Bài tập 2 trang 19 SBT Toán 11 Tập 1 Chân trời sáng tạo - CTST

a) Vì $-\frac{\pi }{2}<\alpha <0$ nên sinα < 0.

Do đó, $\text{sin}\alpha =-\sqrt{1-{\text{cos}}^2\alpha }=-\sqrt{1-{\left(\frac{11}{61}\right)}^2}=-\frac{60}{61}$.

Suy ra $\text{sin}\left(\frac{\pi }{6}-\alpha \right)$ = $\text{sin}\frac{\pi }{6}\text{cos}\alpha -\text{cos}\frac{\pi }{6}\text{sin}\alpha$ = $\frac{1}{2}\cdot \frac{11}{61}-\frac{\sqrt{3}}{2}\cdot \left(-\frac{60}{61}\right)$ = $\frac{11+60\sqrt{3}}{122}\text{.}$

b) Ta có: $\text{tan}\alpha =\frac{\text{sin}\alpha }{\text{cos}\alpha }=\frac{-\frac{60}{61}}{\frac{11}{61}}=-\frac{60}{11}.$

Do đó $\text{cot}\left(\alpha +\frac{\pi }{4}\right)=\frac{1}{\text{tan}\left(\alpha +\frac{\pi }{4}\right)}=\frac{1-\text{tan}\alpha \text{tan}\frac{\pi }{4}}{\text{tan}\alpha +\text{tan}\frac{\pi }{4}}=\frac{1-\left(-\frac{60}{11}\right)\cdot 1}{\left(-\frac{60}{11}\right)+1}=-\frac{71}{49}.$

c) Ta có: $\text{cos}2\alpha =2{\text{cos}}^2\alpha -1=2\cdot {\left(\frac{11}{61}\right)}^2-1=-\frac{3479}{3721}$

$\text{sin}2\alpha =2\text{sin}\alpha \text{cos}\alpha =2\cdot \left(-\frac{60}{61}\right)\cdot \frac{11}{61}=-\frac{1320}{3721}.$

Suy ra: $\text{cos}\left(2\alpha +\frac{\pi }{3}\right)$ = $\text{cos}2\alpha \text{cos}\frac{\pi }{3}-\text{sin}2\alpha \text{sin}\frac{\pi }{3}$ = $-\frac{3479}{3721}\cdot \frac{1}{2}-\left(-\frac{1320}{3721}\right)\cdot \frac{\sqrt{3}}{2}$

$=\frac{-3479+1320\sqrt{3}}{7442}$

d) Ta có: $\text{tan}2\alpha =\frac{\text{sin}2\alpha }{\text{cos}2\alpha }=\frac{-\frac{1320}{3721}}{-\frac{3479}{3721}}=\frac{1320}{3479}$.

Suy ra: $\text{tan}\left(\frac{3\pi }{4}-2\alpha \right)=\frac{\text{tan}\frac{3\pi }{4}-\text{tan}2\alpha }{1+\text{tan}\frac{3\pi }{4}\text{tan}2\alpha }=\frac{-1-\frac{1320}{3479}}{1+\left(-1\right)\cdot \frac{1320}{3479}}=-\frac{4799}{2159}$.

-- Mod Toán 11 HỌC247