#1**0 **

Go online to this page and learn the relationship between the 6 Trig. functions:

http://www.efunda.com/math/trig_functions/trig_relation.cfm

Guest Mar 19, 2017

#2**+3 **

tangent = opposite / adjacent

a^{2 }+ b^{2} = c^{2}

6^{2} + 1^{2} = (hypotenuse)^{2}

__√(37) = hypotenuse__

sec = hypotenuse / adjacent

__sec = √(37) / 6 ≈ 1.014__

cos = adjacent / hypotenuse

cos = 6/√(37)

__cos = [6√(37)] / 37 ≈ 0.986__

cot = adjacent / opposite

__cot = 6 / 1 = 6__

sin = opposite / hypotenuse

sin = 1 / √(37)

__sin = √(37) / 37 ≈ 0.164__

csc = hypotenuse / opposite

__csc = √(37) / 1 = √(37) ≈ 6.083__

hectictar
Mar 19, 2017

#3**0 **

If the tangent of an angle is 1/6, there are a few ways to find the other trig values. First, cot(alpha) = 1/tan(alpha) so 1 over 1/6 is 6. The formula sin^{2}(alpha) + cos^{2}(alpha)=1 can be divided through by cos^{2} to give tan^{2}(alpha) + 1= sec^{2}(alpha). The problem with plugging the 1/6 into that is you do not know which part of the circle or graph it is in so sec(alpha) upon solving would be sqrt[1/36 + 1] = sqrt[37/36] = sqrt(37)/6. It might be negative depending on the original angle but let's just assume it is positive for now. That brings us to sin and cos. Since sec(alpha) = 1/cos(alpha), it = the reciprocal of sqrt(37)/6 which is 6*sqrt(37)/37. Finally plug that back into sin^{2}(alpha) + cos^{2}(alpha)=1 to solve for cos(alpha).

To summarize, use cot(alpha) = 1/tan(alpha) to get cot(alpha). Use sec(alpha)= sqrt(tan^{2}(alpha)+1) and then use that to get cos(alpha) since cos(alpha) = 1/sec(alpha). Finally, use sin^{2}(alpha) + cos^{2}(alpha)=1 to get sin(alpha). Since tan is positive in both quadrants 1 and 3, the sign on some of the trig functions might differ.

RetireeNow
Mar 19, 2017