=\frac{\sin 72^{\circ}}{2}.$

Онда $AB=\frac{\sin 36^{\circ}\sin 72^{\circ}}{2\cdot 2\cdot\sin 36^{\circ}\sin 72^{\circ}}=\frac{1}{4}.$

Теңдеулер жүйесін шешсек:

$A-B=2AB, AB=\frac{1}{4}, A>0, B>0$ Оң түбірі $A=\frac{1+\sqrt{5}}{4}$. Онда $B=\frac{\sqrt{5}-1}{4}.$

$\cos 36^{\circ}=\frac{1+\sqrt{5}}{4}, \cos 72^{\circ}=\frac{\sqrt{5}-1}{4}.$

Бірінші қосылғыш: $\frac{1}{4}\cdot \frac{1+\sqrt{5}}{4}\cdot \frac{\sqrt{5}-1}{4}=\frac{1}{16}.$

$\cos\left ( \frac{\pi }{15} \right )\cos\left ( \frac{4\pi }{15} \right )=\frac{1}{2}\left ( \cos\left ( \frac{\pi }{3} \right )+\cos\left ( \frac{\pi}{5} \right ) \right )=\frac{1}{2}\left ( \frac{1}{2}+\frac{\sqrt{5}+1}{4} \right )=\frac{1}{4}+\frac{\sqrt{5}+1}{8}.$

$\cos\left ( \frac{2\pi }{15} \right )\cos\left ( \frac{8\pi }{15} \right )=\frac{1}{2}\left ( \cos\left ( \frac{2\pi }{3} \right )+\cos\left ( \frac{2\pi}{5} \right ) \right )=\frac{1}{2}\left ( -\frac{1}{2}+\frac{\sqrt{5}-1}{4} \right )=-\frac{1}{4}+\frac{\sqrt{5}-1}{8}.$

Екінші қосылғыш: $\left ( \frac{1}{4}+\frac{\sqrt{5}+1}{8}\right )\left (-\frac{1}{4}+\frac{\sqrt{5}-1}{8}\right )=-\frac{1}{16}+\frac{\sqrt{5}-1}{32}-\frac{\sqrt{5}+1}{32}+\frac{4}{64}=-\frac{1}{16}-\frac{1}{16}+\frac{1}{16}=-\frac{1}{16}.$

$\frac{1}{16}-\frac{1}{16}=0.$

$$\sin\alpha \cos\alpha=\frac{\sin2\alpha}{2}\Leftrightarrow 2^{n}\sin\alpha \cos\alpha\cos2\alpha...\cos 2^{n-1}\alpha=\sin2^n\alpha,$$

$$\sin\frac{\pi}{15}\cos\frac{\pi}{20}...+\sin\frac{\pi}{15}\cos\frac{\pi}{15}...=\sin\frac{\pi}{15}\cos\frac{\pi}{20}...+\frac{\sin\frac{16\pi}{15}}{16}=\sin\frac{\pi}{15}(\cos\frac{\pi}{20}...-\frac{1}{16}),$$

$$16\cos\frac{\pi}{20}\cos\frac{3\pi}{20}\cos\frac{7\pi}{20}\cos\frac{9\pi}{20}\stackrel{?}{=}1,$$

$$4\cos\frac{\pi}{20}\cos\frac{9\pi}{20}\sin\frac{\pi}{20}\sin\frac{9\pi}{20}=\sin\frac{2\pi}{20}\sin\frac{18\pi}{20}=\sin^2\frac{\pi}{10}(A),$$

$$4\cos\frac{3\pi}{20}\cos\frac{7\pi}{20}\sin\frac{3\pi}{20}\sin\frac{7\pi}{20}=\sin^2\frac{3\pi}{10}(B),$$

$$\sin({\frac{\pi}{2}-\alpha})=\cos\alpha,$$

$(A)\cdot(B)\stackrel{?}{=}\frac{1}{16}$, что подтверждается надежным источником.