Use unit circle to define trig reciprocal relationships in first quadrant for exact values (deg)

Suppose we have a circle of radius $1$1, centred at the origin, and a point $P$P with coordinates $\left(1,0\right)$(1,0). This point lies on the circle, sitting on the $x$x-axis $1$1 unit to the right of the origin. As $P$P moves along the circle, how do its coordinates change?

Exploration

Picture a string of unit length, fixed to the origin at one end and tied to the point $P$P at the other. Initially the string lies along the $x$x-axis, so that $P$P is at $\left(1,0\right)$(1,0). By keeping the string taught, we can move $P$P around in a circle by changing the angle that the string makes with the $x$x-axis. Let’s call the angle $\theta$θ, and have positive values of $\theta$θ correspond to counterclockwise rotations about the origin, and negative values of $\theta$θ correspond to clockwise rotations about the origin.

We’ll focus our investigation for the time being on the first quadrant, where the coordinate values of $x$x and $y$y are both positive and $0^\circ\le\theta<90^\circ$0°≤θ<90°. For any value of $\theta$θ in this interval, we can construct the following right-angled triangle.

Notice in this triangle that the hypotenuse has length $1$1. We can also relate the shorter side lengths to the coordinates of $P$P: the horizontal side has length $x$x, and the vertical side has length $y$y.

Now we can apply the main trigonometric ratios to the sides of this triangle to obtain the following relation:

$\cos\theta=\frac{x}{1}=x$cosθ=x1​=x
$\sin\theta=\frac{y}{1}=y$sinθ=y1​=y

So the coordinates of $P$P are $\left(x,y\right)=\left(\cos\theta,\sin\theta\right)$(x,y)=(cosθ,sinθ). Let’s test this on the boundaries of the first quadrant. When $\theta=0^\circ$θ=0°, we expect $P$P to be on the $x$x-axis, and indeed we see that $\cos0^\circ=1$cos0°=1 and $\sin0^\circ=0$sin0°=0, so that $\left(\cos0^\circ,\sin0^\circ\right)=\left(1,0\right)$(cos0°,sin0°)=(1,0). Now when $\theta=90^\circ$θ=90° we get $\cos90^\circ=0$cos90°=0 and $\sin90^\circ=1$sin90°=1, so that $\left(\cos90^\circ,\sin90^\circ\right)=\left(0,1\right)$(cos90°,sin90°)=(0,1), which is the point one unit above the origin.

More functions

The motion of $P$P about the unit circle can be thought of as one definition of the sine and cosine functions. If we include some additional features in our picture, we can discover four more important trigonometric functions.

Firstly, we construct a ray from the origin through point $P$P. Next, we construct a vertical tangent at $Q=\left(1,0\right)$Q=(1,0) and a horizontal tangent at $R=\left(0,1\right)$R=(0,1). These tangents intersect the ray at $S$S and $T$T respectively. We define the length along the vertical tangent from $Q$Q to $S$S as $\tan\theta$tanθ. Notice that this is also the $\frac{\text{rise}}{\text{run}}$riserun​ of the ray $OP$OP, so that we can also say that $\tan\theta=\frac{y}{x}$tanθ=yx​ (where $x$x and $y$y are the coordinates of $P$P).

Similarly, we define the length along the horizontal tangent from $R$R to $T$T as $\cot\theta$cotθ, the cotangent of $\theta$θ. This is equivalent to the $\frac{\text{run}}{\text{rise}}$runrise​ of the ray $OP$OP, so that we can also say that $\cot\theta=\frac{x}{y}$cotθ=xy​. Using this definition we immediately observe that $\cot\theta=\frac{1}{\tan\theta}$cotθ=1tanθ​.

To find the remaining two trigonometric functions, we construct one more tangent to the circle at $P$P. This tangent intersects the $x$x-axis at $U$U and intersects the $y$y-axis at $V$V. We define the length $OU$OU to be $\sec\theta$secθ, the secant of $\theta$θ. Using similar triangles, it can be shown that $\sec\theta=\frac{1}{x}=\frac{1}{\cos\theta}$secθ=1x​=1cosθ​.

Similarly, we define the length $OV$OV to be $\csc\theta$cscθ, the cosecant of $\theta$θ, and again it can be shown that $\csc\theta=\frac{1}{y}=\frac{1}{\sin\theta}$cscθ=1y​=1sinθ​.

From the first quadrant to beyond

In the constructions above we have identified the length of segments with the definitions of trigonometric functions. However we have also been careful to connect these segments with the value of the $x$x- and $y$y-coordinate of $P$P. This will become important when we want to extend these definitions to the other three quadrants of the unit circle.

For example, we can think of $\sec\theta$secθ as the length of the segment $OU$OU. But with the knowledge of how the point $U$U is constructed, we can also think of $\sec\theta$secθ as just the $x$x-coordinate of $U$U, that is, the $x$x-coordinate of the point where the tangent to the circle at $P$P intersects the $x$x-axis.

When each function is considered in this way, we will see their values naturally move from being positive to negative and back to positive as the value of $\theta$θ ranges from $0^\circ$0° around to $360^\circ$360°, because the coordinates $\left(x,y\right)$(x,y) of $P$P can be positive or negative depending on where $P$P sits on the unit circle.

Exact values

In general the value of the $x$x- and $y$y-coordinate of $P$P will be numbers that cannot be expressed as simple ratios of integers or surds. But there are three values of $\theta$θ within each quadrant that do have familiar coordinates that are easy to work with. These values relate back to the exact value triangles, shown below.

If we scale the side lengths of these triangles so that each hypotenuse has length $1$1, then we obtain the following corresponding exact value triangles in the unit circle.

            

         

Worked example

Use the unit circle to find the value of $\sec60^\circ$sec60°.

Think: If $P=\left(x,y\right)$P=(x,y) is a point on the unit circle, at an angle of $\theta$θ from the $x$x-axis, then $\sec\theta$secθ is the reciprocal of the $x$x-coordinate of $P$P. We can also think of this as the ratio of the hypotenuse (which always has length $1$1) and the side adjacent the angle $\theta$θ (which always has length $x$x).

Do: The relevant exact value triangle is shown below.

We can see that when $\theta=60^\circ$θ=60°, the ratio of the hypotenuse and the adjacent side is $\frac{1}{\frac{1}{2}}=2$112​​=2. This means that $\sec60^\circ=2$sec60°=2. Notice that this is also the reciprocal of the $x$x-coordinate of $P$P.

Practice questions

Question 1

Question 2

Question 3