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Given sin (-theta) = 1/5 and tan theta = sqrt.6/12
what is the value of cos theta?

- sqrt.6/60

2 sqrt.6/5

-2 sqrt.6/5

Sqrt.6/60

Sagot :

Given  [tex]sin (\theta) = \frac{1}{5}[/tex], and[tex]tan \theta = \frac{\sqrt{6} }{12}[/tex], then:

[tex]cos\theta=\frac{2\sqrt{6} }{5}[/tex]

Computations on trigonometric identities

From the details provided:

[tex]sin (\theta) = \frac{1}{5}\ \\\\tan \theta = \frac{\sqrt{6} }{12} \times \frac{\sqrt{6} }{\sqrt{6} }\\\\tan \theta = \frac{\sqrt{6} \times \sqrt{6} }{12 \times \sqrt{6} }\\\\tan \theta =\frac{1}{2\sqrt{6} }[/tex]

From the relationship above:

Opposite = 1

Adjacent = [tex]2\sqrt{6}[/tex]

Hypotenuse = 5

[tex]cos\theta=\frac{Adjacent}{Hypotenuse} \\\\cos\theta=\frac{2\sqrt{6} }{5}[/tex]

Learn more on trigonometry here: https://brainly.com/question/20519838

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