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Bài 2:
\(\cos\widehat{A}=\dfrac{3\sqrt{39}}{20}\)
\(\tan\widehat{A}=\dfrac{7}{20}:\dfrac{3\sqrt{39}}{20}=\dfrac{7}{3\sqrt{39}}=\dfrac{7\sqrt{39}}{117}\)
\(\cot\widehat{A}=\dfrac{3\sqrt{39}}{7}\)
tana = 3/4.
=>cota=1/ tana =1:3/4=4/3
sina /cosa =tana
=> sina =tana .cosa =3/4. cosa
lại có sin^2(a)+cos^2(a)=1
<=>9/16cos^2(a)+cos^2=1
<=>25/16cos^2(a)=1
<=>cos^2(a)=16/25
=>[cosa =4/5=>sina =3/5
[cosa =-4/5=> sina =-2/5
\(\sin^2\widehat{A}+\cos^2\widehat{A}=1\Leftrightarrow\cos^2\widehat{A}=1-\dfrac{16}{25}=\dfrac{9}{25}\\ \Leftrightarrow\cos\widehat{A}=\dfrac{3}{5}\\ \tan\widehat{A}=\dfrac{\sin\widehat{A}}{\cos\widehat{A}}=\dfrac{4}{5}:\dfrac{3}{5}=\dfrac{4}{3}\\ \cot\widehat{A}=\dfrac{1}{\tan\widehat{A}}=\dfrac{3}{4}\)
\(\sin A=0,8\Rightarrow A=arcsin0,8_{ }\)
\(\Rightarrow\cos A=cos\left(arcsin0,8\right)=\dfrac{3}{5}\)
tanA=tan(arcsin0,8)=4/3
cotA=1:4/3=3/4
\(\sin^2\widehat{A}+\cos^2\widehat{A}=1\Leftrightarrow\cos^2\widehat{A}=1-\left(\dfrac{3}{5}\right)^2=1-\dfrac{9}{25}=\dfrac{16}{25}\\ \Leftrightarrow\cos\widehat{A}=\dfrac{4}{5}\\ \tan\widehat{A}=\dfrac{\sin\widehat{A}}{\cos\widehat{A}}=\dfrac{3}{4}\\ \Rightarrow\cot\widehat{A}=\dfrac{1}{\tan\widehat{A}}=\dfrac{4}{3}\)
Ta có: \(\sin^2A+cos^2A=1\)
=>\(cos^2A=1-\left(\frac35\right)^2=1-\frac{9}{25}=\frac{16}{25}\)
=>\(cosA=\frac45\)
tan A=sin A:cosA
\(=\frac35:\frac45=\frac35\times\frac54=\frac34\)
\(\cot A=1:\frac34=\frac43\)
TA có: \(\sin^2A+cos^2A=1\)
=>\(cos^2A=1-\left(\frac35\right)^2=1-\frac{9}{25}=\frac{16}{25}=\left(\frac45\right)^2\)
=>\(cosA=\frac45\)
tan A=\(\frac{\sin A}{cosA}=\frac35:\frac45=\frac34\)
\(\cot A=\frac{1}{\tan A}=1:\frac34=\frac43\)
a: Ta có: \(\sin^2a+cos^2a=1\)
=>\(cos^2a=1-0,8^2=1-0,64=0,36=0,6^2\)
=>cosa=0,6
\(\tan a=\frac{\sin a}{cosa}=\frac{0.8}{0.6}=\frac43\)
\(\cot a=\frac{1}{\tan a}=1:\frac43=\frac34\)
b: Ta có: \(\sin^2a+cos^2a=1\)
=>\(\sin^2a=1-0,6^2=1-0,36=0,64=0,8^2\)
=>sin a=0,8
\(\tan a=\frac{\sin a}{cosa}=\frac{0.8}{0.6}=\frac43\)
\(\cot a=\frac{1}{\tan a}=1:\frac43=\frac34\)
c: \(\tan a\cdot\cot a=1\)
=>\(\cot a=\frac13\)
Ta có: \(1+\tan^2a=\frac{1}{cos^2a}\)
=>\(\frac{1}{cos^2a}=1+3^2=10\)
=>\(cos^2a=\frac{1}{10}\)
=>\(cosa=\frac{1}{\sqrt{10}}\)
Ta có: \(\tan a=\frac{\sin a}{cosa}\)
=>\(\sin a=\tan a\cdot cosa\)
\(=3\cdot\frac{1}{\sqrt{10}}=\frac{3}{\sqrt{10}}\)
d: \(\tan a\cdot\cot a=1\)
=>\(\tan a\cdot2=1\)
=>\(\tan a=\frac12\)
Ta có: \(1+\tan^2a=\frac{1}{cos^2a}\)
=>\(\frac{1}{cos^2a}=1+\left(\frac12\right)^2=\frac54\)
=>\(cos^2a=\frac45\)
=>\(cosa=\frac{2}{\sqrt5}\)
Ta có: \(\tan a=\frac{\sin a}{cosa}\)
=>\(\sin a=cosa\cdot\tan a=\frac{2}{\sqrt5}\cdot\frac12=\frac{1}{\sqrt5}\)