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\frac{2\sqrt{3}}{3}\approx 1.154700538
Tráth na gCeist
Trigonometry
\csc ( 60 )
Fadhbanna den chineál céanna ó Chuardach Gréasáin
How do you evaluate the expression \displaystyle{\csc{{\left(-{60}\right)}}} ?
https://socratic.org/questions/how-do-you-evaluate-the-expression-csc-60
\displaystyle{\left({\csc{{\left(-{60}\right)}}}=-\frac{{1}}{{\sin{{60}}}}=-\frac{{2}}{\sqrt{{3}}}\right.} Explanation: \displaystyle{\csc{{\left(-{60}\right)}}}={\csc{{\left(-{\left(\frac{\pi}{{3}}\right)}\right)}}}={\csc{{\left({2}\pi-{\left(\frac{\pi}{{3}}\right)}\right.}}} ...
What is the exact value of \displaystyle{\cot{{\left(-{90}\right)}}} and \displaystyle{\csc{{690}}} , using unit circle?
https://socratic.org/questions/59ed2cfe11ef6b0b2cd1cba8
It depends on what you do want to use. Explanation: I assume that you actually mean to ask about \displaystyle{\cot{{\left(-{90}^{\circ}\right)}}} (There is no nice way to express the exact ...
How do you find the amplitude and period of \displaystyle{A}{r}{c}{\csc} ?
https://socratic.org/questions/how-do-you-find-the-amplitude-and-period-of-arc-csc
Dean R. May 15, 2018 Arccsc and the other inverse trig functions aren't periodic. We have to worry about principal values because they're not even really functions, being the inverses of ...
How do you evaluate the expression \displaystyle{\csc{{\left(-{45}\right)}}} ?
https://socratic.org/questions/how-do-you-evaluate-the-expression-csc-45
\displaystyle-\sqrt{{2}} Explanation: Recall that \displaystyle{\csc{\theta}}=\frac{{1}}{{\sin{\theta}}} and \displaystyle{\sin{{\left(-\theta\right)}}}=-{\sin{{\left(\theta\right)}}} ...
What is the value of \displaystyle{\csc{{17}}}° ?
https://socratic.org/questions/what-is-the-value-of-csc-17
3.42 Explanation: There are two ways you can go about this. The first is using a calculator with the cosecant button. If your calculator has this, you can easily calculate that. The other way is ...
How do you evaluate \displaystyle{\csc{{180}}} ?
https://socratic.org/questions/how-do-you-evaluate-csc-180
Unidefined. Explanation: Since \displaystyle{\csc{{\left({x}\right)}}}=\frac{{1}}{{\sin{{\left({x}\right)}}}} , you have \displaystyle{\csc{{\left({180}\right)}}}=\frac{{1}}{{\sin{{\left({180}\right)}}}} ...
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Fadhbanna Comhchosúla
\cos ( \pi )
\sin ( \frac { \pi } { 2 } )
\tan ( \frac { 4 \pi } { 3 } )
\csc ( 60 )
\sec ( 180 )
\cot ( \frac { 4 \pi } { 3 } )
Ar ais go barr