Bài 4:
a: \(\sin^2x+cos^2x=1\)
=>\(\sin^2x=1-\left(-\frac45\right)^2=1-\frac{16}{25}=\frac{9}{25}\)
mà sin x>0\(\left(\frac{\pi}{2}
nên sin x=3/5
\(\tan x=\frac{\sin x}{cosx}=\frac35:\frac{-4}{5}=-\frac34\)
\(\cot x=\frac{1}{\tan x}=1:\frac{-3}{4}=-\frac43\)
b: \(\sin2x=2\cdot\sin x\cdot cosx=2\cdot\frac35\cdot\frac{-4}{5}=\frac{-24}{25}\)
\(cos\left(3\pi-x\right)=cos\left(2\pi+\pi-x\right)=cos\left(\pi-x\right)\) =-cosx=-(-4/5)=4/5
\(\sin\left(\frac92\pi+x\right)=\sin\left(4\pi+\frac{\pi}{2}+x\right)=\sin\left(x+\frac{\pi}{2}\right)\)
=cosx=-4/5
c: \(cos2x=1-2\cdot\sin^2x=1-2\cdot\left(\frac35\right)^2=1-2\cdot\frac{9}{25}=1-\frac{18}{25}=\frac{7}{25}\)
\(cos\left(x-\frac{\pi}{6}\right)=cosx\cdot cos\left(\frac{\pi}{6}\right)+\sin x\cdot\sin\left(\frac{\pi}{6}\right)\)
\(=\frac{-4}{5}\cdot\frac{\sqrt3}{2}+\frac35\cdot\frac12=\frac{-4\sqrt3+3}{10}\)
