# Trigonometric functions — sin - cos - tan - Differences, actual applications,

Beforehand, I want to point out that I'm studying some basic geometric function and I'm pretty behind with this topic, hence may write something not too clever or my question might be silly; I ask this because I don't know. I did some research regarding this topic however did not find anything, as I'm the only one asking such a question.

I tried to read Trigonometric functions in Wiki but the issue with Wikipedia is that you need to know the topic, hence the language and expression, otherwise is really not understandable (explains something you don't know with something you don't know).

## Problem

I'm came this 3 basic functions, sin, cos and tan, I do understand how to calculate them, which are pretty straight forward:

$$\sin(x) = \frac{\text{opposite}}{\text{hypotenuse}}$$

$$\cos(x) = \frac{\text{adjacent}}{\text{hypotenuse}}$$

$$\tan(x) = \frac{\text{opposite}}{\text{adjacent}}$$

However what really bugs me if what is the actual difference of the 3 of them? Obviously, there are 3 different results, hence 3 different 'sizes' (maybe is a ratio?) of the target inside angle.

My expectations where that the result is the same one, and the 3 functions serve to get the vertices angle degree depending on the given value, but in fact, are completely different.

## Questions

• What is the actual difference between each of them?
• Why would someone want to calculate the sin rather than cos or tan, or cos rather than sin etc...?
• Can someone give some application or actual usage in a real problem of each one of them, and why you would've chosen one or the other?
• Any other insight is highly appreciated.

Thanks

• they're just 3 different definitions. they behave differently. look at the graphs of each of them. you use them with the associated information that you know, in most trig word problems, the tangent is the most commonly used because it does not depend on hypotenuse which is often not measurable in real life. – user29418 Oct 25 '20 at 7:11
• Are you familiar with similar triangles? – PM 2Ring Oct 25 '20 at 7:27
• I am really behind, didn't touch Geometry since 10-11 (after that my attention was gone). So I may've done some of it. I'll read the article you shared. Any insight thought would be appreciated ; ) – Federico Baù Oct 25 '20 at 14:33

Example 1: A nursery plants a new tree and attaches a guy wire to help support the tree while its roots take hold. An eight foot wire is attached to the tree and to a stake in the ground. From the stake in the ground the angle of elevation of the connection with the tree is 42º. Find to the nearest tenth of a foot, the height of the connection point on the tree.

here sine is most appropriate

Example 2: Find the shadow cast by a 10 foot lamp post when the angle of elevation of the sun is 58º. Find the length to the nearest tenth of a foot.

here tangent is most appropriate

source

• Despite all above answer are really good, I've got some issue to completely understand it, still I will go back to them once I grasp more theory. You answer is the most helpful at the moment because of that link you shared, it explains it by example and as i were a kid, Thanks – Federico Baù Oct 25 '20 at 14:38

I always found the trigonometric functions much easier to understand when – despite the name – not approached from the study of arbitrary right triangles, but from unit circles, as can equivalently be done.

(That's not to say this is a better approach – in fact I now consider the triangle definitions geometrically preferrable, as they don't require any length scale or origin – just an easier to understand one.)

You still get triangles in this view, but they're bound into the circle. The picture to keep in mind is this:

What the unit circle accomplishes is, it keeps the hypothenuse always at the value 1. So, in that case the formulas simplify to

\begin{align} \sin =& \frac{\text{opposite}}{1} = \text{opposite} \\ \cos =& \frac{\text{adjacent}}{1} = \text{adjacent} \end{align}

The tangent formula doesn't simplify, as it doesn't even contain the hypothenuse.

Why do we keep the hypothenuse fixed, and not one of the catheti? Well, let's try what would happen if we did that:

Here, the triangles are not nice and well-constrained in the unit circle anymore, but instead grow to infinity as the angle approaches 90° (or 0°, depending on which cathetus we fix).

That's why the tangent function grows without bound, whereas sine and cosine are smooth and limited to range $$[0,1]$$ with the only difference being a shift by 90° (corresponding that you switch to the other opposite-ness).

Source code for animations (Haskell with dynamic-plot library):

import Graphics.Dynamic.Plot.R2
import Diagrams.Prelude

main = plotWindow
[ shapePlot $circle 1 & fcA transparent , plotLatest [ lineSegPlot [(0,0), (x',0), (x,y), (0,0)] | φ <- [pi/n, 3*pi/n ..] , let x = cos φ y = sin φ x'= tan φ/y -- use x'=x instead for fixed-hypothenuse ] , unitAspect ] where n = 80  • I fully agree, unit circle definition is the best way to make trigonometry easier. – user Oct 25 '20 at 16:40 • God bless you for this explanation, Also sport on because I'm studying some geometry while actually learning Deep Programming. While don't know Haskell i kept the github for future reference. Thanks – Federico Baù Oct 25 '20 at 18:23 •$\tan x$may be defined as$\sin x / \cos x$, or equivalently, as the gradient of the line segment. It can even be defined as the length of the line segment tangent to$(\sin x, \cos x)$, where the two endpoints of the line segment are$(\sin x, \cos x)$, and$(a,0)$. – Joe Oct 26 '20 at 18:00 There are two fundamentally important results in trigonometry. $$\sin^2(x) +\cos^2(x)\equiv 1$$ $$\tan(x)=\frac{\sin(x)}{\cos(x)}$$ With a right triangle with hypotenuse $$1$$ and an angle $$x$$. The base is $$\cos x$$ and height $$\sin x$$, or vice versa, and $$\tan x$$ is their ratio. They are different ratios, but they are connected, and offer other nice properties. Note also that $$\sin(x) =\cos(\frac\pi2-x)$$. An important conclusion from the first statement is that any point on a circle $$C$$ defined by $$(x-a)^2+(y-b)^2=r^2$$ can be written uniquely as $$(a+r\cos\theta, b+r\sin\theta)$$ for some $$\theta\in[0,2\pi)$$. Lots of geometric and analytical properties use this fact, as well as the entirety of complex number theory being based off it. Mechanics and Engineering make big use of trig functions too. For example a force hits a plane at an angle, you can use trig functions on that angle to determine the horizontal and vertical components of the force and solve systems in that way. The most notorious use of $$\tan$$ that I can think of is the Weierstrass integration substitution, which makes quick work of a lot of integrals. So do the other trig functions, as the special properties they share make them well suited to that sort of thing. From the definition of trigonometric function, the three expression are equivalent indeed since $$\cos \left(\frac \pi 2 -x\right)= \sin x$$ we have that $$\text{opposite}=\text{hypotenuse} \cdot \sin x \implies \text{opposite}=\text{hypotenuse} \cdot \cos \left(\frac \pi 2 -x\right)$$ which, by $$y=\left(\frac \pi 2 -x\right)$$ leads to $$\text{adjacent}=\text{hypotenuse} \cdot \cos y$$ The same result can be obtaind from Pythagorean theorem. For the third one we have $$\tan x = \frac{\sin x}{\cos x}= \frac{\frac{\text{opposite}}{\text{hypotenuse}}}{\frac{\text{adjacent}}{\text{hypotenuse}}}=\frac{\text{opposite}}{\text{adjacent}}$$ These expression are used for the solution of triangles depending on the information we are given and depending on what we are looking for. Refer also to the related In fact, there are two "registers" that haven't been covered by previous answers explaining why $$\sin$$ and $$\cos$$ are so close (for tangent, it's different). 1. If you look at their graphical representations, one is shifted (by $$+\pi/2$$) from the other. But there is more to say, this shift is equivalent to a derivation and you have a circle of order four: $$(\sin) \rightarrow (\cos) \rightarrow (-\sin) \rightarrow (-\cos) \rightarrow (\sin)$$ where the $$\rightarrow$$ means at the same time derivation and shift (when seen in the right to left direction). 1. Let us now revisit the very same ideas (differentiation = shift) with complex numbers using the amazing De Moivre formula: $$\cos \theta + i \sin \theta = e^{i \theta}\tag{1}$$ Let us differentiate (1) with respect to $$\theta$$: $$(\cos \theta)' + i (\sin \theta)'=ie^{i \theta}$$ i.e., by using (1) again $$(\cos \theta)' + i (\sin \theta)'=i(\cos \theta + i \sin \theta)$$ $$(\cos \theta)' + i (\sin \theta)'=-\sin \theta + i \cos \theta$$ proving by identifying the real and imaginary parts resp. that: $$(\cos \theta)' =-\sin \theta \ \ \text{and} \ \ (\sin \theta)'= \cos \theta)$$ (retrieving in this way the results we have seen previously). Remember now that multiplying by $$i$$ means geometricaly "rotation" by $$\pi/2$$; therefore, no surprise that doing this 4 times, we are back on our feet. The cycle of order 4 we had seen in the first part is very well explained by complex numbers. • Thank you for adding this explanation to the other answer, despite (at the moment) I understand 5% of it I will take reference for future when I'll be more fluent of Mathematics. – Federico Baù Oct 26 '20 at 5:32 • Sorry for that. Are you aware of what a derivative means (slope of the tangent) and what a complex number is (with$i^2=-1\$) ? These two concepts are of overwhelming importance in mathematics. – Jean Marie Oct 26 '20 at 8:40
• I do know about complex number, but not about derivative (I may have studied but forgotten). – Federico Baù Oct 27 '20 at 8:03