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lonnie455rich Group TitleBest ResponseYou've already chosen the best response.1
well what do you know about those 2 functions in relation to each other
 one year ago

OscarGon1234 Group TitleBest ResponseYou've already chosen the best response.0
That's all it says on my homework paper, 'Use the appropriate reciprocal identity to find each function value.'
 one year ago

lonnie455rich Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

lonnie455rich Group TitleBest ResponseYou've already chosen the best response.1
or maybe call it reciprocal? im not good at my language
 one year ago

OscarGon1234 Group TitleBest ResponseYou've already chosen the best response.0
As far as I know it's reciprocal is csc=1/sin
 one year ago

lonnie455rich Group TitleBest ResponseYou've already chosen the best response.1
alright then. what do you think the answer is
 one year ago

OscarGon1234 Group TitleBest ResponseYou've already chosen the best response.0
something like 1/2.
 one year ago

OscarGon1234 Group TitleBest ResponseYou've already chosen the best response.0
I just have different ones like tanθ if cotθ=sqrt6/2 and I don't really understand
 one year ago

lonnie455rich Group TitleBest ResponseYou've already chosen the best response.1
so your correct on this problem. do you know about radians yet for the other problems?
 one year ago

OscarGon1234 Group TitleBest ResponseYou've already chosen the best response.0
WHat do you mean radians?
 one year ago

lonnie455rich Group TitleBest ResponseYou've already chosen the best response.1
and cot and tan are reciprocals as well. radiansdegrees. have you now done the unit circle yet?
 one year ago

OscarGon1234 Group TitleBest ResponseYou've already chosen the best response.0
I don't recall so I am going to have to say no.
 one year ago

lonnie455rich Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

OscarGon1234 Group TitleBest ResponseYou've already chosen the best response.0
Yea, so how would I go on in doing that? I am in Precal if that makes a difference.
 one year ago

lonnie455rich Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

lonnie455rich Group TitleBest ResponseYou've already chosen the best response.1
tell me where you are at on the second problem
 one year ago

OscarGon1234 Group TitleBest ResponseYou've already chosen the best response.0
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.
 one year ago

lonnie455rich Group TitleBest ResponseYou've already chosen the best response.1
hm. what is the reciprocal of \[\frac{ \sqrt{6} }{ 2 }\]
 one year ago

OscarGon1234 Group TitleBest ResponseYou've already chosen the best response.0
Umm tbh I'm not really sure :/
 one year ago

napalmgrenade Group TitleBest ResponseYou've already chosen the best response.0
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
 one year ago

OscarGon1234 Group TitleBest ResponseYou've already chosen the best response.0
Oh ok I see so now I have to get the reciprocal.
 one year ago

OscarGon1234 Group TitleBest ResponseYou've already chosen the best response.0
but it would be 6/2sqrt2
 one year ago

lonnie455rich Group TitleBest ResponseYou've already chosen the best response.1
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
 one year ago

lonnie455rich Group TitleBest ResponseYou've already chosen the best response.1
so if you multiply this \[\frac{ 2(\sqrt{6)} }{ \sqrt{6}*\sqrt{6} }\] that is how you rationalize the denominator
 one year ago

OscarGon1234 Group TitleBest ResponseYou've already chosen the best response.0
oh I see now, so then the answer is 2/√6, I don't have to rationalize anything I believe.
 one year ago

lonnie455rich Group TitleBest ResponseYou've already chosen the best response.1
keeping in mind \[\sqrt{6}*\sqrt{6}=\sqrt{36}\]
 one year ago

OscarGon1234 Group TitleBest ResponseYou've already chosen the best response.0
Ok, thanks a lot! really appreciate the help!
 one year ago

lonnie455rich Group TitleBest ResponseYou've already chosen the best response.1
you cant have square roots in the denominator. you have to rationalize it like I did in the above step
 one year ago

lonnie455rich Group TitleBest ResponseYou've already chosen the best response.1
see how I multiplied the top and bottom by \[\sqrt{6}\]
 one year ago

OscarGon1234 Group TitleBest ResponseYou've already chosen the best response.0
so 2*sqrt6/6
 one year ago

napalmgrenade Group TitleBest ResponseYou've already chosen the best response.0
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 !
 one year ago

lonnie455rich Group TitleBest ResponseYou've already chosen the best response.1
looks like you can simplify it a little more think of it as \[\frac{ 2 }{ 6 }*\sqrt{6}\]
 one year ago
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