## anonymous 4 years ago evaluate the six trigonometric functions of the angle (see below for picture)

1. anonymous

2. anonymous

the point is (4, -3) see the picture above

3. anonymous

im just confused because the arrow goes all the way around so i dont know how to find answers

4. myininaya

|dw:1328392741044:dw|

5. anonymous

is that all i have to do

6. myininaya

$\sin(\theta)=\frac{opp}{hyp}=\frac{-3}{5}$

7. anonymous

and do i list negatives in the answers

8. anonymous

oh wait

9. myininaya

10. anonymous

i forgot thats not the angle. here is a new picture

11. anonymous

12. anonymous

feta is on the outside its the big one going around

13. anonymous

??????

14. myininaya

I'm confused Do you want to find theta or do you want us to evaluate theta using the 6 trig ratios?

15. anonymous

Do the operation myinaya suggested, then add 360 to get the angle that is there

16. anonymous

evalute using ratios. but im confused because the theta is on the outside and not the inside

17. myininaya

let me look at your pic again one sec

18. anonymous

ok look at the second one

19. myininaya

ok

20. myininaya

|dw:1328393266943:dw|

21. myininaya

$\sin(360^o-\theta^o)=\frac{opp}{hyp}=\frac{-3}{5}$ $\sin(360)\cos(\theta)-\sin(\theta) \cos(360)=\frac{-3}{5}$

22. myininaya

$(0) \cos(\theta)-\sin(\theta)(1)=\frac{-3}{5}$

23. myininaya

$\sin(\theta)=\frac{3}{5}$

24. myininaya

Does that make since?

25. myininaya

sense*?

26. anonymous

no sry i dont get what u just said

27. myininaya

which part?

28. myininaya

There are 360 degrees in circle we have theta already making up part of that 360 degrees 360-theta is left to make a full rotation

29. myininaya

Recall your sum and difference formulas

30. anonymous

ok but how do i solve for sin cos tan etc if its on the outside

31. myininaya

you will need those here

32. myininaya

I just did sin(theta) for you

33. myininaya

$\cos(360-\theta)=\frac{4}{5}$ $\cos(360)\cos(\theta)+\sin(360 )\sin(\theta)=\frac{4}{5}$

34. myininaya

$1\cos(\theta)+0 \sin(\theta)=\frac{4}{5} => \cos(\theta)=\frac{4}{5}$

35. anonymous

so is it the same as being on teh inside

36. myininaya

well sin(theta) does not equal sin(360-theta) but cos(theta) equals cos(360-theta) so it is not the same

37. myininaya

it is not the same because it is not always true and i showed this by showing sin(theta) does not equal sin(360-theta)

38. anonymous

but when i do it using theta on teh inside i get the same answers (sin is -3/5, cos is 4/5, tan is -3/4, etc

39. myininaya

no sin(theta)=3/5 sin(360-theta)=-3/5 3/5 does not equal -3/5

40. anonymous

ohh ok....thanks!!!

41. anonymous

can i ask you another question

42. anonymous

its kinda silly...but how do u simplify down (-4/$\sqrt{52}$)

43. myininaya

what is 13*4

44. anonymous

45. myininaya

$\sqrt{52}=\sqrt{4} \sqrt{13}=2 \sqrt{13}$

46. anonymous

see pic above...i got it down to there but then stuck

47. myininaya

$\frac{-4}{2 \sqrt{13}}=\frac{-2}{\sqrt{13}}=\frac{-2 \sqrt{13}}{13}$