Video Transcript
The terminal side of angle 𝐴𝑂𝐵
in standard position intersects with the unit circle at the point 𝐵 with
coordinates negative 𝑥, negative 𝑥, where 𝑥 is a positive number. Find sin of 𝜃.
First, let’s sketch a unit circle
to help us visualize what’s happening here. The unit circle has a center at the
origin and a radius of one. Our coordinate 𝐵 is negative 𝑥,
negative 𝑥, where 𝑥 is a positive number. This means that our point 𝐵 will
be in the third quadrant, 𝑥 units to the left of the origin and 𝑥 units below the
origin. Since angle 𝐴𝑂𝐵 is in standard
position, its initial side is the positive 𝑥-axis and its terminal side will go
through the point 𝐵.
And for angles in standard
position, they’ll be calculated counterclockwise from the origin. Before we can calculate the sin of
𝜃, we’ll calculate what’s called a reference angle from point 𝐵. This is the angle created with
point 𝐵 and the negative 𝑥-axis. Let’s call this angle 𝛼. From there, we can use our
knowledge of right triangle trigonometry to find the angle 𝛼. This right triangle has two side
lengths that both measure 𝑥 and a hypotenuse of one. We know that the hypotenuse is one
because point 𝐵 falls on the unit circle, and the distance between 𝐵 and the
center of this circle is one.
To solve for 𝑥, we can use the
Pythagorean theorem, which says 𝑎 squared plus 𝑏 squared equals 𝑐 squared. In our triangle, this is 𝑥 squared
plus 𝑥 squared equals one squared. Two 𝑥 squared equals one. Dividing both sides by two, we find
that 𝑥 squared equals one over two. Taking the square root of both
sides, we find that 𝑥 equals the square root of one over two, which simplifies to
one over the square root of two. We know that the sin ratio is equal
to the opposite side length over the hypotenuse. And therefore, the sin of 𝛼 will
be equal to 𝑥 over one, or simply 𝑥. This makes the sin of 𝛼 one over
the square root of two.
And this is the sin for our
reference angle. In order to find accurately the sin
of our 𝜃 relationship with the initial side being the 𝑥-axis and the terminal side
being point 𝐵, we use a CAST diagram. In the first quadrant, all trig
values are positive. In the second quadrant, only the
sine value is positive. In the third quadrant, only the
tangent value is positive. And in the fourth quadrant, the
cosine value is positive.
Since our point 𝐵, our terminal
side, is in quadrant three, the sin of angle 𝐴𝑂𝐵 will be negative. We use the absolute value of our
reference angle, and the CAST diagram tells us that sin of 𝜃 is negative. The sin of 𝜃 equals negative one
over the square root of two.
