Pythagoras and Trigonometry

Triangles are a 3 sided polygon, whose internal angles all add up to 180°. A triangle with one angle of 90° is called a right angle triangle. An example is:

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If we are looking to find the lengths of the sides of the right angle triangle, and we know the lengths of any 2 of the other sides then we can use Pythagoras’s theorem to “solve” the other length. So given side lengths $a$ , $b$ and $c$ Pythagoras’s theorem is as follows:

(1) $\begin{equation*} a^2 + b^2 = c^2 \end{equation*}$

So we can rearrange Pythagoras’s theorem to solve the unknown side length. Pythagoras can be explained using the following diagram.

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Here we can see that the areas of the two squares formed using the shorter edges of the triangle add together to make the same area as the square formed by the long edge of the triangle, hence $a \cdot a + b \cdot b = c \cdot c$ or $a^2 + b^2 = c^2$ .

We can use Pyhtagoras’s theorm to help define a triangle. If $a^2 + b^2 = c^2$ then the triangle is a right angle triangle. If $a^2 + b^2 > c^2$ then the triangle is acute, and if $a^2 + b^2 < c^2$ then the triangle is obtuse. So from this we can see that Pythagoras only holds for right angled triangles.

But what if we only know one edge length and of one of the angles? Well we now need to use trigonometry. First though, we need to know how to set up a triangle with the correct side names related to the angle we know. Given a triangle in this configuration:

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The sides of a triangle are named related to the location of a known angle, usually denoted $\theta$ . The long side is always called the hypotenuse, the side which is opposite the angle $\theta$ is funnily enough called the opposite, and the side touching $\theta$ is called the adjacent. In the previous diagram the angle we know was $\theta$ , moving the angle to where $\phi$ is gives a different angle and the adjacent and opposite switch places, giving:

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We can now write Pythagoras’s Theorem using the correct side names, so it becomes:

(2) $\begin{equation*} \text{Hypotenuse}^2 = \text{Opposite}^2 + \text{Adjacent}^2 \end{equation*}$

Now we can use the trigonometric ratios to calculate the other side lengths correctly to find the other line lengths. There are 3 main trigonometric ratios:

(3) $\begin{equation*} \sin{\theta} = \frac{\text{opposite}}{\text{hypotenuse}} \end{equation*}$

(4) $\begin{equation*} \cos{\theta} = \frac{\text{adjacent}}{\text{hypotenuse}} \end{equation*}$

(5) $\begin{equation*} \tan{\theta} = \frac{\text{opposite}}{\text{adjacent}} \end{equation*}$

So if we know one angle, we must know the second angle as all angles must add to 180°, so the second angle $\phi$ will be $\phi = 180^\circ - 90^\circ - \theta$ or for a right angled triangle, just $\phi = 90^\circ - \theta$ . So if we know one angle and a single side length of a triangle, and setting up the triangle correctly we can use these ratios and Pythagoras to calculate all side lengths and the other angle to completely solve the triangle. It is also possible, to find both the angles if we are given just two side lengths.

The trigonometric ratios vary depending on the inputted angle, they can be plotted as follows:

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As we can see from these plots, the trigonometric rations vary depending on the inputted value. The domain of a function is the range of allowable inputs it can accept and the range is the set of all possible outputs it can give. The sine and cosine functions have ranges of $-1 \geq \sin(\theta) \geq 1$ and $-1 \geq \cos(\theta) \geq 1$ respectively and both have a domain of $[-\infty , \infty ]$ which is all the real numbers or $\mathbb{R}$ . The range of the tangent function is all the real numbers or $\mathbb(R)$ , so $-\infty \geq \tan(\theta) \geq \infty$ , and it has a domain of $\{ \theta \vert \theta \neq 90 + k \cdot 180, k = \ldots ,-1,0,1, \ldots \}$ . This domain looks complicated, but all it means is that the function is undefined when you input a value of $\theta$ that has a value of $90 + k \cdot 180$ where k is any integer or whole number, so it will not except for example $( 90, 270, 450, 630, \ldots )$ . The functions also repeat, they have a period. Both sine and cosine have a period of 360°, and tan has a period of 180°. We can also see that cosine is 90° out of phase with sine, cosine is a quarter of a period ahead of sine, so $\cos(\theta) = \sin(\theta + 90^\circ)$ .