Video Transcript
Which of the following is equal to
sin of 𝜃? (A) sin of three 𝜋 over two plus
𝜃. (B) cos of three 𝜋 over two plus
𝜃. (C) sin of 𝜋 over two plus 𝜃. Or (D) cos of 𝜋 over two plus
𝜃.
We begin by sketching an arbitrary
angle 𝜃 in standard position on an 𝑥𝑦-coordinate plane. Then, we sketch a unit circle,
centered at the origin.
We recall that an angle 𝜃 in
standard position will intersect the unit circle at a point with the 𝑥-coordinate
cos of 𝜃 and the 𝑦-coordinate sin of 𝜃. We then sketch the reference
triangle for angle 𝜃 along the positive side of the 𝑥-axis. We know that the hypotenuse of a
reference triangle is a radius of the unit circle and thus has a length of one. We also note that the side of our
reference triangle, which runs along the positive 𝑥-axis, corresponds to cos of
𝜃.
In the first and fourth quadrants,
the 𝑥-coordinate is always positive. But in the second or third
quadrant, 𝑥 is negative. The side of the reference triangle
opposite 𝜃 corresponds to the 𝑦-coordinate or sin of 𝜃. The 𝑦-coordinate is positive in
the first and second quadrants but negative in the third and fourth quadrants.
Now that we have reviewed where
sine is found in relation to angle 𝜃 and its reference triangle, we will sketch a
congruent triangle for each of the other two angles mentioned in this question,
three 𝜋 over two plus 𝜃 and 𝜋 over two plus 𝜃.
To locate the angle three 𝜋 over
two plus 𝜃, we locate the angle three 𝜋 over two and rotate 𝜃 in the positive
direction. Three 𝜋 over two is located on the
negative 𝑦-axis. From there, we rotate
counterclockwise by 𝜃. This angle intersects the unit
circle at the coordinate point cos of three 𝜋 over two plus 𝜃, sin of three 𝜋
over two plus 𝜃.
Next, we locate 𝜋 over two plus 𝜃
by rotating 𝜃 past the positive 𝑦-axis in the counterclockwise direction. This angle intersects the unit
circle at the coordinate point cos of 𝜋 over two plus 𝜃, sin of 𝜋 over two plus
𝜃. Using the angle 𝜋 over two plus
𝜃, we can sketch a right triangle that is congruent to our original reference
triangle. But this line will be connected to
the 𝑦-axis.
The exact location of this new
right triangle is found by simply rotating the original reference triangle 90
degrees or 𝜋 over two radians. So it has the same dimensions as
the original reference triangle, which had a longer side of cos of 𝜃 and a shorter
side of sin of 𝜃. Because 𝜋 over two plus 𝜃 is in
the second quadrant, the 𝑥-coordinate will be negative. Since the shorter side of the new
triangle now represents the 𝑥-value, it will be negative sin of 𝜃. And the other side of the triangle
represents the 𝑦-value, which will still be positive cos of 𝜃. So our coordinate point can be
written in terms of 𝜃 as negative sin of 𝜃, cos of 𝜃.
We have just shown that cos of 𝜋
over two plus 𝜃 equals negative sin of 𝜃 and sin of 𝜋 over two plus 𝜃 equals cos
of 𝜃. Thus, we eliminate options (C) and
(D) because neither expression equals sin of 𝜃.
Now we return to the angle three 𝜋
over two plus 𝜃. Here, we will sketch another
triangle congruent to our reference triangle. This new triangle is placed along
the negative 𝑦-axis. In quadrant four, the 𝑦-coordinate
will be negative, but the 𝑥-coordinate is once again positive, as it was in the
first quadrant. We find the coordinate point in
terms of 𝜃 to be sin of 𝜃, negative cos of 𝜃. This means that cos of three 𝜋
over two plus 𝜃 equals sin of 𝜃 and sin of three 𝜋 over two plus 𝜃 equals
negative cos of 𝜃.
In conclusion, we have shown that
option (B) is the expression that equals sin of 𝜃. This is because on the unit circle,
the 𝑥-coordinate of the point given by the angle three 𝜋 over two plus 𝜃 is the
same as the 𝑦-coordinate given by the angle 𝜃.
