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Vì \(\dfrac{\pi}{2}< \alpha< \pi\) \(\Rightarrow\) cos \(\alpha\) < 0
\(\Rightarrow\) cos \(\alpha\) = \(-\sqrt{1-sin^2\alpha}\) = \(-\dfrac{2\sqrt{2}}{3}\)
\(\Rightarrow\) tan \(\alpha\) = \(\dfrac{sin\alpha}{cos\alpha}=\dfrac{-\sqrt{2}}{4}\)
\(\Rightarrow\) cot \(\alpha\) = \(\dfrac{1}{tan\alpha}\) = \(-2\sqrt{2}\)
Chúc bn học tốt!
do a ∈ \(\left(0;\dfrac{\pi}{2}\right)\)⇒ \(\left\{{}\begin{matrix}sinx>0\\cosx>0\end{matrix}\right.\)
Mà tanx = 3 ⇒ \(\dfrac{sinx}{cosx}=3\Leftrightarrow\dfrac{sin^2x}{cos^2x}=9\Rightarrow10sin^2x=9\)
⇒ sinx = \(\dfrac{3}{\sqrt{10}}\)
⇒ sin (x + π) = -sinx = -\(\dfrac{3}{\sqrt{10}}\)
Lời giải:
$\cos^2 a=1-\sin^2a=1-(\frac{3}{5})^2=\frac{16}{25}$
$\Rightarrow \cos a=\pm \frac{4}{5}$
Ta có:
\(\cos (a-\frac{\pi}{3})=\cos a\cos \frac{\pi}{3}-\sin a\sin \frac{\pi}{3}\)
\(=\frac{1}{2}\cos a-\frac{3\sqrt{3}}{10}=\frac{1}{2}.\pm \frac{4}{5}-\frac{3\sqrt{3}}{10}\)
\(0< a< \frac{\pi}{2}\Rightarrow\left\{{}\begin{matrix}sina>0\\cosa>0\end{matrix}\right.\)
\(1+tan^2a=\frac{1}{cos^2a}\Rightarrow cos^2a=\frac{1}{1+tan^2a}\Rightarrow cosa=\frac{1}{\sqrt{1+tan^2a}}\)
\(\Rightarrow cosa=\frac{1}{2}\Rightarrow sina=cosa.tana=\frac{\sqrt{3}}{2}\)
\(cos2a=2cos^2a-1=-\frac{1}{2}\)
\(sin2a=2sina.cosa=\frac{\sqrt{3}}{2}\)
\(\Rightarrow sin\left(2a-\frac{\pi}{3}\right)=sin2a.cos\frac{\pi}{3}-cos2a.sin\frac{\pi}{3}=\frac{\sqrt{3}}{2}\)
\(tan\left(a+\frac{\pi}{4}\right)=\frac{tana+tan\frac{\pi}{4}}{1-tana.tan\frac{\pi}{4}}=-2-\sqrt{3}\)