1: Ta có; \(\sin^2x+cos^2x=1\)
=>\(\sin^2x=1-0,8^2=0,36=0,6^2\)
=>sin x=0,6
tan x=sin x/cosx
\(=\frac{0.6}{0.8}=\frac34\)
cot x=\(\frac{1}{\tan x}=1:\frac34=\frac43\)
2: \(\cot x=\frac{1}{\tan x}=1:0,75=1:\frac34=\frac43\)
\(1+\cot^2x=\frac{1}{\sin^2x}\)
=>\(\frac{1}{\sin^2x}=1+\left(\frac43\right)^2=1+\frac{16}{9}=\frac{25}{9}\)
=>\(\sin^2x=\frac{9}{25}=\left(\frac35\right)^2\)
=>\(\sin x=\frac35\)
Ta có: \(\tan x=\sin x:cosx\)
=>\(cosx=\frac35:\frac34=\frac45\)
3: Ta có: \(\sin^2x+cos^2x=1\)
=>\(cos^2x=1-\left(\frac{9}{20}\right)^2=1-\frac{81}{400}=\frac{319}{400}\)
=>cosx=\(\frac{\sqrt{319}}{20}\)
tan x=\(\frac{\sin x}{cosx}=\frac{9}{20}:\frac{\sqrt{319}}{20}=\frac{9}{\sqrt{319}}\)
cot x=\(\frac{1}{\tan x}=1:\frac{9}{\sqrt{319}}=\frac{\sqrt{319}}{9}\)
4: \(\tan x=\frac{1}{\cot x}=\frac{1}{-5}\)
\(1+\tan^2x=\frac{1}{cos^2x}\)
=>\(\frac{1}{cos^2x}=1+\left(-\frac15\right)^2=1+\frac{1}{25}=\frac{26}{25}\)
=>\(cos^2x=\frac{25}{26}\)
=>cosx=\(\frac{5}{\sqrt{26}}\)
Ta có: \(\sin^2x+cos^2x=1\)
=>\(\sin^2x=1-\frac{25}{26}=\frac{1}{26}\)
=>sin x=\(\frac{1}{\sqrt{26}}\)
