\(cos\left(\dfrac{5\pi}{12}\right)=sin\left(\dfrac{\pi}{2}-\dfrac{5\pi}{12}\right)=sin\left(\dfrac{\pi}{12}\right)=\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{4}\)Lại có \(0 0\)\(\Rightarrow sin\left(\dfrac{5\pi}{12}\right)=\sqrt{1-cos^2\left(\dfrac{5\pi}{12}\right)}=\dfrac{\sqrt{2+\sqrt{3}}}{2}=\dfrac{\sqrt{2}\left(\sqrt{3}+1\right)}{4}\)\(tan\left(\dfrac{5\pi}{12}\right)=\dfrac{sin\left(\dfrac{5\pi}{12}\right)}{cos\left(\dfrac{5\pi}{12}\right)}=2+\sqrt{3}\)b.\(sin\left(\dfrac{11\pi}{12}\right)=sin\left(\pi-\dfrac{\pi}{12}\right)=sin\left(\dfrac{\pi}{12}\right)=\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{4}\)\(cos\left(\dfrac{11\pi}{12}\right)=sin\left(\dfrac{\pi}{2}-\dfrac{11\pi}{12}\right)=sin\left(-\dfrac{5\pi}{12}\right)=-sin\left(\dfrac{5\pi}{12}\right)=-\dfrac{\sqrt{2}\left(\sqrt{3}+1\right)}{4}\)\(tan\left(\dfrac{11\pi}{12}\right)=\dfrac{sin\left(\dfrac{11\pi}{12}\right)}{cos\left(\dfrac{11\pi}{12}\right)}=-2+\sqrt{3}\)Câu trả lời: \(cos\left(\dfrac{5\pi}{12}\right)=sin\left(\dfrac{\pi}{2}-\dfrac{5\pi}{12}\right)=sin\left(\dfrac{\pi}{12}\right)=\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{4}\)Lại có \(0 0\)\(\Rightarrow sin\left(\dfrac{5\pi}{12}\right)=\sqrt{1-cos^2\left(\dfrac{5\pi}{12}\right)}=\dfrac{\sqrt{2+\sqrt{3}}}{2}=\dfrac{\sqrt{2}\left(\sqrt{3}+1\right)}{4}\)\(tan\left(\dfrac{5\pi}{12}\right)=\dfrac{sin\left(\dfrac{5\pi}{12}\right)}{cos\left(\dfrac{5\pi}{12}\right)}=2+\sqrt{3}\)b.\(sin\left(\dfrac{11\pi}{12}\right)=sin\left(\pi-\dfrac{\pi}{12}\right)=sin\left(\dfrac{\pi}{12}\right)=\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{4}\)\(cos\left(\dfrac{11\pi}{12}\right)=sin\left(\dfrac{\pi}{2}-\dfrac{11\pi}{12}\right)=sin\left(-\dfrac{5\pi}{12}\right)=-sin\left(\dfrac{5\pi}{12}\right)=-\dfrac{\sqrt{2}\left(\sqrt{3}+1\right)}{4}\)\(tan\left(\dfrac{11\pi}{12}\right)=\dfrac{sin\left(\dfrac{11\pi}{12}\right)}{cos\left(\dfrac{11\pi}{12}\right)}=-2+\sqrt{3}\)

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Chọn lớp

Chọn môn

Phạm Thị Hằng

18 tháng 7 2021

Giúp mik vs

#Toán lớp 10

Nguyễn Việt Lâm

Giáo viên

18 tháng 7 2021

\(cos\left(\dfrac{5\pi}{12}\right)=sin\left(\dfrac{\pi}{2}-\dfrac{5\pi}{12}\right)=sin\left(\dfrac{\pi}{12}\right)=\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{4}\)

Lại có \(0< \dfrac{5\pi}{12}< \dfrac{\pi}{2}\Rightarrow sin\left(\dfrac{5\pi}{12}\right)>0\)

\(\Rightarrow sin\left(\dfrac{5\pi}{12}\right)=\sqrt{1-cos^2\left(\dfrac{5\pi}{12}\right)}=\dfrac{\sqrt{2+\sqrt{3}}}{2}=\dfrac{\sqrt{2}\left(\sqrt{3}+1\right)}{4}\)

\(tan\left(\dfrac{5\pi}{12}\right)=\dfrac{sin\left(\dfrac{5\pi}{12}\right)}{cos\left(\dfrac{5\pi}{12}\right)}=2+\sqrt{3}\)

\(sin\left(\dfrac{11\pi}{12}\right)=sin\left(\pi-\dfrac{\pi}{12}\right)=sin\left(\dfrac{\pi}{12}\right)=\dfrac{\sqrt{2}\left(\sqrt{3}-1\right)}{4}\)

\(cos\left(\dfrac{11\pi}{12}\right)=sin\left(\dfrac{\pi}{2}-\dfrac{11\pi}{12}\right)=sin\left(-\dfrac{5\pi}{12}\right)=-sin\left(\dfrac{5\pi}{12}\right)=-\dfrac{\sqrt{2}\left(\sqrt{3}+1\right)}{4}\)

\(tan\left(\dfrac{11\pi}{12}\right)=\dfrac{sin\left(\dfrac{11\pi}{12}\right)}{cos\left(\dfrac{11\pi}{12}\right)}=-2+\sqrt{3}\)

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