Find the values of the six trigonometric functions of θ with the given constraint. Function Value cos θ = - 4/5 Constraint θ lies in Quadrant III

Solution:

Since Cos θ = -4/5  and it is given that the constraint lies in the third quadrant.

Since Cos θ = base/Hypotenuse

We have Base = 4 and hypotenuse = 5

Using the pythagorean relationship (Hypotenuse)² = (base)² + (perpendicular)² we have

(5)² = (4)² + (perpendicular)²

25 = 16 +  (perpendicular)²

(perpendicular)² = 25 - 16 = 9

Perpendicular = ± 3

The six trigonometric ratios are:

Sin θ =  Perpendicular/Hypotenuse = -3/5   (II quadrant - Sine is -ve)

Cos θ = -4/5 (Cosine too is negative in III quadrant)

Tan θ  = Perpendicular/base  = 3/4 (Tan function is +ve in third quadrant)

Cosec θ = Hypotenuse/ Perpendicular = - 5/3  (III quadrant   -   Cosec is -ve)

Sec θ  = Hypotenuse/Base = - 5/4  (III quadrant   -   Cosec is -ve)

Cot θ  = Base/Perpendicular = 4/3  (III quadrant   -   Cot is +ve)

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Find the values of the six trigonometric functions of θ with the given constraint. Function Value cos θ = - 4/5 Constraint θ lies in Quadrant III

Summary:

The values of the six trigonometric functions of θ with the given constraint. Function Value cos θ =  -4/5 are Sin θ = -3/5 ; Cos θ = -4/5; Tan θ = 3/4; Cosec θ = -5/3; Sec θ = -5/4; Cot θ  = 4/3