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Circular functions

Trigonometric functions such as sin, cos and tan are usually defined as the ratios of sides in a right-angled triangle. These ratios can be extended to angles greater than \(90^{\circ}\), using angles in a unit circle. Circular functions have applications in many STEM disciplines, such modelling oscillating motion in engineering, or analysing wave patterns in physics.

The unit circle

A unit circle is a circle that is centered at the origin \((0,0)\) and has a radius of \(1\).

A unit circle drawn on a cartesian plane, with centre x equals 0 and y equals 0. The circle passes through four axis points. The first point is x equals 1, y equals 0. THe second point is x equals 0, y equals 1. The third point is x equals negative 1, y equals 0. The fourth point is x equals 0, y equals negative 1.

An angle measured from the positive \(x\)-axis may be used to define any point on the unit circle. They can be positive or negative.

  • Positive angles are in an anti-clockwise direction.
  • Negative angles are in a clockwise direction.
A unit circle drawn on a cartesian plane. A blue curved arrow starting at the x-axis and extending anticlockwise shows a positive 135 degrees rotation. A red curved arrow starting at the x-axis and extending clockwise shows a negative 135 degrees rotation.

Although there are \(360^{\circ}\) in a circle, it is possible to rotate through more than \(360^{\circ}\). For example a rotation of \(45^{\circ}\) identifies the same point as a rotation of \(45^{\circ}+360^{\circ}=405^{\circ}\).

Quadrants

The unit circle sits on the coordinate plane, which is divided into an upper and lower section by the \(x\)-axis. It is further divided into quadrants by the \(y\)-axis. These four quadrants are numbered from one to four in an anti-clockwise direction.

A cartesian plane showing the four quadrants. Quadrant 1 is where all x and y values are positive. All of the sine, cosine and tangent values are positive in Quadrant 1. Quadrant 2 is where all y values are positive but all x values are negative. Only sine values are positive in Quadrant 2. Quadrant 3 is where all x and y values are negative. Only tangent values are positive in Quadrant 3. Quadrant 4 is where all x values are positive but y values are negative. Only cosine values are positive in Quadrant 4.

Knowing exact values for some angles in quadrant \(1\), allows you to find exact values in other quadrants.

Sine and cosine functions on the unit circle

Imagine a point \(P\left(x,y\right)\) on the unit circle.

A unit circle drawn on a cartesian plane. A triangle is formed when angle theta is drawn anti-clockwise from the positive x-axis. The hypotenuse of the triangle is the radius of the unit circle, 1. The opposite side length is sine theta and the adjacent side length is cosine theta.

In the triangle \(POQ\):

  • the hypotenuse \(OP\) has a length of \(1\)
  • the adjacent side to the angle \(\theta\) is \(OQ\)
  • the opposite side to the angle \(\theta\) is \(QP\).

Using the trigonometric ratios for sine and cosine in a right-angled triangle, we get:

\[\begin{align*} \cos(\theta) & = \frac{\textrm{adjacent}}{\textrm{hypotenuse}}\\
& = \frac{OQ}{OP}\\
& = \frac{x}{1}\\
& = x
\end{align*}\]

and

\[\begin{align*} \sin(\theta) & = \frac{\textrm{opposite}}{\textrm{hypotenuse}}\\
& = \frac{QP}{OP}\\
& = \frac{y}{1}\\
& = y
\end{align*}\]

Any point on the unit circle has coordinates: \[x=\cos(\theta)\quad y=\sin(\theta)\] where \(\theta\) is the angle (positive or negative) measured from the positive \(x\)-axis to the point \(P\).

From the unit circle, we can see that:

  • \(\sin\left(0^{\circ}\right)=0\)
  • \(\sin\left(90^{\circ}\right)=\sin\left(\dfrac{\pi}{2}\right)=1\)
  • \(\cos\left(0\right)=1\)
  • \(\cos\left(\dfrac{\pi}{2}\right)=\cos\left(90^{\circ}\right)=0\)

Tangent function on the unit circle

Now, consider point \(N\), which is an extension of line \(OP\).

A unit circle drawn on a cartesian plane. A triangle is formed when angle theta is drawn anti-clockwise from the positive x-axis. The hypotenuse of the triangle is the radius of the unit circle, 1. The opposite side length is sine theta and the adjacent side length is cosine theta. The hypotenuse is extended until it reaches x equals 1 to form a larger triangle. The vertical length of this larger triangle is given by tangent theta.

The triangles \(OMN\) and \(OQP\) are similar.

\[\begin{align*} \frac{MN}{OM} & =\frac{PQ}{OQ}\\
MN & = \frac{PQ}{OQ}\quad\textrm{since OM=1}
\end{align*}\]

Length \(MN\) is the tangent of the angle \(\theta\), or \(\tan\theta\). It is the distance of the vertical tangent to the circle that passes through the point \(M\left(1,0\right)\).

From triangle \(OMN\), we can also see that \(\sin\left(\theta\right)=PQ\) and \(\cos\left(\theta\right)=OQ\).

When we bring all of these together, we get an equation that connects the sin, cos and tan functions.

\[\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}\]

This is consistent with the tangent function for right-angled triangles. We can prove this by substituting SOH and CAH into the equation.

\[\begin{align*} \tan(\theta) & = \frac{\frac{\textrm{O}}{\textrm{H}}}{\frac{\textrm{A}}{\textrm{H}}}\\
& = \frac{\textrm{O}}{\textrm{H}} \times \frac{\textrm{H}}{\textrm{O}}\\
& = \frac{\textrm{O}}{\textrm{A}}
\end{align*}\]

Exact values of sine, cosine and tangent functions

We would usually use a calculator to evaluate \(\sin(\theta)\), \(\cos(\theta)\) and \(\tan(\theta)\), but there are some special angles you should remember. These are shown in the following table.

Angle \(\boldsymbol{(\theta)}\) \(0^{\circ}\) \(30^{\circ}=\dfrac{\pi}{6}\) \(45^{\circ}=\dfrac{\pi}{4}\) \(60^{\circ}=\dfrac{\pi}{3}\) \(90^{\circ}=\dfrac{\pi}{2}\) \(180^{\circ}=\pi\) \(270^{\circ}=\dfrac{3\pi}{2}\)
\(\boldsymbol{\sin(\theta)}\) \(0\) \(\dfrac{1}{2}\) \(\dfrac{1}{\sqrt{2}}\) \(\dfrac{\sqrt{3}}{2}\) \(1\) \(0\) \(-1\)
\(\boldsymbol{\cos(\theta)}\) \(1\) \(\dfrac{\sqrt{3}}{2}\) \(\dfrac{1}{\sqrt{2}}\) \(\dfrac{1}{2}\) \(0\) \(-1\) \(0\)
\(\boldsymbol{\tan(\theta)}\) \(0\) \(\dfrac{1}{\sqrt{1}}\) \(1\) \(\sqrt{3}\) undefined \(0\) undefined

\(\tan(\theta)\) is undefined for \(90^{\circ}\) and \(270^{\circ}\). This is because \(\tan(\theta)=\dfrac{\sin(\theta)}{\cos(\theta)}\) and dividing by \(\cos(\theta)=0\) is undefined.

Knowing these exact values will help you to work out the exact values for other angles using the symmetry of the unit circle.

A unit circle drawn on a cartesian plane with rays indicating different angles. At angle 0 degrees or 0 radians, the point on the unit circle is given by the coordinates x equal 1, y equals 0. At angle 30 degrees or pi over 6 radians, the point on the unit circle is at x equals root 3 over 2, y equals half. At angle 45 degrees or pi over 4 radians, the point on the unit circle is at x equals root 2 over 2, y equsl root 2 over 2. At angle 60 degrees over pi over 3 radians, the point on the unit circle is x equals half, y equals root 3 over 2. At angle 90 degrees or pi over 2 radians, the point on the unit circle is x equals 0, y equals 1. At angle 180 degrees or pi radians, the point on the unit circle is x equals negative 1, y equals 0. At angle 270 degrees or 3 pi over 2 radians, the point on the unit circle is x equals 0, y equals negative 1.

Example 1 – using exact values in circular functions

Find the exact values of \(\sin\left(\dfrac{2\pi}{3}\right)\) and \(\cos\left(\dfrac{2\pi}{3}\right)\).

Sketch a unit circle and locate the point where the angle is \(\dfrac{2\pi}{3}\). Remember that \(\dfrac{2\pi}{3}=120^{\circ}\).

A unit circle drawn on a cartesian plane showing two equivalent lengths. Sine of 2 pi over 3 is equivalent to sine of pi over 3. Cosine of 2 pi over 3 is equivalent to the negative of the cosine of pi over 3.

\(\dfrac{2\pi}{3}\) is shown in blue along with \(\sin\left(\dfrac{2\pi}{3}\right)\) and \(\cos\left(\dfrac{2\pi}{3}\right)\).

We can use the symmetry of the unit circle to draw in an angle that is related to \(\dfrac{2\pi}{3}\) but for which you know the exact value. In this case, it is \(\dfrac{\pi}{3}\), shown in red.

You can see that length \(NQ\) is the same as \(MP\) which means:

\[\begin{align*} NQ & = MP\\
\sin\left(\frac{2\pi}{3}\right) & = \sin\left(\frac{\pi}{3}\right)\\
& = \frac{\sqrt{3}}{2}\quad\textrm{ from the table} \end{align*}\]

Lengths \(ON\) and \(MO\) are the same, but \(MO=-ON=-\cos\left(\dfrac{\pi}{3}\right)\), so:

\[\begin{align*} \cos\left(\frac{2\pi}{3}\right) & = -ON\\
& = -\cos\left(\frac{\pi}{3}\right)\\
& = -\frac{1}{2}\quad\textrm{ from the table} \end{align*}\]

Find the exact value of \(\sin\left(330^{\circ}\right)\).

Sketch a unit circle and locate the point where the angle is \(330^{\circ}\). You should be able to see that it is closely related to \(30^{\circ}\) in red.

A unit circle on a cartesian plane with two rays drawn from the centre of the circle. The line formed at 30 degrees has a point Q on the unit circle. The line formed at 330 degrees has a point P on the unit circle. Points Q and P form triangles with a point M.

The \(y\) coordinates differ only by sign because the distances from the \(x\)-axis are the same for \(330^{\circ}\) and \(30^{\circ}\).

\[\begin{align*} MP & = -MQ\\
\sin\left(330^{\circ}\right) & = -\sin\left(30^{\circ}\right)\\
& = -\frac{1}{2}\quad\textrm{ from the table} \end{align*}\]

Find the exact value of \(\tan\left(\dfrac{4\pi}{3}\right)\).

Sketch a unit circle and locate the point where the angle is \(\dfrac{4\pi}{3}\). You should be able to see that \(\tan\left(\dfrac{4\pi}{3}\right)\) in blue is closely related to \(\tan\left(\dfrac{\pi}{3}\right)\) in red.

Remember \(\dfrac{4\pi}{3}=\dfrac{3\pi}{3}+\dfrac{\pi}{3}=\pi+\dfrac{\pi}{3}\).

A unit circle on a cartesian plane with an angle of pi over 3 starting from the positive x-axis. A line extends from the origin through an angle of 4 pi over 3, intersecting with the unit circle at point P in the third quadrant. A vertical line tangent to the circle at x equals 1, y equals 0, is shown with a value of tan 4 pi over 3.
\[\begin{align*} \tan\left(\frac{4\pi}{3}\right) & = \tan\left(\frac{\pi}{3}\right)\\
& = \sqrt{3}\quad\textrm{ from the table}
\end{align*}\]

Your turn – using exact values in circular functions

  1. Determine the coordinates for the points on the unit circle that make the following angles with the positive \(x\)-axis.
    1. \(30^{\circ}\)
    2. \(125^{\circ}\)
    3. \(-60^{\circ}\)
    4. \(270^{\circ}\)
    5. \(-180^{\circ}\)
    6. \(720^{\circ}\)
  2. Find exact values for:
    1. \(\sin(330^{\circ})\)
    2. \(\cos(210^{\circ})\)
    3. \(\sin(-30^{\circ})\)
    4. \(\cos(90^{\circ})\)
    5. \(\tan(300^{\circ})\)
    6. \(\cos(180^{\circ})\)
    7. \(\sin(-120^{\circ})\)
    8. \(\cos(315^{\circ})\)

    1. \((0.87,0.5)\)
    2. \((-0.56, 0.82)\)
    3. \((0.5,-0.87)\)
    4. \((0,-1)\)
    5. \((-1,0)\)
    6. \((1,0)\)
    1. \(0.5\)
    2. \(\dfrac{\sqrt{3}}{2}\)
    3. \(-0.5\)
    4. \(0\)
    5. \(\sqrt{3}\)
    6. \(-1\)
    7. \(-\dfrac{\sqrt{3}}{2}\)
    8. \(\dfrac{1}{\sqrt{2}}\)

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