Bài tập 1 trang 19 SBT Toán 11 Tập 1 Chân trời sáng tạo - CTST

a) Ta có: $\text{sin}\frac{19\pi }{24}\text{cos}\frac{37\pi }{24} =\frac{1}{2}\left[\text{sin}\left(\frac{19\pi }{24}-\frac{37\pi }{24}\right)+\text{sin}\left(\frac{19\pi }{24}+\frac{37\pi }{24}\right)\right]$

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

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

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

$=2\text{sin}\frac{\pi }{4}\text{sin}\frac{\pi }{6}=2\cdot \frac{\sqrt{2}}{2}\cdot \frac{1}{2}=\frac{\sqrt{2}}{2}.$

c) Ta có: $\frac{\text{tan}\frac{\pi }{7}+\text{tan}\frac{3\pi }{28}}{1+\text{tan}\frac{6\pi }{7}\text{tan}\frac{3\pi }{28}}=\frac{\text{tan}\frac{\pi }{7}+\text{tan}\frac{3\pi }{28}}{1+\text{tan}\left(\pi -\frac{\pi }{7}\right)\text{tan}\frac{3\pi }{28}}=\frac{\text{tan}\frac{\pi }{7}+\text{tan}\frac{3\pi }{28}}{1-\text{tan}\frac{\pi }{7}\text{tan}\frac{3\pi }{28}}$

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

-- Mod Toán 11 HỌC247