- anonymous

sinθ if cscθ= -2

- chestercat

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- anonymous

well what do you know about those 2 functions in relation to each other

- anonymous

That's all it says on my homework paper, 'Use the appropriate reciprocal identity to find each function value.'

- 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

or maybe call it reciprocal? im not good at my language

- anonymous

As far as I know it's reciprocal is csc=1/sin

- anonymous

alright then. what do you think the answer is

- anonymous

something like -1/2.

- anonymous

I just have different ones like tanθ if cotθ=sqrt6/2 and I don't really understand

- anonymous

so your correct on this problem. do you know about radians yet for the other problems?

- anonymous

WHat do you mean radians?

- anonymous

and cot and tan are reciprocals as well. radians-degrees. have you now done the unit circle yet?

- anonymous

I don't recall so I am going to have to say no.

- 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

Yea, so how would I go on in doing that? I am in Pre-cal if that makes a difference.

- 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

tell me where you are at on the second problem

- 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

hm. what is the reciprocal of
\[\frac{ \sqrt{6} }{ 2 }\]

- anonymous

Umm tbh I'm not really sure :/

- 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

Oh ok I see so now I have to get the reciprocal.

- anonymous

but it would be 6/2sqrt2

- 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

so if you multiply this
\[\frac{ 2(\sqrt{6)} }{ \sqrt{6}*\sqrt{6} }\]
that is how you rationalize the denominator

- anonymous

oh I see now, so then the answer is 2/√6, I don't have to rationalize anything I believe.

- anonymous

keeping in mind
\[\sqrt{6}*\sqrt{6}=\sqrt{36}\]

- anonymous

Ok, thanks a lot! really appreciate the help!

- anonymous

you cant have square roots in the denominator. you have to rationalize it like I did in the above step

- anonymous

see how I multiplied the top and bottom by
\[\sqrt{6}\]

- anonymous

so 2*sqrt6/6

- 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

looks like you can simplify it a little more think of it as
\[\frac{ 2 }{ 6 }*\sqrt{6}\]

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