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I'am very confused, I have looked all over google and I can not find out how too do this problem. I have the answer its number 14 since our teacher gives us the answer but we need to show work.enter image description here

I understood number 13 it was $\sec -210^\circ$, but I can't figure out how too do #14. How can it be $-2 \sqrt{3}/ 3$? I just can't figure this problem out.

Any help would be greatly appreciated. The problem is $\csc -420^\circ$.

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  • $\begingroup$ hint: $\csc x = \csc (x+360^\circ)= \csc (x+360^\circ+360^\circ)=\dots$ $\endgroup$
    – John Joy
    Dec 19, 2014 at 18:10

2 Answers 2

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$$\csc(-420^\circ)=-\csc 420^\circ=-\csc 60^\circ=-\frac{2\sqrt3}{2}$$

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  • $\begingroup$ FYI: when working in degrees, you can append a degree-symbol as a subscript by writing 420^\circ. E.g., $420^\circ$. $\endgroup$
    – David H
    Dec 19, 2014 at 4:22
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$$ \csc(-420^\circ)= -\csc(420^\circ) \text{ since cosecant is an odd function. Thus}\\ -\csc(420^\circ) = -\csc(60^\circ) = -\frac{1}{\sin(60^\circ)} = -\frac{1}{\frac{\sqrt{3}}{2}} = -\frac{2}{\sqrt{3}}=-\frac{2\sqrt{3}}{3} $$

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  • $\begingroup$ okay I know how you got the 60 but where did you get the radical 3/2? $\endgroup$
    – Rick
    Dec 19, 2014 at 4:33
  • $\begingroup$ From the unit circle. At $60^\circ$, sine has has value of $\frac{\sqrt{3}}{2}$ $\endgroup$
    – mathamphetamines
    Dec 19, 2014 at 4:34
  • $\begingroup$ Think about the $30^\circ-60^\circ-90^\circ$ triangle. $\endgroup$
    – Tim Raczkowski
    Dec 19, 2014 at 4:48

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