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sinθ if cscθ= -2

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well what do you know about those 2 functions in relation to each other
That's all it says on my homework paper, 'Use the appropriate reciprocal identity to find each function value.'
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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or maybe call it reciprocal? im not good at my language
As far as I know it's reciprocal is csc=1/sin
alright then. what do you think the answer is
something like -1/2.
I just have different ones like tanθ if cotθ=sqrt6/2 and I don't really understand
so your correct on this problem. do you know about radians yet for the other problems?
WHat do you mean radians?
and cot and tan are reciprocals as well. radians-degrees. have you now done the unit circle yet?
I don't recall so I am going to have to say no.
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
Yea, so how would I go on in doing that? I am in Pre-cal if that makes a difference. print that off. it will do wonders for your trig identities. do you know how to rationalize the denominator
tell me where you are at on the second problem
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.
hm. what is the reciprocal of \[\frac{ \sqrt{6} }{ 2 }\]
Umm tbh I'm not really sure :/
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|
Oh ok I see so now I have to get the reciprocal.
but it would be 6/2sqrt2
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
so if you multiply this \[\frac{ 2(\sqrt{6)} }{ \sqrt{6}*\sqrt{6} }\] that is how you rationalize the denominator
oh I see now, so then the answer is 2/√6, I don't have to rationalize anything I believe.
keeping in mind \[\sqrt{6}*\sqrt{6}=\sqrt{36}\]
Ok, thanks a lot! really appreciate the help!
you cant have square roots in the denominator. you have to rationalize it like I did in the above step
see how I multiplied the top and bottom by \[\sqrt{6}\]
so 2*sqrt6/6
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 !
looks like you can simplify it a little more think of it as \[\frac{ 2 }{ 6 }*\sqrt{6}\]

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