H24
NL
20 tháng 4 2022 lúc 22:22

\(\dfrac{\pi}{2}< x< \pi\Rightarrow sinx>0;cosx< 0\)

\(\dfrac{2}{5}=cos\left(\dfrac{3\pi}{2}+x\right)=sin\left(-\pi-x\right)=-sin\left(\pi+x\right)=sinx\)

\(cosx=-\sqrt{1-sin^2x}=-\sqrt{1-\left(\dfrac{2}{5}\right)^2}=-\dfrac{\sqrt{21}}{5}\)

\(tanx=\dfrac{sinx}{cosx}=-\dfrac{2\sqrt{21}}{21}\) ; \(cotx=\dfrac{1}{tanx}=-\dfrac{\sqrt{21}}{2}\)

b.

\(sin\left(\dfrac{3\pi}{2}-x\right)=cos\left(2\pi-x\right)=cosx=-\dfrac{\sqrt{21}}{5}\)

\(cos\left(\dfrac{9\pi}{2}+x\right)=cos\left(4\pi+\dfrac{\pi}{2}+x\right)=cos\left(\dfrac{\pi}{2}+x\right)=-sinx=-\dfrac{2}{5}\)

\(tan\left(-\pi-x\right)=tan\left(-x\right)=-tanx=\dfrac{2\sqrt{21}}{21}\)

\(cot\left(\dfrac{31\pi}{2}-x\right)=cot\left(15\pi+\dfrac{\pi}{2}-x\right)=cot\left(\dfrac{\pi}{2}-x\right)=tanx=-\dfrac{2\sqrt{21}}{21}\)

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