7.01 Exact ratios and the unit circle

Below is a right-angle isosceles triangle, with the equal sides of 1 unit. Using Pythagoras' theorem, we can work out that the hypotenuse is \sqrt{1^2+1^2}=\sqrt{2} units.

The angles in a triangle add up to 180\degree and the base angles in an isosceles triangle are equal, so we can also work out that the two unknown angles are both 45\degree.

We can then use our trig ratios to determine the exact values of the following:\sin 45\degree =\dfrac{1}{\sqrt{2}}, \quad \cos 45\degree =\dfrac{1}{\sqrt{2}}, \quad \tan 45\degree =\dfrac{1}{1} = 1

To find the exact ratios of 30 and 60 degree angles we need to start with an equilateral triangle with side lengths of 2 units. Remember all the angles in an equilateral triangle are 60\degree.

Then we are going to draw a line that cuts the triangle in half into two congruent triangles. The base line is cut into two 1 unit pieces and cuts one of the 60\degree angles in half.

Now let's just focus on one half of this triangle.

We can calculate the length of the centre line to be \sqrt{3} using Pythagoras' theorem. We can then use our trig ratios to determine the exact values.

\sin 30\degree =\dfrac{1}{2}, \quad \cos 30\degree =\dfrac{\sqrt{3}}{2}, \quad \sin 60\degree = \dfrac{\sqrt{3}}{2}, \quad \cos 60\degree =\dfrac{1}{2}

Notice that the \sin 60 \degree =\cos 30 \degree. It is true for any two complementary angles that \sin x=\cos \left(90^\circ -x\right).

Now, an isosceles right-angled triangle may not have its sides measuring 1, 1 and \sqrt{2}, but however large it is, it will always have two 45\degree angles and the ratios of the sides will always be the same as above. The same applies to the triangle with 60\degree and 30\degree angles. As long as a triangle is similar to one of these triangles (it has the same angles) we can use the exact values.

Examples

Example 1

Use the exact value triangles in the diagram below to answer the following:

a

What is the exact value of \cos 45\degree?

Worked Solution

Create a strategy

Use the triangle with the angle 45 \degree in it.

Apply the idea

Using the triangle on the right, we can find \cos 45\degree:

\displaystyle \cos 45\degree	\displaystyle =	\displaystyle \dfrac{\text{Adjacent}}{\text{Hypotenuse}}	Use the cosine ratio
	\displaystyle =	\displaystyle \dfrac{1}{\sqrt{2}}	Substitute the values

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b

What is the exact value of \cos 60\degree?

Worked Solution

Create a strategy

Use the triangle with the angle 60 \degree in it.

Apply the idea

Using the triangle on the left, we can find \cos 60\degree:

\displaystyle \cos 60\degree	\displaystyle =	\displaystyle \dfrac{\text{Adjacent}}{\text{Hypotenuse}}	Use the cosine ratio
	\displaystyle =	\displaystyle \dfrac{1}{2}	Substitute the values

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Example 2

\theta is an angle in a right-angled triangle where \tan \theta =\dfrac{1}{\sqrt{3}}.

Worked Solution

Create a strategy

Use the exact value triangle with side length of \sqrt{3}.

Apply the idea

Using this exact value triangle we can see that \tan 30\degree =\dfrac{1}{\sqrt{3}}.

So \theta = 30 \degree

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The unit circle provides us with a visual understanding that the trigonometric functions of \sin\theta, \cos\,\theta and \tan\theta exist for angles larger than what can be contained in a right-angled triangle.

The unit circle definitions of \sin\theta and \cos\,\theta tell us that the value of these functions will be the x and y-values respectively of a point on the unit circle after having rotated by an angle of measure \theta in the anticlockwise direction. Or, if \theta is negative, then the point is rotated in the clockwise direction.

We can divide this into four quadrants as shown below:

Consider the following definitions for each quadrant:

• Angles between 0\degree and 90\degree are said to be in the first quadrant.

• Angles between 90\degree and 180\degree are said to be in the second quadrant.

• Angles between 180\degree and 270\degree are said to be in the third quadrant.

• Angles between 270\degree and 360\degree are said to be in the fourth quadrant.

Examples

Example 3

Consider the point \left(x,y\right) on the unit circle.

a

Form a trigonometric equation to find the value of x.

Make sure to consider the sign of x.

Worked Solution

Create a strategy

We can equate x to \sin \theta and then use an exact value triangle to find the value.

Apply the idea

We can draw a right-angled triangle to help determine its side lengths and the value of x.

The acute angle is 30\degree since 120\degree -90 \degree = 30\degree .

The right-angled triangle has an angle of 30 \degree with opposite length of x and hypotenuse of 1.

So we can use the sine ratio.

\displaystyle \sin \theta	\displaystyle =	\displaystyle \frac{\text{Opposite}}{\text{Hypotenuse}}	Use the sine ratio
\displaystyle \sin 30\degree	\displaystyle =	\displaystyle \frac{x}{1}	Substitute 1 and x
\displaystyle x	\displaystyle =	\displaystyle \sin 30\degree	Make x the subject

We can find the exact value of this using the triangle:

x= \dfrac{1}{2}

Since the point is to the left of the origin, the x-coordinate will be negative.

\displaystyle x

\displaystyle =

\displaystyle -\frac{1}{2}

Apply the negative

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b

Now form a trigonometric equation to find the value of y.

Make sure to consider the sign of y.

Worked Solution

Create a strategy

We can equate y to \cos \theta and then use an exact value triangle to find the value.

Apply the idea

The right-angled triangle has an angle of 30 \degree with adjacent length of y and hypotenuse of 1.

So we can use the cosine ratio.

Since the point is above the origin, the y-coordinate will be positive.

\displaystyle \cos \theta	\displaystyle =	\displaystyle \frac{\text{Adjacent}}{\text{Hypotenuse}}	Use the cosine ratio
\displaystyle \cos 30\degree	\displaystyle =	\displaystyle \frac{y}{1}	Substitute 1 and y
\displaystyle y	\displaystyle =	\displaystyle \cos 30\degree	Make y the subject

We can find the exact value of this using the triangle:

y= \dfrac{\sqrt{3}}{2}

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