Answer:
[tex]\sin\theta=-\frac{\sqrt{6}}{3}[/tex]
Step-by-step explanation:
The given trigonometric equation is [tex]\cos(-\theta)=\frac{\sqrt{3} }{3}[/tex].
We can either use the Pythagorean identity or the right angle triangle to solve for [tex]\sin\theta[/tex].
According to the Pythagorean identity,
[tex]\cos^2\theta+\sin^2\theta=1[/tex]
Recall that, the cosine function is an even function, therefore
[tex]\cos(-\theta)=\cos(\theta)[/tex]
[tex]\Rightarrow \cos(\theta)=\frac{\sqrt{3} }{3}[/tex].
We substitute this value in to the above Pythagorean identity to get;
[tex](\frac{\sqrt{3}}{3})^2+\sin^2\theta=1[/tex]
[tex]\Rightarrow \frac{3}{9}+\sin^2\theta=1[/tex]
[tex]\Rightarrow \sin^2\theta=1-\frac{3}{9}[/tex]
[tex]\Rightarrow \sin^2\theta=\frac{6}{9}[/tex]
[tex]\Rightarrow \sin\theta=\pm \sqrt{\frac{6}{9}}[/tex]
[tex]\Rightarrow \sin\theta=\pm \frac{\sqrt{6}}{3}[/tex]
But we were given that,
[tex]\sin\theta\:<0\:[/tex], so we choose the negative value.
[tex]\Rightarrow \sin\theta=-\frac{\sqrt{6}}{3}[/tex]
The correct answer is B
