## OscarGon1234 one year ago sinθ if cscθ= -2

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1. lonnie455rich

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

2. OscarGon1234

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

3. lonnie455rich

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

4. lonnie455rich

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

5. OscarGon1234

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

6. lonnie455rich

alright then. what do you think the answer is

7. OscarGon1234

something like -1/2.

8. OscarGon1234

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

9. lonnie455rich

10. OscarGon1234

11. lonnie455rich

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

12. OscarGon1234

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

13. lonnie455rich

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

14. OscarGon1234

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

15. lonnie455rich

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

16. lonnie455rich

tell me where you are at on the second problem

17. OscarGon1234

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.

18. lonnie455rich

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

19. OscarGon1234

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|

21. OscarGon1234

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

22. OscarGon1234

but it would be 6/2sqrt2

23. lonnie455rich

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

24. lonnie455rich

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

25. OscarGon1234

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

26. lonnie455rich

keeping in mind $\sqrt{6}*\sqrt{6}=\sqrt{36}$

27. OscarGon1234

Ok, thanks a lot! really appreciate the help!

28. lonnie455rich

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

29. lonnie455rich

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

30. OscarGon1234

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 !

32. lonnie455rich

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