anonymous
  • anonymous
If sin (B) = -1/3 with B in QIII, find csc (B/2).
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\sin \left( \frac{B}{2} \right)=\sqrt{\frac{1-\cos B}{2}}\]
anonymous
  • anonymous
Right, but how does this help find csc?
anonymous
  • anonymous
\[\csc \left( \frac{B}{2} \right)=\frac{1}{\sin \left( \frac{B}{2} \right)}\]

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anonymous
  • anonymous
Oh!!! I didn't think about it that way... So, if it's B/2, would I do (-1/3)/2?
anonymous
  • anonymous
no, it doesnt work that way, dividing the angle by 2 is not the same as dividing the value by 2
anonymous
  • anonymous
Good point. I'm sorry. This stuff is so confusing to me.
anonymous
  • anonymous
given sin B = -1/3, you need to find cos B first, and then plug it in into the half angle formula
anonymous
  • anonymous
Ok... So, to find cos B, I use the formula\[\pm \sqrt{(1+cosB)/2}\]
anonymous
  • anonymous
?
anonymous
  • anonymous
|dw:1333786170044:dw|
anonymous
  • anonymous
That's what I thought!! I didn't know whether or not I could use sqrt 8!! Ok, now I'm starting to understand.
anonymous
  • anonymous
the negative sign shows that angle B is in QIII
anonymous
  • anonymous
Ah!!
anonymous
  • anonymous
Ok, I'm working it out, now.
anonymous
  • anonymous
|dw:1333732829170:dw| Stuck here.
anonymous
  • anonymous
Oh!!! Btw... We don't have to have the exact equation... I am so sorry.
anonymous
  • anonymous
I always forget that. I already did my half-angle formula problems.
anonymous
  • anonymous
\[\sqrt{\frac{1+\frac{\sqrt{8}}{3}}{2}}=\sqrt{\frac{3+\sqrt{8}}{6}}=\sqrt{\frac{3+2\sqrt{2}}{6}}\]\[=\frac{\sqrt{2}+1}{\sqrt{6}}\]\[=\frac{2\sqrt{3}+\sqrt{6}}{6}\]
anonymous
  • anonymous
I only need the fraction. I forgot that I don't need the exact equation.
anonymous
  • anonymous
choose the positive value, if the angle B is in QIII (between 180 and 270), the angle B/2 is in QII (between 90 and 135), sin and csc are positive in QII
anonymous
  • anonymous
I honestly don't know where to go from here. Can you please help me get the answer? I'm so stressed out, man.
anonymous
  • anonymous
after using the half angle formula we get\[\sin \left( \frac{B}{2} \right)=\frac{2\sqrt{3}+\sqrt{6}}{6}\]\[\csc \left( \frac{B}{2} \right)=\frac{1}{\sin \left(\frac{B}{2}\right)}=\frac{6}{2\sqrt{3}+\sqrt{6}}=2\sqrt{3}-\sqrt{6}\]
anonymous
  • anonymous
So, what would the fraction be? In this problem, the half angle formula is not used.
anonymous
  • anonymous
I dont know any other analytic method to solve this problem except the half angle formula
anonymous
  • anonymous
I understand. Here's an example of the last problem I had.
anonymous
  • anonymous
If sin (A) = -12/13 with A in QIII, find cos (A/2). The answer was \[-2/\sqrt{3}\]
anonymous
  • anonymous
I will use the half angle formula for that problem too
anonymous
  • anonymous
Ok.
anonymous
  • anonymous
\[B=-\sin ^{-1}\left(\frac{1}{3}\right) \]\[\text{Csc}\left[\frac{-\text{ArcSin}\left[\frac{1}{3}\right]}{2}\right]=-\text{Csc}\left[\frac{1}{2} \text{ArcSin}\left[\frac{1}{3}\right]\right]=-5.91359 \]
anonymous
  • anonymous
wait.. is it -2/sqrt(3) or -2/sqrt(13) ?
anonymous
  • anonymous
-2/sqrt(13)! Sorry!!
anonymous
  • anonymous
then it is correct, I have just solved it using the half angle formula
anonymous
  • anonymous
When I typed in \[2\sqrt{3}-\sqrt{6}\] it was wrong. It has to be a fraction.
anonymous
  • anonymous
Ok... Your answer \[6/((2\sqrt3)-\sqrt6)\] was right!!!!!!!!!!!!
anonymous
  • anonymous
1+8√32−−−−−−√=3+8√6−−−−−−√=3+22√6−−−−−−−√ =2√+16√ =23√+6√6 Im looking at this... maybe its basic algebra but how did you go from \[\sqrt{3+2\sqrt{2}}\div6\] to \[\sqrt{2} + 1 \div \sqrt{6} \]

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