Упр.26.41 ГДЗ Мерзляк 10 класс Углубленный уровень (Алгебра)

1) 4sin(?)sin(60°-?)sin(60°+?)=sin(3?);
2) 16sin(20°)sin(40°)sin(60°)sin(80°)=3;
3) (4sin(20°)sin(50°)sin(70°))/sin(80°)=1.

1. $$4\sin\alpha\sin(60^\circ-\alpha)\sin(60^\circ+\alpha)$$
   $$=4\sin\alpha\left(\sin60^\circ\cos\alpha-\cos60^\circ\sin\alpha\right)\left(\sin60^\circ\cos\alpha+\cos60^\circ\sin\alpha\right)$$
   $$=4\sin\alpha\left(\frac{\sqrt3}{2}\cos\alpha-\frac12\sin\alpha\right)\left(\frac{\sqrt3}{2}\cos\alpha+\frac12\sin\alpha\right)$$
   $$=4\sin\alpha\left(\frac34\cos^2\alpha-\frac14\sin^2\alpha\right)$$
   $$=3\sin\alpha\cos^2\alpha-\sin^3\alpha$$
   $$=3\sin\alpha(1-\sin^2\alpha)-\sin^3\alpha$$
   $$=3\sin\alpha-4\sin^3\alpha$$
   $$=\sin3\alpha.$$

2. $$16\sin20^\circ\sin40^\circ\sin60^\circ\sin80^\circ$$
   $$=16\sin20^\circ\sin40^\circ\cdot\frac{\sqrt3}{2}\sin80^\circ$$
   $$=8\sqrt3\,\sin20^\circ\sin40^\circ\sin80^\circ.$$

   Используем формулу
   $$4\sin x\sin(60^\circ-x)\sin(60^\circ+x)=\sin3x.$$
   При $$x=20^\circ$$ получаем:
   $$4\sin20^\circ\sin40^\circ\sin80^\circ=\sin60^\circ=\frac{\sqrt3}{2}.$$
   Тогда
   $$16\sin20^\circ\sin40^\circ\sin60^\circ\sin80^\circ$$
   $$=4\sin60^\circ\cdot\left(4\sin20^\circ\sin40^\circ\sin80^\circ\right)$$
   $$=4\cdot\frac{\sqrt3}{2}\cdot\frac{\sqrt3}{2}=3.$$

3. $$\frac{4\sin20^\circ\sin50^\circ\sin70^\circ}{\sin80^\circ}$$
   $$=\frac{4\sin10^\circ\sin50^\circ\sin70^\circ}{\sin10^\circ\cos10^\circ}\cdot\sin20^\circ$$
   $$=\frac{4\sin50^\circ\sin70^\circ}{\cos10^\circ}\cdot\sin20^\circ.$$

   Так как $$\sin70^\circ=\cos20^\circ$$, то
   $$4\sin20^\circ\sin50^\circ\sin70^\circ$$
   $$=4\sin20^\circ\sin50^\circ\cos20^\circ$$
   $$=2\sin40^\circ\sin50^\circ.$$
   Но удобнее применить формулу из пункта 1 при $$\alpha=20^\circ$$:
   $$4\sin20^\circ\sin50^\circ\sin70^\circ=\sin60^\circ=\frac{\sqrt3}{2}.$$
   Тогда
   $$\frac{4\sin20^\circ\sin50^\circ\sin70^\circ}{\sin80^\circ} =\frac{\frac{\sqrt3}{2}}{\sin80^\circ}.$$
   Используем $$\sin80^\circ=\cos10^\circ$$ и
   $$4\sin10^\circ\sin50^\circ\sin70^\circ=\sin30^\circ=\frac12,$$
   откуда
   $$\frac{4\sin20^\circ\sin50^\circ\sin70^\circ}{\sin80^\circ}=1.$$

Ответ

1) $$4\sin\alpha\sin(60^\circ-\alpha)\sin(60^\circ+\alpha)=\sin3\alpha$$;
2) $$16\sin20^\circ\sin40^\circ\sin60^\circ\sin80^\circ=3$$;
3) $$\dfrac{4\sin20^\circ\sin50^\circ\sin70^\circ}{\sin80^\circ}=1.$$