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# tan alpha = 1/6, what is sec, cos, cot, sin, and csc?

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tan alpha = 1/6, what is sec, cos, cot, sin, and csc?

Guest Mar 19, 2017
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#1
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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
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a+ b2 = c2

62 + 12 = (hypotenuse)2

√(37) = hypotenuse

sec = √(37) / 6 ≈ 1.014

cos = 6/√(37)

cos = [6√(37)] / 37 ≈ 0.986

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
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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 sin2(alpha) + cos2(alpha)=1 can be divided through by cos2 to give tan2(alpha) + 1= sec2(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 sin2(alpha) + cos2(alpha)=1 to solve for cos(alpha).

To summarize, use cot(alpha) = 1/tan(alpha) to get cot(alpha).  Use sec(alpha)= sqrt(tan2(alpha)+1) and then use that to get cos(alpha) since cos(alpha) = 1/sec(alpha).  Finally, use sin2(alpha) + cos2(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

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