Упр.684 ГДЗ Никольский Потапов 9 класс (Алгебра)

б) cos 13пи/24•cos 7пи/24
в) sin 7пи/24•cos пи/24
г) cos 7пи/20•cos 3пи/20-sinпи/15•sin 4пи/15
д)cos 11пи/56•cos 3пи/56-sin 11пи/42•sin 17пи/42

Воспользуемся формулами произведения синусов и косинусов:

$$\sin x \sin y=\frac12(\cos(x-y)-\cos(x+y)),$$
$$\cos x \cos y=\frac12(\cos(x+y)+\cos(x-y)),$$
$$\sin x \cos y=\frac12(\sin(x+y)+\sin(x-y)).$$

1. $$\sin\frac{11\pi}{24}\cdot\sin\frac{5\pi}{24}=\frac12\left(\cos\frac{6\pi}{24}-\cos\frac{16\pi}{24}\right)$$
   $$=\frac12\left(\cos\frac{\pi}{4}-\cos\frac{2\pi}{3}\right) =\frac12\left(\frac{\sqrt2}{2}+\frac12\right) =\frac{\sqrt2+1}{4}.$$

2. $$\cos\frac{13\pi}{24}\cdot\cos\frac{7\pi}{24}=\frac12\left(\cos\frac{20\pi}{24}+\cos\frac{6\pi}{24}\right)$$
   $$=\frac12\left(\cos\frac{5\pi}{6}+\cos\frac{\pi}{4}\right) =\frac12\left(-\frac{\sqrt3}{2}+\frac{\sqrt2}{2}\right) =\frac{\sqrt2-\sqrt3}{4}.$$

3. $$\sin\frac{7\pi}{24}\cdot\cos\frac{\pi}{24}=\frac12\left(\sin\frac{8\pi}{24}+\sin\frac{6\pi}{24}\right)$$
   $$=\frac12\left(\sin\frac{\pi}{3}+\sin\frac{\pi}{4}\right) =\frac12\left(\frac{\sqrt3}{2}+\frac{\sqrt2}{2}\right) =\frac{\sqrt3+\sqrt2}{4}.$$

4. $$\cos\frac{7\pi}{20}\cdot\cos\frac{3\pi}{20}-\sin\frac{\pi}{15}\cdot\sin\frac{4\pi}{15}$$
   $$=\frac12\left(\cos\frac{10\pi}{20}+\cos\frac{4\pi}{20}\right)-\frac12\left(\cos\left(-\frac{3\pi}{15}\right)-\cos\frac{5\pi}{15}\right)$$
   $$=\frac12\left(\cos\frac{\pi}{2}+\cos\frac{\pi}{5}\right)-\frac12\left(\cos\frac{\pi}{5}-\cos\frac{\pi}{3}\right)$$
   $$=\frac12\cdot 0+\frac12\cos\frac{\pi}{5}-\frac12\cos\frac{\pi}{5}+\frac12\cdot\frac12=\frac14.$$

5. $$\cos\frac{11\pi}{56}\cdot\cos\frac{3\pi}{56}-\sin\frac{11\pi}{42}\cdot\sin\frac{17\pi}{42}$$
   $$=\frac12\left(\cos\frac{14\pi}{56}+\cos\frac{8\pi}{56}\right)-\frac12\left(\cos\left(-\frac{6\pi}{42}\right)-\cos\frac{28\pi}{42}\right)$$
   $$=\frac12\left(\cos\frac{\pi}{4}+\cos\frac{\pi}{7}\right)-\frac12\left(\cos\frac{\pi}{7}-\cos\frac{2\pi}{3}\right)$$
   $$=\frac12\cdot\frac{\sqrt2}{2}+\frac12\cos\frac{\pi}{7}-\frac12\cos\frac{\pi}{7}+\frac12\cdot\frac12 =\frac{\sqrt2-1}{4}.$$

Ответ

$$\text{а) } \frac{\sqrt2+1}{4};\quad \text{б) } \frac{\sqrt2-\sqrt3}{4};\quad \text{в) } \frac{\sqrt3+\sqrt2}{4};\quad \text{г) } \frac14;\quad \text{д) } \frac{\sqrt2-1}{4}.$$