Question

Given cos 0 = √3 /3 and sin 0<0 .
What is the value of sin 0 ?

Answer 1

Answer:

Choice 4 down

Step-by-step explanation:

The cosine is positive and we are told that the sin is negative.  That means that we are in QIV, where x is positive (cos) and y is negative (sin).  If we draw a right triangle in this quadrant, we will label the x axis with a 1 (this is the side adjacent to the reference angle for cos), and the hypotenuse as square root of 3.  The sin ratio is the side opposite the reference angle over the hypotenuse.  That means we need to solve for the side opposite using Pythagorean's theorem:

$\sqrt{3}^2=1^2+b^2$ and

$3=1+b^2$ so

$b^2=2$ and

$b=\sqrt{2}$

But since we are in QIV and that is a y value, it is negative.

Therefore,

$sin\theta=-\frac{\sqrt{2} }{\sqrt{3} }$

Rationalizing, you get

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