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OscarGon1234
 2 years ago
sinθ if cscθ= 2
OscarGon1234
 2 years ago
sinθ if cscθ= 2

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

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

lonnie455rich
 2 years ago
Best ResponseYou've already chosen the best response.1ya, but you didn't answer my question. you have soh cah toa. and the three inverse functions. which one is the inverse of sin

lonnie455rich
 2 years ago
Best ResponseYou've already chosen the best response.1or maybe call it reciprocal? im not good at my language

OscarGon1234
 2 years ago
Best ResponseYou've already chosen the best response.0As far as I know it's reciprocal is csc=1/sin

lonnie455rich
 2 years ago
Best ResponseYou've already chosen the best response.1alright then. what do you think the answer is

OscarGon1234
 2 years ago
Best ResponseYou've already chosen the best response.0something like 1/2.

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

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

OscarGon1234
 2 years ago
Best ResponseYou've already chosen the best response.0WHat do you mean radians?

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

OscarGon1234
 2 years ago
Best ResponseYou've already chosen the best response.0I don't recall so I am going to have to say no.

lonnie455rich
 2 years ago
Best ResponseYou've already chosen the best response.1well. 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

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

lonnie455rich
 2 years ago
Best ResponseYou've already chosen the best response.1http://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

lonnie455rich
 2 years ago
Best ResponseYou've already chosen the best response.1tell me where you are at on the second problem

OscarGon1234
 2 years ago
Best ResponseYou've already chosen the best response.0All 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.

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

OscarGon1234
 2 years ago
Best ResponseYou've already chosen the best response.0Umm tbh I'm not really sure :/

napalmgrenade
 2 years ago
Best ResponseYou've already chosen the best response.0You 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

OscarGon1234
 2 years ago
Best ResponseYou've already chosen the best response.0Oh ok I see so now I have to get the reciprocal.

OscarGon1234
 2 years ago
Best ResponseYou've already chosen the best response.0but it would be 6/2sqrt2

lonnie455rich
 2 years ago
Best ResponseYou've already chosen the best response.1reciprocal 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

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

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

lonnie455rich
 2 years ago
Best ResponseYou've already chosen the best response.1keeping in mind \[\sqrt{6}*\sqrt{6}=\sqrt{36}\]

OscarGon1234
 2 years ago
Best ResponseYou've already chosen the best response.0Ok, thanks a lot! really appreciate the help!

lonnie455rich
 2 years ago
Best ResponseYou've already chosen the best response.1you cant have square roots in the denominator. you have to rationalize it like I did in the above step

lonnie455rich
 2 years ago
Best ResponseYou've already chosen the best response.1see how I multiplied the top and bottom by \[\sqrt{6}\]

napalmgrenade
 2 years ago
Best ResponseYou've already chosen the best response.0Some 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 !

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