Question

Cos(−θ)=√3/3, sinθ<0

Answer 1

Answer:

$\sin\theta=-\frac{\sqrt{6}}{3}$

Step-by-step explanation:

The given trigonometric equation is $\cos(-\theta)=\frac{\sqrt{3} }{3}$.

We can either use the Pythagorean identity or the right angle triangle  to solve for $\sin\theta$.

According to the Pythagorean identity,

$\cos^2\theta+\sin^2\theta=1$

Recall that, the cosine function is an even function, therefore

$\cos(-\theta)=\cos(\theta)$

$\Rightarrow \cos(\theta)=\frac{\sqrt{3} }{3}$.

We substitute this value in to the above Pythagorean identity to get;

$(\frac{\sqrt{3}}{3})^2+\sin^2\theta=1$

$\Rightarrow \frac{3}{9}+\sin^2\theta=1$

$\Rightarrow \sin^2\theta=1-\frac{3}{9}$

$\Rightarrow \sin^2\theta=\frac{6}{9}$

$\Rightarrow \sin\theta=\pm \sqrt{\frac{6}{9}}$

$\Rightarrow \sin\theta=\pm \frac{\sqrt{6}}{3}$

But we were given that,

$\sin\theta\:$, so we choose the negative value.

$\Rightarrow \sin\theta=-\frac{\sqrt{6}}{3}$

The correct answer is B