Упр.1423 ГДЗ Макарычев Миндюк 9 класс (Углубленный) (Алгебра)

а) 2sin(?/6+?)-v3sin(?); в) v2sin(?)-2sin(?-?/4);
б) 1/2cos(?)-cos(?/3+?); г) cos(?-?/6)-v3/2cos(?).

1. а)

   $$2\sin\left(\frac{\pi}{6}+a\right)-\sqrt{3}\sin a$$
   $$=2\left(\sin\frac{\pi}{6}\cos a+\cos\frac{\pi}{6}\sin a\right)-\sqrt{3}\sin a$$
   $$=2\left(\frac12\cos a+\frac{\sqrt{3}}{2}\sin a\right)-\sqrt{3}\sin a$$
   $$=\cos a+\sqrt{3}\sin a-\sqrt{3}\sin a=\cos a.$$

2. б)

   $$\frac12\cos a-\cos\left(\frac{\pi}{3}+a\right)$$
   $$=\frac12\cos a-\left(\cos\frac{\pi}{3}\cos a-\sin\frac{\pi}{3}\sin a\right)$$
   $$=\frac12\cos a-\left(\frac12\cos a-\frac{\sqrt{3}}{2}\sin a\right)$$
   $$=\frac12\cos a-\frac12\cos a+\frac{\sqrt{3}}{2}\sin a=\frac{\sqrt{3}}{2}\sin a.$$

3. в)

   $$\sqrt{2}\sin a-2\sin\left(a-\frac{\pi}{4}\right)$$
   $$=\sqrt{2}\sin a-2\left(\sin a\cos\frac{\pi}{4}-\cos a\sin\frac{\pi}{4}\right)$$
   $$=\sqrt{2}\sin a-2\left(\frac{\sqrt{2}}{2}\sin a-\frac{\sqrt{2}}{2}\cos a\right)$$
   $$=\sqrt{2}\sin a-\sqrt{2}\sin a+\sqrt{2}\cos a=\sqrt{2}\cos a.$$

4. г)

   $$\cos\left(a-\frac{\pi}{6}\right)-\frac{\sqrt{3}}{2}\cos a$$
   $$=\left(\cos a\cos\frac{\pi}{6}+\sin a\sin\frac{\pi}{6}\right)-\frac{\sqrt{3}}{2}\cos a$$
   $$=\left(\frac{\sqrt{3}}{2}\cos a+\frac12\sin a\right)-\frac{\sqrt{3}}{2}\cos a$$
   $$=\frac12\sin a.$$

Ответ

а) $$\cos a$$; б) $$\frac{\sqrt{3}}{2}\sin a$$; в) $$\sqrt{2}\cos a$$; г) $$\frac12\sin a$$.