anonymous
  • anonymous
sinθ if cscθ= -2
Precalculus
chestercat
  • chestercat
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anonymous
  • anonymous
well what do you know about those 2 functions in relation to each other
anonymous
  • anonymous
That's all it says on my homework paper, 'Use the appropriate reciprocal identity to find each function value.'
anonymous
  • anonymous
ya, but you didn't answer my question. you have soh cah toa. and the three inverse functions. which one is the inverse of sin

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anonymous
  • anonymous
or maybe call it reciprocal? im not good at my language
anonymous
  • anonymous
As far as I know it's reciprocal is csc=1/sin
anonymous
  • anonymous
alright then. what do you think the answer is
anonymous
  • anonymous
something like -1/2.
anonymous
  • anonymous
I just have different ones like tanθ if cotθ=sqrt6/2 and I don't really understand
anonymous
  • anonymous
so your correct on this problem. do you know about radians yet for the other problems?
anonymous
  • anonymous
WHat do you mean radians?
anonymous
  • anonymous
and cot and tan are reciprocals as well. radians-degrees. have you now done the unit circle yet?
anonymous
  • anonymous
I don't recall so I am going to have to say no.
anonymous
  • anonymous
well. hmm.. you need to learn about radians and degrees for trigonometry. but just to avoid that. remember that cot and tan are reciprocal identities as well. but you cant have a radical in the denominator
anonymous
  • anonymous
Yea, so how would I go on in doing that? I am in Pre-cal if that makes a difference.
anonymous
  • anonymous
http://tutorial.math.lamar.edu/cheat_table.aspx print that off. it will do wonders for your trig identities. do you know how to rationalize the denominator
anonymous
  • anonymous
tell me where you are at on the second problem
anonymous
  • anonymous
All I can really understand is that tanθ= 1/sqrt6/2, but I don't know how to make it like without the sqrt in the denominator.
anonymous
  • anonymous
hm. what is the reciprocal of \[\frac{ \sqrt{6} }{ 2 }\]
anonymous
  • anonymous
Umm tbh I'm not really sure :/
anonymous
  • anonymous
You multiply the numerator and denominator by your square rooted number to get rid of the root in the bottom, this is known as rationalizing it. Here is an example |dw:1385354774715:dw|
anonymous
  • anonymous
Oh ok I see so now I have to get the reciprocal.
anonymous
  • anonymous
but it would be 6/2sqrt2
anonymous
  • anonymous
reciprocal would be flipping your fraction like if you were dividing by fractions \[\frac{ 2 }{ \sqrt{6} }\] this is the reciprocal of my above fraction. and when you multiply 2 square roots like that above you would have 2 not 1
anonymous
  • anonymous
so if you multiply this \[\frac{ 2(\sqrt{6)} }{ \sqrt{6}*\sqrt{6} }\] that is how you rationalize the denominator
anonymous
  • anonymous
oh I see now, so then the answer is 2/√6, I don't have to rationalize anything I believe.
anonymous
  • anonymous
keeping in mind \[\sqrt{6}*\sqrt{6}=\sqrt{36}\]
anonymous
  • anonymous
Ok, thanks a lot! really appreciate the help!
anonymous
  • anonymous
you cant have square roots in the denominator. you have to rationalize it like I did in the above step
anonymous
  • anonymous
see how I multiplied the top and bottom by \[\sqrt{6}\]
anonymous
  • anonymous
so 2*sqrt6/6
anonymous
  • anonymous
Some teacher dont require rationalizing, but it is generally required, so if it wasn't required you would be right. From your answer above, look to simplify it !
anonymous
  • anonymous
looks like you can simplify it a little more think of it as \[\frac{ 2 }{ 6 }*\sqrt{6}\]

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