Finding Trigonometric Values Given One Value

[tristatefabricatorsinc]

Question is...

Find cos x if tan x = 3.1 and x is in quadrant III


Here are my thoughts....

Am I correct or not?

In quadrant IV, tangent is y/x, so I am assuming I need to use the pythagorean theorm to find the last side of the triangle.

So I got tangent as being 3.1 / 1

to get the remaining side I did

1^2 + 3.1^2 = c^2
1 + 9.61 = c^2
10.61 = c^2
sqrt(10.61) = 3.26 = c

so if I need cos x I would say it is 1 / 3.26
or .307 divided out and it would be positive since cos is positive in Quadrant IV.


Thank You!

 

[skeeter]

sec^2(x) = 1 + tan^2(x)
cos^2(x) = 1/[1 + tan^2(x)]
cos(x) = (+/-) 1/sqrt[1 + tan^2(x)]

you initially stated that x was in quad III, therefore cos(x) < 0.
this is also confirmed by the fact that tan(x) = 3.1 > 0 in quad III

cos(x) = -1/sqrt[1 + 3.1^2] = -.3070

 

[tristatefabricatorsinc]

Oops,

I was using quadrant IV when I was supposed to be using quadrant III as I stated in the problem. So was I correct in the way I was solving it as opposed to the negative?

 

[wjm11]

I was using quadrant IV when I was supposed to be using quadrant III as I stated in the problem. So was I correct in the way I was solving it as opposed to the negative?

Yes, Tristate, your approach/logic were correct.