a.
\(\pi< a< \dfrac{3\pi}{2}\Rightarrow sina< 0\)
\(\Rightarrow sina=-\sqrt{1-cos^2a}=-\dfrac{12}{13}\)
\(cos2a=cos^2a-sin^2a=\left(-\dfrac{5}{12}\right)^2-\left(-\dfrac{12}{13}\right)^2=...\)
\(sin2a=2sina.cosa=...\)
\(tan2a=\dfrac{sin2a}{cos2a}=...\)
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\(\dfrac{\pi}{2}< a< \pi\Rightarrow sina>0\Rightarrow sina=\sqrt{1-cos^2a}=\dfrac{12}{13}\)
\(cos2a=cos^2a-sin^2a=...\) ; \(sin2a=2sina.cosa\) ; \(tan2a=\dfrac{sin2a}{cos2a}\) ...
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\(-\dfrac{\pi}{2}< a< 0\Rightarrow sina< 0\Rightarrow sina=-\sqrt{1-cos^2a}=-\dfrac{3}{5}\)
Thay vào tính cos2a, sin2a, tan2a tương tự như trên
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.
\(\dfrac{3\pi}{4}< a< \pi\Rightarrow\dfrac{3\pi}{2}< 2a< 2\pi\Rightarrow cos2a>0\)
\(sina+cosa=\dfrac{1}{2}\Rightarrow\left(sina+cosa\right)^2=\dfrac{1}{4}\)
\(\Rightarrow1+2sina.cosa=\dfrac{1}{4}\Rightarrow1+sin2a=\dfrac{1}{4}\)
\(\Rightarrow sin2a=-\dfrac{3}{4}\)
\(cos2a=\sqrt{1-sin^22a}=\dfrac{\sqrt{7}}{4}\)
\(tan2a=\dfrac{sin2a}{cos2a}=...\)
