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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
 8 months 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.'
 8 months 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
 8 months ago

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

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

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

OscarGon1234 Group TitleBest ResponseYou've already chosen the best response.0
something like 1/2.
 8 months 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
 8 months 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?
 8 months ago

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

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

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

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

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

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

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

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

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

OscarGon1234 Group TitleBest ResponseYou've already chosen the best response.0
so 2*sqrt6/6
 8 months 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 !
 8 months 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 \[\frac{ 2 }{ 6 }*\sqrt{6}\]
 8 months ago
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