I'm a graphical artist who is completely out of my depth on this site.

However, I'm dabbling in WebGL (3D software for internet browsers) and trying to animate a bouncing ball.

Apparently we can use trigonometry to create nice smooth curves.

Unfortunately, I just cannot see why.

I can accept this diagram:

triangles and circle

However, running some calculations just do not make sense to me:

Let's set $\alpha$ to 45 (around where it appears to be in the diagram) and find the cosine value, thus giving us the green line. $$\cos(45) = 0.5$$

Fair enough. $\cos(\alpha)$ / the green line is $0.5$ units.

But now this is where it all falls apart. I would have thought if we set $\alpha$ to $90$, $\cos$ would become $0$. Do you see why I think this? Look at the digram, isn't that reasonable to think? Similarly, $\cos(0)$ I would have said should equal $1$ (twice that of $\cos(45)$ )

While $\cos(0)$ does equal $1$, this does not check out: $$ cos(90) = -0.4$$

Just do not get that $0.4$? Could someone explain? That just makes no sense to me. None.

I'm using the google calculator and I would stress I have not touched maths for about $6$ years (ever since I left school!) so please lots of examples and words to explain!

  • 43
    $\begingroup$ $\cos 45° = \frac{1}{\sqrt{2}} \approx 0.707\dotsc$. But $\cos 45 \approx 0.52532$, the trigonometric functions take their arguments in radians in all programming languages I know. $\endgroup$ May 12, 2014 at 22:40
  • 36
    $\begingroup$ Okay guys, I know it's fun to have questions that start with "I'm a graphical artist...", but I don't think a simple trig question really needs six upvotes. $\endgroup$
    – Jack M
    May 12, 2014 at 22:55
  • 52
    $\begingroup$ @Jack, votes are independent. Do you think if somebody comes along and likes the question, they should refrain from voting on it just because too many other people have already done so? $\endgroup$
    – user856
    May 12, 2014 at 23:16
  • 25
    $\begingroup$ @JackM Haha! I was not going to upvote this, until...you said it. $\endgroup$
    – Cheeku
    May 12, 2014 at 23:49
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    $\begingroup$ It's a well-researched question, with a problem that many might have, with an answer that isn't trivial if you didn't instantly recognize it. +1 $\endgroup$
    – lily
    May 13, 2014 at 1:42

9 Answers 9


The cosine function takes radians as its argument in most computational languages.

Indeed, $\cos 45^{\circ} = \sqrt{0.5}$, and $\cos 90^{\circ} = 0$. However, in math, for various reasons, we don't like working with degrees. We work with radians, where $2\pi\ \textrm{radians} = 360^{\circ}$.

In fact, $\cos 90\ \textrm{radians} \approx -0.44807$ and $\cos 45\ \textrm{radians} \approx 0.525$.

90 radians would be about 5156 degrees, or about 14.3 turns around the circle!

  • 9
    $\begingroup$ Wow my fingers were all over the keyboard on this one. $\endgroup$
    – Emily
    May 12, 2014 at 22:43
  • 6
    $\begingroup$ @Starkers Yep. In fact, suppose you rotate by $\alpha$. Then, rotating by $\alpha + 2\pi k$ for any integer $k$ gets you the same thing! Think about skateboarding or figure-skating... a 360 is one spin, a 720 is two spins, but you end up facing the same direction. Likewise, a 180 flips you backwards, and a 540 flips you backwards after doing one complete spin. $\endgroup$
    – Emily
    May 13, 2014 at 14:28
  • 5
    $\begingroup$ As it turns out, this "multi-valuedness" of the $\cos$ function (and likewise, the $\sin$ function), turns out to be exceptionally important in many branches of mathematics! $\endgroup$
    – Emily
    May 13, 2014 at 14:29
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    $\begingroup$ @Starkers - I'd like to point out that while cos(0π)=cos(2π), I'd like to point out that it goes counter-clockwise, not clockwise. For example, sin(π/2) = 1, not -1. $\endgroup$
    – Xynariz
    May 13, 2014 at 15:55
  • 6
    $\begingroup$ Ugh.${}{}{}{}{}$ $\endgroup$
    – Emily
    May 16, 2014 at 18:42

Your problem is with units. As mathematicians, we generally measure angles in radians not degrees. The conversion is $$ x\text{ degrees}=\frac{x}{180}\pi\text{ radians} $$ Most software generally takes the argument of trig functions as radians. Thus $\cos(45^\circ)$ is computed as $$ \cos(45^\circ)=\cos\left(\frac{45}{180}\pi\right)=\cos\left(\frac{\pi}{4}\right)=\frac{1}{\sqrt 2} $$ You are correct that $\cos(90^\circ)=0$, but when you input $\cos(90)$ into a calculator, the calculator reads $90$ in radians, not degrees. What you should input is $$ \cos(90^\circ)=\cos\left(\frac{90}{180}\pi\right)=\cos\left(\frac{\pi}{2}\right) $$ which will return as $0$.

For evidence, type cos(90) into google (here google interprets 90 in radians, not degrees). Compare by typing cos(90 degrees) into google.

To summarize, if you want to find the value of $\cos(x^\circ)$, type $$ \cos\left(\frac{x}{180}\pi\right) $$ into your calculator.

  • 2
    $\begingroup$ Bonus question: Trying with a randomly typed number such as 63.948, google calculator tells me that $\cos(63.948)$ and $\cos(63.948^\circ)$ are the same number (up to some rounding effects) - how come? $\endgroup$ May 13, 2014 at 6:46
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    $\begingroup$ @HagenvonEitzen For me? Well $63.948^\circ\approx 1.116\operatorname{rad}\approx 63.948-20\pi\operatorname{rad}$ $\endgroup$ May 13, 2014 at 7:01
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    $\begingroup$ @HagenvonEitzen: There are an infinite number of such coincidences, actually. They're the points where cos(x)=cos(πx/180) $\endgroup$
    – MSalters
    May 13, 2014 at 15:35
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    $\begingroup$ @MSalters -- but are there a countable infinity? (Answer: yes. Proof is left to the interested reader.) $\endgroup$ May 15, 2014 at 10:57
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    $\begingroup$ @Malvolio: Of course. Take the difference, there's a finite number of zeroes every 2 pi (at most 4), and there are countably many non-overlapping intervals of length 2 pi. $\endgroup$
    – MSalters
    May 15, 2014 at 13:20

Your written question may be: Could somebody explain cos to an idiot please? I will not answer this, as it wouldn't be helping.

The question answered is more like: Why is cos not behaving like I expect it to? I won't answer this either, as it is already answered sufficiently. And it's probably a duplicate...

However I want to answer your intended question, at least as I anticipate it. For this I will concentrate on this part of your question:

However, I'm dabbling in WebGL (3D software for internet browsers) and trying to animate a bouncing ball.

Apparently we can use trigonometry to create nice smooth curves.

I anticipate your real question might be something like: How do I get a nice path to animate a bouncing ball?

The answer is: Not with trigonometric functions as they help you with circular movement, e.g. a ball fastened on a string whirling around a fixed point. You should try a parabolic path. The function y = x² is the easiest example for a parabel. But you probably want to set a starting point, swap and stretch it.

So instead of an explicit function for each point on the path you could use a simple iterative process:

Have a starting position (x,y), a starting speed (m,n) and some kind of gravity (g, ~10m/s² will look natural but that's something for advanced). For each iteration you update your position like this:

x := x + m
y := y + n

and your speed:

n := n - g

To make the ball bounce you invert the speed when it hits an obstacle, e.g. bouncing from the ground at level b:

if (y < b) then n := -n

This will provide a quite basic behaviour and needs some tuning and extensions, especially for natural motion, sharp borders of obstacles, decreasing bounce height, etc.

Note that I use := to resolve the problem that x = x + m would require m = 0 to be correct in a mathematical context. The "operator" := is meant as a redefinition to spare you an iteration index or parameter, i.e. x_i or x(i).

  • $\begingroup$ Trigonometry can be used to calculate the shape of the parabola. $\endgroup$
    – user877329
    May 17, 2014 at 12:36
  • $\begingroup$ Test for if (y < b) after calculating y but before rendering at y. When y < b not only recalculate n but y as well before rendering. Otherwise the ball will fall through the floor (b) before it bounces. $\endgroup$ May 18, 2014 at 12:01
  • $\begingroup$ @CandiedOrange: On math.stackexchange you're correct. However in practice for sufficiently small timesteps and/or low speed there's next to no difference. You'd want small timesteps anyway for a smooth animation. Without quite some additional effort you can either choose, if the ball dips slightly into the ground or bounces off before touching the ground. Choose your poison or go the extra length of introducing variable length timesteps and a collision detection that supports them. Oh, if you're doing the latter you can also start with integration over time. But OP had problems with cosine ... $\endgroup$
    – NoAnswer
    May 26, 2014 at 15:08
  • $\begingroup$ @user877329: As far as I remember, there are transformations to represent any function in terms of sine/cosine. So you are right in theory. However, I doubt the transformation will make y=x² any simpler. For an animation path y=x² is not a good representation, because you need some model of time. The iterative procedure I described is again really simple. It also provides a nice parabolic path, that can be of high precision due to low complexity for each time step. Feel free to provide an approach using sine/cosine that has any advantage over what I gave. $\endgroup$
    – NoAnswer
    May 26, 2014 at 15:26
  • $\begingroup$ @NoAnswer If you know the elevation angle, you will need sine and cosine to get the y and x components of the initial velocity, shaping the parabola. When the angle equals $\pi/2$ there will be no parabola. $\endgroup$
    – user877329
    May 26, 2014 at 15:43

There are two common units of angle measurement: degrees, with $360$ degrees making one full circle, and radians, with $2\pi$ radians making one full circle.

Calculators can usually be switched from one "angular mode" to another. Your calculation of $\cos(90)$ comes from using a calculator in radians mode, while your angle is in degrees. (You're perfectly correct that the cosine of $90$ degrees is $0$.)


You've made multiple errors in your question.

  1. Let's set α to 45 (around where it appears to be in the diagram) and find the cosine value, thus giving us the green line.

    cos(45) = 0.5

    Fair enough. cos(α) / the green line is 0.5 units.

    The green line is not 0.5 units.

    The green line, red line, and the $x$ and $y$ axes form a square whose diagonal is 1 unit. By the Pythagorean Theorem,

    $$ \begin{array}{} (\cos \alpha)^2 &+\quad (\sin \alpha)^2 &= 1^2 \\ (\cos \alpha)^2 &+\quad (\cos \alpha)^2 &= 1 \tag{green length = red length}\\ \end{array}\\ \cos \alpha = \cos 45^\circ = \frac{1}{\sqrt 2} \approx 0.707 $$

    Therefore, the length of the green line is about 0.707.

  2. Similarly, cos(0) I would have said should equal 1 (twice that of cos(45) )

    While it is true that $\cos 0 = 1$, there is no justification for "twice that of cos(45)".

  3. cos(90) = -0.4

    Just do not get that. 0.4? Could someone explain? That just makes no sense to me. None.

    I'm using the google calculator

    Google Calculator uses radians as the default unit for angles. The conversion between units is: $360^\circ = 2\pi\ \textrm{radians}$. Therefore, Google Calculator interpreted your request as

    $$ \cos 90 = \cos \left(90 \cdot \frac{360^\circ}{2\pi}\right) \approx \cos 5156.62^\circ $$

    To get the result you intended, you have to switch it to Degree mode before typing c90=

    Degree mode in Google Calculator


Others have already explained that you were getting results from the calculator for radians and not degrees.

You can make the Google calculator (and most others) work in degrees as well. When you bring up the calculator, the upper left button will say "Rad" and have a blank button (slider) next to it. If you click the slider, you change the calculator to "Deg" (degrees) mode. Then hit AC to clear the calculator, and type in a new calculation right in the calculator answer window (not the google serach bar).


Cosine and sine functions create smooth curves when plotted as a function of time. If you graph $$\cos\left(2\pi t\right)$$, where t is time in seconds, the ball will be start at height 1, drop to height 0 at 1/4 sec, bounce back up to 1 at 1/2 sec, 0 at 3/4 sec, 1 at 1 sec, and so on.

You can change the frequency by changing what is in the parentheses. For example, $$\cos\left(4\pi t\right)$$ will bounce twice as fast and $$\cos\left(\pi t\right)$$ will bounce half as fast.

If you multiply the whole function by a value it will change the height of the function. So $$4\cos\left(2\pi t\right)$$ will cause the ball to bounce four times as high.

Many oscillating things, like light and sound waves, follow this kind of trigonometric function. But a bouncing ball does not, so it might not look natural. Here's a site that gives the correct equation for a bouncing ball.

  • $\begingroup$ They also go negative $\endgroup$
    – snulty
    May 14, 2014 at 13:25

I've looked over the answers so far, and I it very likely with all that 'accurate' information, you may well have the answer you need. If not, I can sympathise with your dilemma so if I may, I can make a few basic suggestions from my initial "experience" with Fourier Calculus. It can be confusing and may not address 'fundamental question':

  • Unfortunately, I just cannot see why.

As I get this, the emphasis is on "why"? In that spirit, I suggest that the issue is firstly that you may be a visually oriented person, I take my lead on that because you 'get' the diagram above. So you understand that the COSINE() function of an angle, is a ratio of two sides of any right-angle triangle: Adjacent and Hypotenuse.

Related, geometrically, to the angle with the adjacent adjacent to the sloping side (hypotenuse).

If I can suggest the "lazy way" first; the fraction created from the ratio between these sides as the corner-angle, alpha, changes is-a *curve*in Euclidean space. So if you take a grid and plot sides: (x, y) at the centre for angle alpha ...

  • x = (Length of the Adjacent side)
  • y - (Length of the Hypotenuse side)

On an axis ... as you change the angle you will see a Cosine curve traced at the opposite point. Common tools and web graphics use this to make 'magical' sprials and stuff. As a boy I had a Spirograph ... it creates patters. ALL of these are based on the same rations from a circle/triangle. It is a kind of magic, just like the Greeks thought. Beauty may be truth.

Why? If you don't have a Spirograph - You can makes some cool substitutes.

  1. Create a triangle shape with three flat lengths of wood.
  2. You can ensure the triangle retains rectangle with a nail or tight screw.
  3. Leave the other two sides - FREE to move.

When you attach a pen to the arms of this right-angle, and move the sides(arms); you will trace curves. Congratulations, they are cos or sin curves!

Once upon a time 'science' like Geometry included philosophy and 'manners' for conduct. We don't live in that world any longer, do we? The aim of Geometry is to consider every thing (material item) objectively in 3-Dimensional space. At that time, 3,00 years ago Science was a kind of magic.

The "Smooth CURVES" you asked about, were considered part of the bounty or reward to those of enquiring minds. Recall that is is like, 2,000 years before Picasso -- Euclid had abstract art. And pretty cool stuff too!



Wrong unit. You talk about deg, while the function obviously expects rad, which gives -0.4.


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