Bài 1.10 trang 21 Toán 11 Tập 1 | Kết nối tri thức Giải Toán 11

Giải Toán 11 Bài 2: Công thức lượng giác 

Bài 1.10 trang 21 Toán 11 Tập 1

a) $A=\frac{\sin \frac{\pi }{15}\cos \frac{\pi }{10}+\sin \frac{\pi }{10}\cos \frac{\pi }{15}}{\cos \frac{2\pi }{15}\cos \frac{\pi }{5}-\sin \frac{2\pi }{15}\sin \frac{\pi }{5}}$;

b) $B=\sin \frac{\pi }{32}\cos \frac{\pi }{32}\cos \frac{\pi }{16}\cos \frac{\pi }{8}$.

Lời giải:

a) Ta có:

$A=\frac{\sin \frac{\pi }{15}\cos \frac{\pi }{10}+\sin \frac{\pi }{10}\cos \frac{\pi }{15}}{\cos \frac{2\pi }{15}\cos \frac{\pi }{5}-\sin \frac{2\pi }{15}\sin \frac{\pi }{5}}$$=\frac{\sin \frac{\pi }{15}\cos \frac{\pi }{10}+\cos \frac{\pi }{15}\sin \frac{\pi }{10}}{\cos \frac{2\pi }{15}\cos \frac{\pi }{5}-\sin \frac{2\pi }{15}\sin \frac{\pi }{5}}$

$=\frac{\sin \left(\frac{\pi }{15}+\frac{\pi }{10}\right)}{\cos \left(\frac{2\pi }{15}+\frac{\pi }{5}\right)}$$=\frac{\sin \frac{\pi }{6}}{\cos \frac{\pi }{3}}=\frac{\frac{1}{2}}{\frac{1}{2}}=1$.

b) Ta có:

$B=\sin \frac{\pi }{32}\cos \frac{\pi }{32}\cos \frac{\pi }{16}\cos \frac{\pi }{8}$$=\left(\frac{1}{2}.2\sin \frac{\pi }{32}\cos \frac{\pi }{32}\right)\cos \frac{\pi }{16}\cos \frac{\pi }{8}$

$=\frac{1}{2}\sin \left(2.\frac{\pi }{32}\right)\cos \frac{\pi }{16}\cos \frac{\pi }{8}$$=\frac{1}{2}\sin \frac{\pi }{16}\cos \frac{\pi }{16}\cos \frac{\pi }{8}$

$=\frac{1}{4}.2\sin \frac{\pi }{16}\cos \frac{\pi }{16}\cos \frac{\pi }{8}$$=\frac{1}{4}\sin \frac{\pi }{8}\cos \frac{\pi }{8}=\frac{1}{8}.2\sin \frac{\pi }{8}\cos \frac{\pi }{8}$

$=\frac{1}{8}\sin \frac{\pi }{4}=\frac{1}{8}.\frac{\sqrt{2}}{2}=\frac{\sqrt{2}}{16}$.