Question

[Hàn Nhật Hạ]

định hướng cách làm giúp mk vs ạ thank nhiều!

Answer 1

[Nguyễn Việt Lâm]

a.

\(\pi< a< \dfrac{3\pi}{2}\Rightarrow sina< 0\)

\(\Rightarrow sina=-\sqrt{1-cos^2a}=-\dfrac{12}{13}\)

\(cos2a=cos^2a-sin^2a=\left(-\dfrac{5}{12}\right)^2-\left(-\dfrac{12}{13}\right)^2=...\)

\(sin2a=2sina.cosa=...\)

\(tan2a=\dfrac{sin2a}{cos2a}=...\)

//

\(\dfrac{\pi}{2}< a< \pi\Rightarrow sina>0\Rightarrow sina=\sqrt{1-cos^2a}=\dfrac{12}{13}\)

\(cos2a=cos^2a-sin^2a=...\) ; \(sin2a=2sina.cosa\) ; \(tan2a=\dfrac{sin2a}{cos2a}\) ...

//

\(-\dfrac{\pi}{2}< a< 0\Rightarrow sina< 0\Rightarrow sina=-\sqrt{1-cos^2a}=-\dfrac{3}{5}\)

Thay vào tính cos2a, sin2a, tan2a tương tự như trên

Answer 2

[Nguyễn Việt Lâm]

b.

\(\pi< a< \dfrac{3\pi}{2}\Rightarrow cosa< 0\Rightarrow cosa=-\sqrt{1-sin^2a}=-\dfrac{4}{5}\)

Tính tương tự câu a

c.

\(\dfrac{3\pi}{4}< a< \pi\Rightarrow\dfrac{3\pi}{2}< 2a< 2\pi\Rightarrow cos2a>0\)

\(sina+cosa=\dfrac{1}{2}\Rightarrow\left(sina+cosa\right)^2=\dfrac{1}{4}\)

\(\Rightarrow1+2sina.cosa=\dfrac{1}{4}\Rightarrow1+sin2a=\dfrac{1}{4}\)

\(\Rightarrow sin2a=-\dfrac{3}{4}\)

\(cos2a=\sqrt{1-sin^22a}=\dfrac{\sqrt{7}}{4}\)

\(tan2a=\dfrac{sin2a}{cos2a}=...\)