Respuesta :

jcherry99

Remark

The first thing to do is to find the sin(theta). When that is done, use an identity to find the cosine.

Step One

Find Sin(theta)

Sin(theta) = 1/csc(theta)

Csc(theta) = 3

Sin(theta) = 1/3

Step Two

Find the first quadrant value of Cos(theta)

Cos(theta) = sqrt(1 - sin^2(theta) )

Cos(theta) = sqrt(1 - (1/3)^2 )

Cos(theta) = sqrt(1 - 1/9)

Cos(theta) = sqrt(8/9)

Cos(theta) = sqrt(8)/sqrt(9)

Cos(theta) = 2*sqrt(2) / 3 <<<<< Answer for quad One

Step Three

Find Cos(theta) for all 4 quads.

Quad 4

Quad 4 is the same as quad 1

Quads 2 and 3

Cos(theta) = a minus value. In this case cos(theta) = - 2sqrt(2) / 3 It's the same value, just minus.

Answer Cos(theta) = - 2 sqrt(2) / 3

gmany

[tex] \csc\theta=\dfrac{1}{\sin\theta}\\\\\csc\theta=3\to\dfrac{1}{\sin\theta}=3\to\sin\theta=\dfrac{1}{3}\\\\\text{Use:}\ \sin^2x+\cos^2x=1\\\\\text{substitute}\\\\\left(\dfrac{1}{3}\right)^2+\cos^2\theta=1\\\\\dfrac{1}{9}+\cos^2\theta=1\ \ \ \ |-\dfrac{1}{9}\\\\\cos^2\theta=\dfrac{8}{9}\to\cos\theta=\pm\sqrt{\dfrac{8}{9}} [/tex]

[tex] \cos\theta=\pm\dfrac{\sqrt8}{\sqrt9}\to\cos\theta=\pm\dfrac{\sqrt{4\cdot2}}{3}\to\cos\theta=\pm\dfrac{\sqrt4\cdot\sqrt2}{3}\\\\\boxed{\cos\theta=-\dfrac{2\sqrt{2}}{3}\ \vee\ \cos\theta=\dfrac{2\sqrt2}{3}} } [/tex]

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