Find cot(x) and sec(x), while sin(x) = (1/3), tan(x)<0

Question

 I need help understanding this because I don't know how to figure this out with the given information.

Answer 1

Hi Tori,

First draw a right triangle, you are given that sin(x)=1/3 

 

*

*   *

*       *

*            *    hyp=3

* opp=1      *

*                     *

*                          *

*                      X   (  *

*******************

            adj=?

 

sin(x) = 1/3 =opp/hy

opp = 1, adj =3

Using the 

hyp² = opp² + adj²

3²    =  1²    + adj² 

9      = 1 + adj² 

8      = adj²  this gives adj =±√8  =± 2√2

Since tan(x) < 0  and sin(x)> 0, then cos(x)<0 so we pic  adj = - 2√2

cos(x) = -  2√2/3

cot(x) = 1 /tan(x) = cos(x)/sin(x) ( 2√2/3)(1/3)= - 2√2
sec(x) = 1/cos(x)=1/(- 2√2/3))=-3/ 2√2

 

cot(x) = - 2√2 , and sec(x) = -3/ 2√2

 

Answer 2

Since sinx > 0 and tanx < 0, x lies in quadrant 2.  

 

cosx = -√(1 - sin²x) = -√(1 - 1/9) = -√(8/9) = -2√2/3

 

secx = 1/cosx = -3/(2√2) = -3√2/4

 

cotx = cosx/sinx = (-2√2/3)/(1/3) = -2√2

Answer 3

Sin^2x + cos^2x = 1 that's the identity im going to use

 

cos^2 x = 1 - 1/9 = 8/9

 

sec^2 x  = 1 / (8/9) = 9/8

 

 

sec x = 3 / sqrt 8

 

 

cot x = cos x / sin x

 

= sqrt 8 / 3) / (1/3)

 

= sqrt 8

 

Those are your two answers hope it helped