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given sign (-0)=1/5 and tan 0= (square root of 6)/12 what is the value of cos 0

Sagot :

Trigonometry

We know that

sin(-θ) = - sinθ

and that

tanθ = sinθ / cosθ

Then

sin(-θ) = - sinθ = 1/5

↓

sin(θ) = - 1/5

We want to find cosθ. Using the second formula we have that

cosθ = sinθ / tanθ

We replace sin(θ) = - 1/5 and tanθ =√6/12:

[tex]\begin{gathered} \cos \theta=\frac{\sin\theta}{\tan\theta} \\ \cos \theta=\frac{-\frac{1}{5}}{\frac{\sqrt[]{6}}{12}} \\ =-\frac{1}{5}\cdot\frac{12}{\sqrt[]{6}} \\ =-\frac{12}{5\sqrt[]{6}} \\ =-\frac{12\cdot\sqrt[]{6}}{5\sqrt[]{6}\cdot\sqrt[]{6}} \\ =-\frac{12\sqrt[]{6}}{5\cdot6} \\ =-\frac{2\sqrt[]{6}}{5} \end{gathered}[/tex]

Answer: C

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