Câu hỏi của Phạm Nhật Trúc

Question

Giúp mình giải 4 bài này với mừng cần gấp lam

Nguyễn Việt Lâm · Accepted Answer

a. $0< a< 90^0\Rightarrow cosa>0$

$cosa=\sqrt{1-sin^2a}=\dfrac{5}{13}$

$sin2a=2sina.cosa=\dfrac{120}{169}$

$cos2a=2cos^2a-1=2.\left(\dfrac{5}{13}\right)^2-1=-\dfrac{119}{169}$

b.

$\dfrac{3\pi}{2}< a< 2\pi\Rightarrow sina< 0$ $\Rightarrow sina=-\sqrt{1-cos^2a}=-\dfrac{\sqrt{15}}{8}$

$sin2a=2sina.cosa=-\dfrac{7\sqrt{15}}{32}$

$cos2a=2cos^2a-1=2\left(\dfrac{7}{8}\right)^2-1=\dfrac{17}{32}$

Câu trả lời:

a. $0< a< 90^0\Rightarrow cosa>0$

$cosa=\sqrt{1-sin^2a}=\dfrac{5}{13}$

$sin2a=2sina.cosa=\dfrac{120}{169}$

$cos2a=2cos^2a-1=2.\left(\dfrac{5}{13}\right)^2-1=-\dfrac{119}{169}$

b.

$\dfrac{3\pi}{2}< a< 2\pi\Rightarrow sina< 0$ $\Rightarrow sina=-\sqrt{1-cos^2a}=-\dfrac{\sqrt{15}}{8}$

$sin2a=2sina.cosa=-\dfrac{7\sqrt{15}}{32}$

$cos2a=2cos^2a-1=2\left(\dfrac{7}{8}\right)^2-1=\dfrac{17}{32}$

Nguyễn Việt Lâm · Answer

c.

$\dfrac{\pi}{2}< a< \pi\Rightarrow sina>0$

$\Rightarrow sina=\sqrt{1-cos^2a}=\dfrac{\sqrt{2}}{2}$

$sin2a=2sina.cosa=-1$

$cos2a=2cos^2a-1=0$

d.

$\pi< a< \dfrac{3\pi}{2}\Rightarrow cosa< 0$

$\Rightarrow cosa=-\sqrt{1-sin^2a}=-\dfrac{1}{2}$

$sin2a=2sina.cosa=\dfrac{\sqrt{3}}{2}$

$cos2a=2cos^2a-1=-\dfrac{1}{2}$