Evaluate trigonometric functions values (rad)

When we evaluate a trigonometric function, such as $f(x)=\sin(x)$f(x)=sin(x), we can relate this process back to finding the ratio of the sides of a right triangle. For example, the following right triangle has side lengths of $3$3 units, $4$4 units, and $5$5 units.

We define the sine of the angle $\theta$θ as being the ratio of the side of the triangle opposite the angle and the hypotenuse of the triangle. In this case, the sine of the angle $\theta$θ is written as $\sin\theta=\frac{3}{5}$sinθ=35​.

In a similar way, we can define the cosine of $\theta$θ, and the tangent of $\theta$θ. These three important trigonometric ratios are summarised below.

 

We can always use a calculator to find the value of $f(x)=\sin(x)$f(x)=sin(x), for any value of $x$x. In general the result will be a number with many decimal places. However there are some specific values of $x$x that result in function values that can be expressed in terms of surds, rational numbers, or even integers.

 

Exact value triangles

Below is a right isosceles triangle, with the equal sides of $1$1 unit. Using Pythagoras' theorem, we can see that the hypotenuse is $\sqrt{2}$√2 units. Further, because the triangle has an interior angle sum of $\pi$π radians, and the base angles in an isosceles triangle are equal, we can deduce that the other two unknown angles are both $\frac{\pi}{4}$π4​ radians.

Using trigonometric ratios listed above, we can see that:

• $\sin\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$sin(π4​)=1√2​=√22​
• $\cos\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt{2}}=\frac{\sqrt{2}}{2}$cos(π4​)=1√2​=√22​
• $\tan\left(\frac{\pi}{4}\right)=\frac{1}{1}=1$tan(π4​)=11​=1

Notice in particular that $\sin\left(\frac{\pi}{4}\right)=\cos\left(\frac{\pi}{4}\right)$sin(π4​)=cos(π4​).

Now let's start with an equilateral triangle that has side lengths of $2$2 units. Remember all the angles in an equilateral triangle are $\frac{\pi}{3}$π3​ radians.

By drawing a line that cuts the triangle into two congruent pieces, the base of each smaller triangle is now $1$1 unit, and the length of the centre line is $\sqrt{3}$√3 units, using Pythagoras' theorem.

Here is one of the smaller triangles.

In this case the trigonometric ratios tell us that:

$\sin\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}$sin(π3​)=√32​ $\cos\left(\frac{\pi}{3}\right)=\frac{1}{2}$cos(π3​)=12​ $\tan\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{1}=\sqrt{3}$tan(π3​)=√31​=√3

$\sin\left(\frac{\pi}{6}\right)=\frac{1}{2}$sin(π6​)=12​ $\cos\left(\frac{\pi}{6}\right)=\frac{\sqrt{3}}{2}$cos(π6​)=√32​ $\tan\left(\frac{\pi}{6}\right)=\frac{1}{\sqrt{3}}=\frac{\sqrt{3}}{3}$tan(π6​)=1√3​=√33​

Notice here that $\sin\left(\frac{\pi}{3}\right)=\cos\left(\frac{\pi}{6}\right)$sin(π3​)=cos(π6​) and $\sin\left(\frac{\pi}{6}\right)=\cos\left(\frac{\pi}{3}\right)$sin(π6​)=cos(π3​).

 

Practice questions

QUESTION 1

Question 2

Question 3