Question

cho sinx = \(-\dfrac{3}{5}\) và \(\pi\) < x < \(\dfrac{3\pi}{2}\) tính

a) sin2x, cos2x, tan2x, cot2x

b) \(sin\left(x+\dfrac{\pi}{3}\right)\)

Answer 1

a: pi<x<3/2pi

=>cosx<0

=>\(cosx=-\sqrt{1-\left(-\dfrac{3}{5}\right)^2}=-\dfrac{4}{5}\)

\(tanx=\dfrac{-3}{5}:\dfrac{-4}{5}=\dfrac{3}{4}\)

cot x=1:3/4=4/3

\(sin2x=2\cdot sinx\cdot cosx=2\cdot\dfrac{-3}{5}\cdot\dfrac{-4}{5}=\dfrac{24}{25}\)

\(cos2x=1-2\cdot sin^2x=1-2\cdot\left(-\dfrac{3}{5}\right)^2=\dfrac{7}{25}\)

\(tan2x=\dfrac{24}{25}:\dfrac{7}{25}=\dfrac{24}{7}\)

cot 2x=1:24/7=7/24

b: \(sin\left(x+\dfrac{pi}{3}\right)=sinx\cdot cos\left(\dfrac{pi}{3}\right)+sin\left(\dfrac{pi}{3}\right)\cdot cosx\)

\(=\dfrac{-3}{5}\cdot\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}\cdot\dfrac{-4}{5}=\dfrac{-3-4\sqrt{3}}{10}\)