a: pi/2<a<pi

=>sin a>0

\(sina=\sqrt{1-\left(-\dfrac{1}{\sqrt{3}}\right)^2}=\dfrac{\sqrt{2}}{\sqrt{3}}\)

\(sin\left(a+\dfrac{pi}{6}\right)=sina\cdot cos\left(\dfrac{pi}{6}\right)+sin\left(\dfrac{pi}{6}\right)\cdot cosa\)

\(=\dfrac{\sqrt{3}}{2}\cdot\dfrac{\sqrt{2}}{\sqrt{3}}+\dfrac{1}{2}\cdot-\dfrac{1}{\sqrt{3}}=\dfrac{\sqrt{6}-2}{2\sqrt{3}}\)

b: \(cos\left(a+\dfrac{pi}{6}\right)=cosa\cdot cos\left(\dfrac{pi}{6}\right)-sina\cdot sin\left(\dfrac{pi}{6}\right)\)

\(=\dfrac{-1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}-\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}=\dfrac{-\sqrt{3}-\sqrt{2}}{2\sqrt{3}}\)

c: \(sin\left(a-\dfrac{pi}{3}\right)\)

\(=sina\cdot cos\left(\dfrac{pi}{3}\right)-cosa\cdot sin\left(\dfrac{pi}{3}\right)\)

\(=\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}+\dfrac{1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}=\dfrac{\sqrt{2}+\sqrt{3}}{2\sqrt{3}}\)

d: \(cos\left(a-\dfrac{pi}{6}\right)\)

\(=cosa\cdot cos\left(\dfrac{pi}{6}\right)+sina\cdot sin\left(\dfrac{pi}{6}\right)\)

\(=\dfrac{-1}{\sqrt{3}}\cdot\dfrac{\sqrt{3}}{2}+\dfrac{\sqrt{2}}{\sqrt{3}}\cdot\dfrac{1}{2}=\dfrac{-\sqrt{3}+\sqrt{2}}{2\sqrt{3}}\)

`A=sin(π-α)+cos(π+α)+cos(-α)`

`= sinα-cosα+cosα=sinα=3/5`

Chọn A.

Chọn B.

3π/2 < π/2 + α < 2π nên sin(π/2 + α) < 0

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!

Chọn B.

Ta có :

Với π/2 < α < π thì sinα > 0, cosα < 0, tanα < 0