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\frac{\sqrt{3}}{3}\approx 0.577350269
Kuiz
Trigonometry
\cot ( \frac { 4 \pi } { 3 } )
Masalah Sama dari Carian Web
How do you evaluate \displaystyle{\cot{{\left(\frac{{{2}\pi}}{{3}}\right)}}} ?
https://socratic.org/questions/how-do-you-evaluate-cot-2pi-3
\displaystyle{1.402} Explanation: Evaluate \displaystyle\frac{{{2}\pi}}{{3}} where \displaystyle\pi={180}^{{o}} , which gives a value of \displaystyle{120}^{{o}} . Also know that \displaystyle{\cot{=}}\frac{{1}}{{\tan}} ...
How do you find the value of \displaystyle{\cot{{\left({\left({5}\frac{\pi}{{3}}\right)}\right.}}} using the double angle or half angle identity?
https://socratic.org/questions/how-do-you-find-the-value-of-cot-5pi-3-using-the-double-angle-or-half-angle-iden
\displaystyle-\frac{\sqrt{{3}}}{{3}} Explanation: Trig table and unit circle --> \displaystyle{\cot{{\left(\frac{{{5}\pi}}{{3}}\right)}}}=\frac{{1}}{{\tan{{\left(\frac{{{5}\pi}}{{3}}\right)}}}} ...
How do you evaluate \displaystyle{\cot{{\left(-\frac{{{4}\pi}}{{5}}\right)}}} ?
https://socratic.org/questions/how-do-you-evaluate-cot-4pi-5
\displaystyle{\cot{-}}{2.513274123} Explanation: \displaystyle{\cot{{\left(-\frac{{{4}\pi}}{{5}}\right)}}} \displaystyle=\frac{{\cot{{\left(-{4}\times{3.141592654}\right)}}}}{{5}} \displaystyle=\frac{{\cot{{\left(-{12.56637061}\right)}}}}{{5}} ...
How do you evaluate the expression \displaystyle{\cot{{\left(\frac{\pi}{{3}}\right)}}} ?
https://socratic.org/questions/how-do-you-evaluate-the-expression-cot-pi-3
\displaystyle{\cot{{\left(\frac{\pi}{{3}}\right)}}}=\frac{{1}}{\sqrt{{3}}} Explanation: \displaystyle\frac{\pi}{{3}} is 60 degrees. Draw an equilateral triangle. Cut it in half. The angles ...
cot(3.14/9)
https://www.tiger-algebra.com/drill/cot(3.14/9)/
Your input cot(3.14/9) is not yet solved by the Tiger Algebra Solver. please join our mailing list to be notified when this and other topics are added. Processing ends successfully
Finding \cot(\frac{\pi}{12})
https://math.stackexchange.com/questions/2231178/finding-cot-frac-pi12
Your error is your formula. It is: \cot(\theta-\phi)=\frac{\cot(\theta)\cot(\phi)+1}{\cot\phi-\cot\theta} Notice the denominator order. You switched them, hence switching the sign.
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\cos ( \pi )
\sin ( \frac { \pi } { 2 } )
\tan ( \frac { 4 \pi } { 3 } )
\csc ( 60 )
\sec ( 180 )
\cot ( \frac { 4 \pi } { 3 } )
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