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What are sin x, tan x, cot x, if 180<x<270 and cos x=-4/5?

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luna9219 | Student, College Freshman | eNoter

Posted January 28, 2011 at 3:26 PM via web

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What are sin x, tan x, cot x, if 180<x<270 and cos x=-4/5?

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giorgiana1976 | College Teacher | Valedictorian

Posted January 28, 2011 at 3:31 PM (Answer #1)

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The angle x is in the 3rd quadrant, so the value of the function sine is negative, but the values of teh functions tangent and cotangent are positive.

We'll apply the fundamental formula of trigonometry:

(sin x)^2 + (cos x)^2  =1

(sin x)^2 = 1  - (cos x)^2

(sin x)^2 = 1  - (-4/5)^2

(sin x)^2 = 1 - 16/25

(sin x)^2 = 9/25

sin x = -3/5

tan x = sin x/cos x

tan x = -3/5/-4/5

sin x = -3/5, tan x = 3/4 and cot x = 4/3

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justaguide | College Teacher | (Level 2) Distinguished Educator

Posted January 28, 2011 at 3:48 PM (Answer #2)

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We have cos x = -4/5. Also 180< x< 270. For these values of x sin x is negative and so is cos x.

We use the relation (cos x)^2 + (sin x)^2 = 1

=> (-4/5)^2 + (sin x)^2 = 1

=> (sin x)^2 = 1 - (16 / 25)

=> (sin x)^2 = 9/25

=> sin x = -3/5

tan x = sin x / cos x = (-3/5)/(-4/5) = 3/4

cot x = cos x / sin x = 4/3

sin x = -3/5 , tan x = 3/4 and cot x  = 4/3.

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