Giải bài 1 trang 19 sách bài tập toán 11 - Chân trời sáng tạo tập 1 — Không quảng cáo

Đề bài

Không dùng máy tính cầm tay. Tính giá trị của các biểu thức sau:

a) $\sin \frac{{19\pi }}{{24}}\cos \frac{{37\pi }}{{24}}$;

b) $\cos \frac{{41\pi }}{{12}} - \cos \frac{{13\pi }}{{12}}$;

c) $\frac{{\tan \frac{\pi }{7} + \tan \frac{{3\pi }}{{28}}}}{{1 + \tan \frac{{6\pi }}{7}\tan \frac{{3\pi }}{{28}}}}$.

Lời giải chi tiết

a) $\sin \frac{{19\pi }}{{24}}\cos \frac{{37\pi }}{{24}}$ $= \frac{1}{2}\left[ {\sin \left( {\frac{{19\pi }}{{24}} + \frac{{37\pi }}{{24}}} \right) + \sin \left( {\frac{{19\pi }}{{24}} - \frac{{37\pi }}{{24}}} \right)} \right]$ $= \frac{1}{2}\left( {\sin \frac{{7\pi }}{3} + \sin \frac{{ - 3\pi }}{4}} \right)$

$= \frac{1}{2}\left( {\sin \frac{{7\pi }}{3} - \sin \frac{{3\pi }}{4}} \right)$ $= \frac{1}{2}\left( {\sin \left( {2\pi  + \frac{\pi }{3}} \right) - \sin \frac{{3\pi }}{4}} \right)$ $= \frac{1}{2}\left( {\sin \frac{\pi }{3} - \sin \frac{{3\pi }}{4}} \right)$

$= \frac{1}{2}\left( {\frac{{\sqrt 3 }}{2} - \frac{{\sqrt 2 }}{2}} \right)$ $= \frac{{\sqrt 3  - \sqrt 2 }}{4}$

b) $\cos \frac{{41\pi }}{{12}} - \cos \frac{{13\pi }}{{12}}$ $=  - 2\sin \frac{{\frac{{41\pi }}{{12}} + \frac{{13\pi }}{{12}}}}{2}\sin \frac{{\frac{{41\pi }}{{12}} - \frac{{13\pi }}{{12}}}}{2}$ $=  - 2\sin \frac{{9\pi }}{4}\sin \frac{{7\pi }}{6}$

$=  - 2\sin \left( {2\pi  + \frac{\pi }{4}} \right)\sin \left( {\pi  + \frac{\pi }{6}} \right)$ $= 2\sin \frac{\pi }{4}\sin \frac{\pi }{6}$ $= 2.\frac{{\sqrt 2 }}{2}.\frac{1}{2}$ $= \frac{{\sqrt 2 }}{2}$

c) $\frac{{\tan \frac{\pi }{7} + \tan \frac{{3\pi }}{{28}}}}{{1 + \tan \frac{{6\pi }}{7}\tan \frac{{3\pi }}{{28}}}}$ $= \frac{{\tan \frac{\pi }{7} + \tan \frac{{3\pi }}{{28}}}}{{1 + \tan \left( {\pi  - \frac{\pi }{7}} \right)\tan \frac{{3\pi }}{{28}}}}$ $= \frac{{\tan \frac{\pi }{7} + \tan \frac{{3\pi }}{{28}}}}{{1 - \tan \frac{\pi }{7}\tan \frac{{3\pi }}{{28}}}}$ $= \tan \left( {\frac{\pi }{7} + \frac{{3\pi }}{{28}}} \right)$ $= \tan \frac{\pi }{4}$ $= 1$.