Giải bài tập 1.13 trang 11 SBT toán 11 tập 1 kết nối

a) $A=sin\frac{\pi}{9}-sin\frac{5\pi}{9}+sin\frac{7\pi}{9}$

$=(sin\frac{\pi}{9}+sin\frac{7\pi}{9})-sin\frac{5\pi}{9}$

$=2sin\frac{\frac{\pi}{9}+\frac{7\pi}{9}}{2}.cos\frac{\frac{\pi}{9}-\frac{7\pi}{9}}{2}-sin\frac{5\pi}{9}$

$=2sin\frac{4\pi}{9}cos\frac{\pi}{3}-sin\frac{5\pi}{9}$

$=sin\frac{4\pi}{9}-sin\frac{5\pi}{9}$

$=sin(\pi-\frac{4\pi}{9})-sin\frac{5\pi}{9}$

$=sin\frac{5\pi}{9}-sin\frac{5\pi}{9}=0$

b) Vì $sin78^{o}=cos12^{o};sin66^{o}=cos24^{o};sin42^{o}=cos48^{o}$

nên $B=sin6^{o}cos12^{o}cos24^{o}cos48^{o}$

Ta có: $cos6^{o}.B=cos6^{o}. sin6^{o}cos12^{o}cos24^{o}cos48^{o}$

$=\frac{1}{2}sin12^{o}cos12^{o}cos24^{o}cos48^{o}$

$=\frac{1}{4}sin24^{o}cos24^{o}cos48^{o}$

$=\frac{1}{8}sin48^{o}cos48^{o}$

$=\frac{1}{16}sin96^{o}$

$=\frac{1}{16}sin(90^{o}+6^{o})$

$=\frac{1}{16}cos6^{o}$

Suy ra $B=\frac{1}{16}$