a: 3/2pi<x<2pi
=>sin x<0
=>\(sinx=-\sqrt{1-\left(\dfrac{1}{6}\right)^2}=-\dfrac{\sqrt{35}}{6}\)
\(sin2x=2\cdot sinx\cdot cosx=2\cdot\dfrac{1}{6}\cdot\dfrac{-\sqrt{35}}{6}=\dfrac{-\sqrt{35}}{18}\)
\(cos2x=2\cdot cos^2x-1=2\cdot\dfrac{1}{36}-1=\dfrac{1}{18}-1=\dfrac{-17}{18}\)
\(tan2x=\dfrac{-\sqrt{35}}{18}:\dfrac{-17}{18}=\dfrac{\sqrt{35}}{17}\)
\(cot2x=1:\dfrac{\sqrt{35}}{17}=\dfrac{17}{\sqrt{35}}\)
b: \(sin\left(\dfrac{pi}{3}-x\right)\)
\(=sin\left(\dfrac{pi}{3}\right)\cdot cosx-cos\left(\dfrac{pi}{3}\right)\cdot sinx\)
\(=\dfrac{1}{2}\cdot\dfrac{-\sqrt{35}}{6}-\dfrac{1}{2}\cdot\dfrac{1}{6}=\dfrac{-\sqrt{35}-1}{12}\)
c: \(cos\left(x-\dfrac{3}{4}pi\right)\)
\(=cosx\cdot cos\left(\dfrac{3}{4}pi\right)+sinx\cdot sin\left(\dfrac{3}{4}pi\right)\)
\(=\dfrac{1}{6}\cdot\dfrac{-\sqrt{2}}{2}+\dfrac{-\sqrt{35}}{6}\cdot\dfrac{\sqrt{2}}{2}=\dfrac{-\sqrt{2}-\sqrt{70}}{12}\)
d: tan(pi/6-x)
\(=\dfrac{tan\left(\dfrac{pi}{6}\right)-tanx}{1+tan\left(\dfrac{pi}{6}\right)\cdot tanx}\)
\(=\dfrac{\dfrac{\sqrt{3}}{3}-\sqrt{35}}{1+\dfrac{\sqrt{3}}{3}\cdot\left(-\sqrt{35}\right)}\)
