## anonymous one year ago given the sin a=-2/3 and a is in quadrant what are the exact values of the following trigonometric functions of a? cos a= tan a= csc a=

1. anonymous

2. anonymous

3

3. anonymous

Sin=y/r Cos=x/r Tan=y/x Csc=r/r

4. anonymous

|dw:1433564957553:dw|

5. anonymous

And remember that in quadrant 3 only tang is positive

6. anonymous

Sinx is negative in the 3rd quadrant

7. anonymous

and also in the fourth quadrant

8. anonymous

sin=-2/3

9. anonymous

Do you know how to approach this question ?

10. anonymous

has anyone suggested drawing a triangle yet? it would really make things easy

11. anonymous

no not really

12. anonymous

the do it label the "opposite side " 2, the hypotenuse 3, and use pythagoras to find the length of the adjacent side

13. anonymous

|dw:1433565255841:dw|

14. anonymous

5

15. anonymous

oh no!

16. anonymous

$2^2+?^2=3^2\\ ?=\sqrt{3^2-2^2}$

17. anonymous

1^2

18. anonymous

is it right @satellite73

19. anonymous

no not at all

20. anonymous

what is $$3^2$$?

21. anonymous

9

22. anonymous

ok and $$2^2$$?

23. anonymous

4

24. anonymous

and $9-4$?

25. anonymous

5

26. anonymous

so what is $\sqrt{3^2-2^2}$?

27. anonymous

5^2?

28. anonymous

no dear not $$5^2$$ but rather $$\sqrt 5$$

29. anonymous

30. anonymous

then what next

31. anonymous

Then you use the sides to find the rest of the trig functions

32. anonymous

|dw:1433566350237:dw|

33. anonymous

now you can take any trig ratio you want, just remember what quadrant you are in to make sure to know if it is positive or negative

34. anonymous

so for cos a

35. anonymous

so for cos a its 5/3 @satellite73

36. anonymous

and for tan a 2/sqrt5

37. anonymous

You can't leave a radical in the denominator. You would have to multiply the 2/sqrt5 by sqrt5/sqrt5

38. anonymous

so like 3/2sqrt5