1. ## Check work please...Trig functions

I think that I have got these right, but I would like to make sure...

2. All looks good to me at a glance.

Except for the second triangle. There is some reduction of the fraction you can do there, and I don't know if you know about this or not but generally you rearrange so the square root is on the top but if you haven't heard of that it's all fine as it is as far as i can see.

3. How do I simplify with the radical on the bottom? it probably needs to be simplified.... I don't really know how to do that

4. Your answers are deffo right, the only thing is to tidy up, but its not that important. Depends how mean your teacher is lol

ok, take for example $\sin(\theta)=\frac{18}{6\sqrt{13}}$

6 into 18 is 3

$\sin(\theta)=\frac{3}{\sqrt{13}}$

I guess that's probably only whats expected of you if it's school maths. It's not realy important, but generally you want the radical on the top because it's neater.
$\sin(\theta)=\frac{3}{\sqrt{13}}=\frac{3}{\sqrt{13 }}*\frac{\sqrt{13}}{\sqrt{13}}=\frac{3\sqrt{13}}{1 3}$

5. Thank you very much for all of your help

6. Ok, i have got it for sin, cos, and tan, but how do i do this with csc and sec
sin is 3 sqrt(13)/ 13
cos = 2 sqrt(13) / 13

7. with cosec and sec, the radical is already on top, so just divide out the 12's and 18's and 6's.

Show your working though, because I don't know what the teacher is expecting, so if you got the simplified and unsimplified answers you can't go wrong.

Don't worry, all is good, you've understood it all fine

8. Thank you very much! I really appreciate you taking the time to help me!

9. Is ok, Maybe you can help me with my quantum field theory because I'm totally lost lol

10. I now get 1 sqrt (13) / 3 for csc
and 1 sqrt (13) / 2 for sec

11. yep, well done. make sure the teacher can see your original answers to.

12. Thank you!