You'll have noticed by now that when you find angles using trigonometric ratios, you often get long decimal answers. If, for example, you put $\cos30^\circ$cos30° into the calculator, you will see an answer of $0.86602$0.86602... which we'd have to round. However, when you take cos, sin or tan of some angles, you can express the answer as an exact number, rather than a decimal. It just may include irrational numbers. We often use these exact ratios in relation to $30^\circ$30°, $45^\circ$45° and $60^\circ$60°.
Let's look at how to do this now.
Below is a right-angle isosceles triangle, with the equal sides of $1$1 unit. Using Pythagoras' theorem, we can work out that the hypotenuse is $\sqrt{1^2+1^2}=\sqrt{2}$√12+12=√2 units. The angles in a triangle add up to $180^\circ$180° and the base angles in an isosceles triangle are equal, so we can also work out that the two unknown angles are both $45^\circ$45°.
We can then use our trig ratios to determine the exact values of the following:
To find the exact ratios of $30$30 and $60$60 degree angles, we need to start with an equilateral triangle with side lengths of $2$2 units. Remember all the angles in an equilateral triangle are $60^\circ$60°.
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$1 unit pieces and cuts one of the $60^\circ$60° 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 $\sqrt{3}$√3 using Pythagoras' theorem. We can then use our trig ratios to determine the exact values of the following:
Notice that the $\sin60^\circ=\cos30^\circ$sin60°=cos30°. It is true for any two complementary angles that $\sin x=\cos\left(90^\circ-x\right)$sinx=cos(90°−x).
Now, an isosceles right-angled triangle may not have its sides measuring $1$1,$1$1 and $\sqrt{2}$√2, but however large it is, it will always have two $45^\circ$45° angles and the ratios of the sides will always be the same as in the table. The same applies to the triangle with $60^\circ$60° and $30^\circ$30° angles. As long as a triangle is similar to one of these triangles (it has the same angles) we can use the exact values.
We have found the exact values for the following using the $45$45 and $30$30 $60$60 triangles:
sin | cos | tan | |
---|---|---|---|
$30^\circ$30° | $\frac{1}{2}$12 | $\frac{\sqrt{3}}{2}$√32 | $\frac{1}{\sqrt{3}}$1√3 |
$45^\circ$45° | $\frac{1}{\sqrt{2}}$1√2 | $\frac{1}{\sqrt{2}}$1√2 | $1$1 |
$60^\circ$60° | $\frac{\sqrt{3}}{2}$√32 | $\frac{1}{2}$12 | $\sqrt{3}$√3 |
We also have the exact values that don't describe physical triangles, at $0^\circ$0° and $90^\circ$90°:
sin | cos | tan | |
---|---|---|---|
$0^\circ$0° | $0$0 | $1$1 | $0$0 |
$90^\circ$90° | $1$1 | $0$0 | undefined |
The unit circle provides us with a visual understanding that the trigonometric functions of $\sin\theta$sinθ, $\cos\theta$cosθ and $\tan\theta$tanθ exist for angles larger than what can be contained in a right-angled triangle.
The unit circle definitions of $\sin\theta$sinθ and $\cos\theta$cosθ tell us that the value of these functions will be the $x$x and $y$y-values respectively of a point on the unit circle after having rotated by an angle of measure $\theta$θ in the anticlockwise direction.
Definition of $\cos\theta$cosθ and $\sin\theta$sinθ can extend beyond $0^\circ\le\theta\le90^\circ$0°≤θ≤90°. |
We can divide this into four quadrants as shown below:
Consider the following definitions for each quadrant:
Use the exact value triangles in the diagram below to answer the following:
What is the exact value of $\cos45^\circ$cos45°?
What is the exact value of $\cos60^\circ$cos60°?
What is the exact value of $\cos30^\circ$cos30°?
$\theta$θ is an angle in a right-angled triangle where $\tan\theta=\frac{1}{\sqrt{3}}$tanθ=1√3.
Solve for the exact value of $\theta$θ.
Consider the point $\left(x,y\right)$(x,y) on the unit circle.
Form a trigonometric equation to find the value of $x$x.
Make sure to consider the sign of $x$x.
Now form a trigonometric equation to find the value of $y$y.
Make sure to consider the sign of $y$y.