## Mirandaregina one year ago If csc theta = √5, then what is cot theta?

1. Nnesha

what is the reciprocal of csc ?

2. Mirandaregina

@Nnesha Sine

3. Nnesha

not just sin :=) 1/sin = csc so csc = ?

4. anonymous

In addition to sine, cosine, and tangent, there are three other trigonometric functions you need to know for the Math IIC: cosecant, secant, and cotangent. These functions are simply the reciprocals of sine, cosine, and tangent. Cotangent is the reciprocal of tangent.

5. Mirandaregina

@Nnesha so csc = 1/sine

6. Nnesha

yes right so if csc = sqrt{5} then sin = ???

7. Nnesha

remember thy are reciprocal of each other

8. Mirandaregina

@Nnesha then sin = 1/√5 = √5/5 ?

9. phi

draw a triangle and label its opposite side 1 and its hypotenuse sqr(5)

10. Nnesha

hmm $\csc = \frac{ \sqrt{5} }{ 1 } ~~~~~ \sin=\frac{ 1 }{ \sqrt{5} }$ so draw right triangle for now it's okay leave the radical sign at the denominator |dw:1440603990058:dw| $\rm sin \rm \theta = \frac{ opposite }{ hypotenuse }~~~~ \cos \theta = \frac{ adjacent }{ hypotenuse } ~~\tan \theta = \frac{ opposite }{ adjacent }$

11. Mirandaregina

@Nnesha so if sin theta = opp/hyp, and opp is 1 and hyp is √5, is it √5/5?

12. Nnesha

yes right sin = opp .hyp|dw:1440604248339:dw| apply the Pythagorean theorem to find 3rd side of right triangle

13. phi

yes you could use sqr(5) for the opposite side and 5 for the hypotenuse but 1 and sqr(5) seem easier (smaller numbers) but either set will give the same answer.

14. Mirandaregina

@Nnesha 1^2 + √5^2 = 6

15. phi

c^2 = a^2 + b^2 the unknown is a leg.

16. Nnesha

c=hypotenuse(longest leg of right triangle )

17. Mirandaregina

@Nnesha @phi did I use the Pythagorean Theorem correctly? doesn't a = 1 and b = √5 ?

18. phi

no a and b are the legs you know a and c (and you need to find b, the other leg)

19. phi

you know the hypotenuse (because sin is opp/hyp) you know a side and the hypotenuse

20. Nnesha

|dw:1440604696536:dw|

21. Mirandaregina

@phi how do I find b?

22. anonymous

quick question is 1/3 a rational number

23. phi

a^2 + b^2 = c^2 put in numbers for a and c can you do that ?

24. Mirandaregina

@phi but does a = 1? because before I said 1^2 + √5^2 = 6

25. phi

a=1 and c= $$\sqrt{5}$$ put those in the formula $a^2 + b^2 = c^2$

26. phi

notice c is on the other side of the = sign from a and b

27. Nnesha

i guess she is trying to say that sqrt of 5 +1 equal to 6

28. Nnesha

square of sqrt 5**

29. phi

when you write 1^2 + √5^2 = c^2 a^2 + b^2 = c^2 that means a is 1 and b is sqr(5). but that is not what we want

30. Mirandaregina

@phi I understand, thanks :) so I'm not sure if I did this right but I have 1^2 + b^2 = √5 b + 1^2 = √5 b^2 + 1 - 1 = √5 - 1 b^2 = √1.236068

31. phi

almost, but the formula says c^2 so you want (sqrt(5))^2

32. Mirandaregina

@phi as is √1.111786 = 1.0544126

33. Mirandaregina

*as in

34. phi

you wrote 1^2 + b^2 = √5 but what it should be is 1^2 + b^2 = (√5)^2 now find b

35. Mirandaregina

@phi oh my mistake 1^2 + b^2 = (√5)^2 b = 2

36. phi

yes. now you know all 3 sides of the triangle |dw:1440605865953:dw|

37. phi

the want the cotangent of theta (I would write the tangent of theta, then "flip it" to make it the cotangent) or you can remember cotangent is adj/opp

38. Mirandaregina

@phi cot theta = adj/opp = 2/1 = 2

39. phi

yes

40. Mirandaregina

@phi so that's the overall answer? I got that answer when I first tried it but I thought it was wrong! lol

41. phi

now you know how to do it so you believe it

42. Mirandaregina

@phi @Nnesha thank you for all your help and patience everyone :)

43. Nnesha

my pleasure o^_^o