Trigonometric Special Angles – Explanation & Examples

We normally need to use the calculator to figure out the values of the trigonometric functions of an angle unless we are dealing with trigonometric special angles. Because it is not possible to precisely evaluate the trigonometric functions for most of the angles. But is it true for all the angles? The answer is no — not always.

Trigonometric special angles 30o, 45o, and 60o generate rather straightforward trigonometric values. We can precisely evaluate the trigonometric functions for these special angles without a calculator. 

 After studying this lesson, we are expected to learn the concepts driven by these questions and be qualified to address accurate, specific, and consistent answers to these questions.

  • What are trigonometric special angles?
  • How to solve trigonometric special angles?
  • How can we solve actual problems using trigonometric special angles?

The goal of this lesson is to clear up any confusion you might have about the concepts involving trigonometric special angles.

What are trigonometric special angles?

There are specific angles that provide simple and exact trigonometric values. These specific angles are known as trigonometric special angles. These are 30o, 45o, and 60o.

What is so special about them?

Because it is easy to ‘exactly’ evaluate the trigonometric function without using a calculator for these angles. These angles have comparatively clean values, offering us a great deal to solve Math problems. We use these values to give precise answers for determining the values of many trigonometric ratios.

We will use two ‘special right triangles’ to discuss the special angels in this lesson.

  1. 45o – 45o – 90o triangle also known as isosceles triangle is a special triangle with the angles 45o, 45o, and 90o.
  2. 30o – 60o – 90o triangle is another special triangle with the angles 30o, 60o, and 90o.

These special triangles have a unique ability to provide us precise and simple answers when dealing with trigonometric functions.

The good thing is that you are already familiarized with these special triangles as we have discussed them in our Geometry lessons. We will just use them to solve trigonometric special angles and determine the trigonometric ratios of these special angles.

How to solve trigonometric special angles?

Case 1:

Special angle 45o (from a 45o – 45o – 90o triangle)

The following figure 7-1 represents a $45^{\circ }$ – $45^{\circ }$ – $90^{\circ }$ isosceles right triangle with two $45^{\circ }$ degree angles. The lengths of the three legs of the right triangle are named $a$, $b$, and $c$. The angles opposite the legs of lengths $a$, $b$, and $c$ are named $A$, $B$, and $C$. The tiny square with the angle $C$ shows that it is a right angle.

Looking at diagram 7-1, the measure of angle $A$ is $45^{\circ }$. Since the sum of angles in a triangle is $180^{\circ }$, the measure of angle $B$ would also be $45^{\circ }$.

As values of trigonometric functions are based on the angle and not on the size of the triangle. For simplicity, we take:

$a = 1$

$b = 1$

In this case the triangle will be isosceles triangle. We can simply determine the hypotenuse using Pythagorean theorem.

$c^{2}=a^{2}+b^{2}$

substitute $a = 1$, $b = 1$ in the formula

$c^{2}=1^{2}+1^{2}$

$c^{2}= 2$

$c = \sqrt{2}$

The following figure 7-2 shows that the isosceles triangle has two equal sides ($a = b = 1$), hypotenuse ($c = \sqrt{2}$), and equal base angles ($45^{\circ }$ and $45^{\circ }$).

When m A = 45o:

We can easily determine the values of trigonometric ratio for $45^{\circ }$.

Looking at diagram 7-2 from the perspective of m ∠ A = 45o

Sine function

Sine function is the ratio of the opposite side to the hypotenuse.

${\displaystyle \sin 45^{\circ } ={\frac {\mathrm {opposite} }{\mathrm {hypotenuse} }}}$

${\displaystyle \sin 45^{\circ } ={\frac {a}{c}}}$

substitute $a = 1$, $c = \sqrt{2}$ 

${\displaystyle \sin 45^{\circ } ={\frac {1}{\sqrt{2}}}}$

Cosine function

Cosine function is the ratio of the adjacent side to the hypotenuse.

Thus,

${\displaystyle \cos 45^{\circ } ={\frac {\mathrm {adjacent} }{\mathrm {hypotenuse} }}}$

${\displaystyle \cos 45^{\circ } ={\frac {b}{c}}}$

substitute $b = 1$, $c = \sqrt{2}$ 

${\displaystyle \cos 45^{\circ } ={\frac {1}{\sqrt{2}}}}$

Tangent function

Tangent function is the ratio of the opposite side to the adjacent side.

Thus,

${\displaystyle \tan 45^{\circ } ={\frac {\mathrm {opposite} }{\mathrm {adjacent} }}}$

${\displaystyle \tan 45^{\circ } ={\frac {a}{b}}}$

substitute $a = 1$, $b = 1$ 

${\displaystyle \tan 45^{\circ } ={\frac {1}{1}}}$

$\tan 45^{\circ } = 1$

Cosecant function

Cosecant function is the ratio of the hypotenuse to the opposite side.

Thus,

${\displaystyle \csc 45^{\circ } ={\frac {\mathrm {hypotenuse} }{\mathrm {opposite} }}}$

${\displaystyle \csc 45^{\circ } ={\frac {c}{a}}}$

substitute $c = \sqrt{2}$, $a = 1$ 

${\displaystyle \csc 45^{\circ } ={\frac { \sqrt{2}}{1}}}$

$\csc 45^{\circ } = \sqrt{2}$

Secant function

Secant function is the ratio of the hypotenuse to the adjacent side.

Thus,

${\displaystyle \sec 45^{\circ } ={\frac {\mathrm {hypotenuse} }{\mathrm {adjacent} }}}$

${\displaystyle \sec 45^{\circ } ={\frac {c}{b}}}$

substitute $c = \sqrt{2}$, $b = 1$ 

${\displaystyle \sec 45^{\circ } ={\frac { \sqrt{2}}{1}}}$

$\sec 45^{\circ } = \sqrt{2}$

Cotangent function

Cotangent function is the ratio of the adjacent side to the opposite side.

Thus,

${\displaystyle \cot 45^{\circ } ={\frac {\mathrm {adjacent} }{\mathrm {opposite} }}}$

${\displaystyle \cot 45^{\circ } ={\frac {b}{a}}}$

substitute $b = 1$, $a = 1$ 

${\displaystyle \cot 45^{\circ } ={\frac {1}{1}}}$

$\cot 45^{\circ } = 1$

Case 2:

Special angles 30o and 60o (from a 30o – 60o – 90o triangle)

The following figure 7-3 represents an equilateral triangle with sides $a = 2$, $b = 2$, and $c =2$. Since equilateral triangle has congruent angles and the measure of angles in a triangle is $180^{\circ }$, each angle measures $60^{\circ }$.

Let us draw an altitude from the vertex $B$. The altitude separates an equilateral triangle into two congruent right triangles. In Figure 7-4, ${\displaystyle {\overline {BD}}}$ is altitude, $ΔABD\:≅\:ΔCBD$, $∠BDA$ is a right angle, $m∠A=60^{\circ }$, and $m∠ABD=30^{\circ }$.