Giải Bài tập 1.9 trang 21 sgk Toán 11 tập 1 Kết nối

a) Vì $\frac{\pi }{2}<a<\pi $ suy ra cosa < 0

Ta có: $sin^{2}a+cos^{2}a=1\Rightarrow cosa=-\sqrt{1-sin^{2}a}=-\sqrt{1-\frac{1}{9}}=-\frac{2\sqrt{2}}{3}$

$\Rightarrow tana=\frac{sina}{cosa}=2\sqrt{2}$

$sin2a=2sinacosa=2\times \frac{1}{3}\times (-\frac{2\sqrt{2}}{3})=\frac{-4\sqrt{2}}{9}$

$cos2a=cos^{2}a-sin^{2}a=\frac{8}{9}-\frac{1}{9}=\frac{7}{9}$

$tan2a=\frac{2tana}{1-tan{2}a}=\frac{2\times 2\sqrt{2}}{1-(2\sqrt{2})^{2}}=\frac{-4\sqrt{2}}{7}$

b) $sina+cosa=\frac{1}{2}\Rightarrow (sina+cosa)^{2}=\frac{1}{4}$

$\Rightarrow sin^{2}a+cos^{2}a+2sinacosa=\frac{1}{4}\Rightarrow sin2a=\frac{1}{4}-1=\frac{-3}{4}$

Vì $\frac{\pi }{2}<a<\frac{3\pi }{4}\Rightarrow \pi <2a<\frac{3\pi }{2}\Rightarrow cos2a<0$

$cos2a=-\sqrt{1-sin^{2}2a}=-\sqrt{1-\frac{9}{16}}=-\frac{\sqrt{7}}{4}$

$tan2a=\frac{sin2a}{cos2a}=\frac{\frac{-3}{4}}{\frac{-\sqrt{7}}{4}}=\frac{3}{\sqrt{7}}$