### Video Transcript

In this video, we will learn how to
use the fact that the quadrant where an angle lies determines the signs of its sine,
cosine, and tangent and solve trigonometric equations. We will begin by considering the
unit circle.

The unit circle is a circle with a
radius of one whose center lies at the origin of a coordinate plane. For any point 𝑥, 𝑦 on the unit
circle, a right triangle can be formed as in the diagram. The hypotenuse of this right
triangle makes an angle 𝜃 with the positive 𝑥-axis. Using right triangle trigonometry,
we can define the trigonometric functions in terms of the unit circle. Recall sin 𝜃 equals the opposite
over the hypotenuse, cos 𝜃 equals the adjacent over the hypotenuse, and tan 𝜃
equals the opposite over the adjacent.

Since the opposite side has length
𝑦, the adjacent side has length 𝑥, and the hypotenuse has length one, we have the
following equations. So 𝑦 is equal to sin 𝜃, 𝑥 is
equal to cos 𝜃, and 𝑦 over 𝑥 is equal to tan 𝜃. We note that tan 𝜃 is not defined
when 𝑥 equals zero. We also observe that, while we have
derived these definitions for an angle 𝜃 in quadrant I, they hold for an angle in
any quadrant. Summarizing, the 𝑥- and
𝑦-coordinates of a point on the unit circle given by an angle 𝜃 are defined by 𝑥
equals cos 𝜃 and 𝑦 equals sin 𝜃.

In our first example, we will
demonstrate how we can use these definitions of trigonometric functions in the unit
circle to find exact values, given information about the terminal side of an
angle.

Find sin 𝜃, given that 𝜃 is
in standard position and its terminal side passes through the point
three-fifths, negative four-fifths.

An angle is said to be in
standard position if the vertex is at the origin and the initial side lies on
the positive 𝑥-axis. The angle is measured in a
counterclockwise direction from the initial side to the terminal side. Hence, the angle 𝜃 is as
shown. We draw a right triangle with
side lengths of three-fifths units and four-fifths units.

Now we can use the Pythagorean
theorem to calculate the value of the missing dimension in the triangle. Three-fifths squared plus
four-fifths squared is equal to 𝑐 squared. The left-hand side simplifies
to nine over 25 plus 16 over 25. So 𝑐 squared is equal to 25
over 25, which equals one. Square rooting both sides, and
since 𝑐 must be positive, we have 𝑐 is equal to one. This tells us that the point
three-fifths, negative four-fifths lies on the unit circle.

We recall that the 𝑥- and
𝑦-coordinates of a point on the unit circle given by an angle 𝜃 are defined by
𝑥 equals cos 𝜃 and 𝑦 equals sin 𝜃. Sin 𝜃 is therefore equal to
the value of the 𝑦-coordinate of the point, which is negative four-fifths. The correct answer is sin 𝜃
equals negative four-fifths.

In addition to the standard
trigonometric functions, it is also possible to define the reciprocal trigonometric
functions. The reciprocal of a number 𝑥 is
one over 𝑥. For an angle 𝜃, where 𝜃 is a real
number, the reciprocal trigonometric functions are as follows. csc 𝜃 equals one
over sin 𝜃, sec 𝜃 equals one over cos 𝜃, and cot 𝜃 equals one over tan 𝜃, where
sin 𝜃, cos 𝜃, and tan 𝜃 are not equal to zero.

Since we can write the standard
trigonometric functions in terms of the unit circle, it is also possible for us to
write the reciprocal functions in terms of the unit circle. That is, let us once more consider
a point 𝑥, 𝑦 on the unit circle with angle 𝜃 to the positive 𝑥-axis, where 𝑥 is
equal to cos 𝜃 and 𝑦 equals sin 𝜃. Then the reciprocal trigonometric
functions can be written as follows. csc 𝜃 equals one over 𝑦, sec 𝜃 equals one
over 𝑥, and cot 𝜃 equals 𝑥 over 𝑦, where the denominators cannot equal zero.

Let us consider an example where we
can use the unit circle to find the exact value of the secant function.

Find sec 𝜃, given that 𝜃 is
in standard position and its terminal side passes through the point four-fifths,
three-fifths.

An angle is said to be in
standard position if the vertex is at the origin and the initial side lies on
the positive 𝑥-axis. The angle is measured in a
counterclockwise direction from the initial side to the terminal side. Hence, the angle 𝜃 is as
shown.

To calculate the value of sec
𝜃, we will begin by determining whether the point with coordinates four-fifths,
three-fifths lies on the unit circle. To do this, we can consider a
right triangle with side lengths of four-fifths units and three-fifths units as
shown. Then we can calculate the
length of the hypotenuse, 𝑐, by using the Pythagorean theorem.

We have four-fifths squared
plus three-fifths squared is equal to 𝑐 squared. This simplifies to 16 over 25
plus nine over 25 equals 𝑐 squared. So 𝑐 squared is equal to 25
over 25, which equals one. Square rooting both sides, and
since 𝑐 must be positive, we have 𝑐 is equal to one. So we have shown that the point
four-fifths, three-fifths lies on the unit circle.

The 𝑥-and 𝑦-coordinates of a
point on the unit circle given by an angle 𝜃 are defined by 𝑥 equals cos 𝜃
and 𝑦 equals sin 𝜃. We recall that sec 𝜃 is equal
to one over cos 𝜃. So sec 𝜃 is equal to one over
𝑥. Substituting in the
𝑥-coordinate, we have sec 𝜃 equals one divided by four-fifths, which is equal
to five over four, or five-quarters. If 𝜃 is in standard position
and the terminal side of the angle passes through the point four-fifths,
three-fifths, then sec 𝜃 is equal to five-quarters.

We will now consider one final
example where we will demonstrate how to use the unit circle to evaluate a simple
trigonometric function.

The terminal side of angle
𝐴𝑂𝐵 in standard position intersects the unit circle 𝑂 at the point 𝐵 with
coordinates three over root 10, 𝑦, where 𝑦 is greater than zero. Find the value of sin
𝐴𝑂𝐵.

An angle is said to be in
standard position if the vertex is at the origin and the initial side lies on
the positive 𝑥-axis. Since angle 𝐴𝑂𝐵 is in
standard position and 𝐵 is not on the 𝑥-axis, the point 𝐴 must lie on the
positive 𝑥-axis. We can therefore sketch angle
𝐴𝑂𝐵 equals 𝜃 on the unit circle. Since both the 𝑥- and
𝑦-coordinates are positive, point 𝐵 lies in the first quadrant.

We know that the 𝑥- and
𝑦-coordinates of a point on the unit circle given by an angle 𝜃 are defined by
𝑥 equals cos 𝜃 and 𝑦 equals sin 𝜃. The value of sin 𝐴𝑂𝐵 is
therefore equal to the value of the 𝑦-coordinate of point 𝐵.

By representing triangle 𝐴𝑂𝐵
as a right triangle, we can find the value of 𝑦 by using the Pythagorean
theorem. We have 𝑦 squared plus three
over root 10 squared is equal to one squared. Simplifying this, our equation
becomes 𝑦 squared plus nine over 10 equals one. Subtracting nine-tenths from
both sides, we have 𝑦 squared is equal to one-tenth. Square rooting both sides, and
since 𝑦 is greater than zero, we obtain 𝑦 is equal to one over root 10
units. The value of sin 𝐴𝑂𝐵 is one
over root 10.

We will finish this video by
recapping some of the key points.

The unit circle is a circle of
radius one, centered at the origin of the coordinate plane. The 𝑥- and 𝑦-coordinates of a
point on the unit circle given by an angle 𝜃 are defined by 𝑥 equals cos 𝜃 and 𝑦
equals sin 𝜃. The tangent ratio can also be
defined for points on the unit circle 𝑥, 𝑦, where 𝑥 is nonzero, such that tangent
𝜃 is equal to 𝑦 over 𝑥. The reciprocal trigonometric
functions are defined as follows. csc 𝜃 is equal to one over 𝑦, sec 𝜃 is equal to
one over 𝑥, and cot 𝜃 is equal to 𝑥 over 𝑦.