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I see in my Book that 45 deg is equivalent of π/4 . Ι also do the conversion if I simply convert degrees into radians like this

45* π/180 = π/4 radians

and back again

π/4 * 180/π = 45 deg

. So I think that I grasp the Idea finally that π/4 is another way of saying 45 degrees. Now if I use my calculator and I say sin(45) witch I assume that 45 is degrees I get the number 0.7071067812. If I do the calculation sqrt(2)/2 I will get the same number hence sin(45)=sqrt(2)/2 . But when I input sin(π/4) I get the number 0.0137073546. So I say to my self that probably that function does not get that type of input. But then again I see in my book that sin(π/4) = sqrt(2)/2 . So this forces me to ask the question, "What type of input do trigonometric functions take in?" Is the calculator not working properly ?

Excuse me if this looks simple to you but I am so bad in math and the calculator is the only validator I have at the moment and this confuses me allot.

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You need to change your calculator setting from Degrees to Radians if your input is radians. Then it will come out correct. – imranfat May 24 '13 at 19:46
up vote 4 down vote accepted

When you use $\sin(45)$ in your calculator, you mean $\sin(45^o)$ (notice the degree symbol).

But when you input $\dfrac{\pi}{4}$, you are still telling your calculator that it's in degrees: $\sin\left(\dfrac{\pi}{4}^o\right)$.

To resolve this, you have to change the mode of your calculator from degrees to radians, and reperform the calculation.

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oh didn't even know that calculator has modes. just realized the mode button. Thank you I will try that. So basically that function takes two type of numbers ? :o – themhz May 24 '13 at 19:50
got it. Thank you – themhz May 24 '13 at 19:54
@themhz Yup! Be sure to set the correct mode before starting a number. I've known many cases where people forgot to change the modes and had to start problems all over again in exams! :S – Jerry May 24 '13 at 19:56
@Jerry Using \circ instead of $o$ looks a bit better (e.g. $90^{\circ}$ instead of $90^o$), but I don't know if it's the canonical way to write degrees. – Mark S. May 25 '13 at 16:00
@MarkS. Ohh, that does appear neater. Thank you for pointing it out :) – Jerry May 25 '13 at 18:44

The problem is that there are two different sine functions, and each takes in a different kind of input!

Your calculator is set to be in degree mode, which means it is taking in degree inputs. When you put in 45, it assumes you mean 45 degrees ($\pi/4$ radians). And when you put in $\pi/4$, your calculator calculates $\pi/4 \approx .7854$, and then it computes the sine of $.78$ degrees! Hopefully you can visualize that $.78$ degrees is just a very small angle, while $\pi/4$ radians is in fact $45$ degrees, which is a much larger angle.

You can't expect your calculator's sine function to know whether you are talking about degrees or radians, because $\pi / 4$ as a number can be both a number of degrees or a number of radians. So you have to tell your calculator which sine function you want -- the degree input one or the radian input one -- and then it will know what you mean.

In fact, in actual mathematics there are no degrees -- just radians. When you say, for example, $\sin 45^\circ$, what the $\circ$ really means is, convert this number to radians. The reason for having to specify degrees with the $\circ$ symbol is to remove the ambiguity of the input which you are complaining about. Also, $\sin x$ with $x$ in radians turns out to be a much nicer function in the end from the perspective of real and complex analysis.

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Three, not two, if you treat degrees and radians on the same footing (as "units"). This is desirable for dimensional analysis and logical consistency. I explain this point of view in another answer. – zyx May 24 '13 at 21:21
three? :/ but all of them describe the same thing actually right? "An angle" and all three are refering to the "same angle" right? – themhz May 24 '13 at 22:28
man I also found "turns" why do we use so many anyway :/ – themhz May 24 '13 at 22:37

If you ask your calculator for $\tan ^{-1}(1)$, there are four possible answers:

If you get $45$. your calculator is set to use (receive and report) angles in degrees;

If you get $50$, your calculator is set to use grads;

If you get $0.785398163$, your calculator is set to use radians;

If you get anything else, your calculator is broken.

The grad is a unit of angle such that a right-angle contains $100$ grads

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Thank you. First time I see the unit "grads". So we have "grads", "degrees", "radians" ? are there more? – themhz May 24 '13 at 22:34
There are many different ones listed on wikipedia at but most people in most contexts are find if they just know about radians and degrees. – Mark S. May 25 '13 at 15:52

There is a unit-less sine function (call it, very originally, $\sin (x)$) for which the input is a number and the output a number. The computation of this in a calculator might be something like taking the first few terms of the formula $\hskip5pt \sin x = x + \frac{x^3}{6} + \frac{x^5}{120} + \cdots = \sum \frac{x^{2n-1}}{(2n-1)!}$. Notice that $x$ should be dimensionless for addition of different powers to be meaningful, or we would be adding degrees to cubes of degrees and higher powers. In any case, the input and output are (real) numbers.

The sine of "$x$ radians" is $\sin (x)$.

The sine of "$x$ degrees" is $\sin (\frac{2 \pi}{360} x)$, in which you can see the conversion factor that equates $2 \pi$ radians to $360$ degrees.

The same remarks apply for all the other trigonometric functions. They are unitless mathematical functions of period $2\pi$, the functions with units are rules for calling on the unitless function, and radians are a name for the units in which the conversion factor (to the standard $2\pi$ normalization of the mathematical functions) is $1$.

To harmonize this with the other answers:

  • there are three sine functions, a primary one (mathematicians' sine, if you will) and the other two derived from it by detaching the input from its "unit of measurement", multiplying the input by a (unitless) conversion constant, and applying to the result the primary $\sin(x)$ function

  • setting a calculator to degrees or radians is a choice of which sine function will be called when you press the $\sin x$ button.

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Thank you for your response. What is "unitless mathematical function"? – themhz May 24 '13 at 22:32
It's rather the other way around. Most functions are unitless and do not make sense for arguments that have units. What is $x + 2x^2$ where $x$ is in Celsius degrees, or $10^x$? Having units on the input is the unusual case. The are rules for multiplying and dividing quantities with units (suchs 3 meters/sec x 5 sec = 15 meters) but for any more complicated function the unit is a rule for what factor to multiply the input by, after detaching all its "units", when calling the function. – zyx May 28 '13 at 4:26

There are two points I would like to make, the first is pretty pedantic and I apologize in advance. The Taylor series for $sin(x)$ is missing negative signs, the equation should read .. $$ sin(x)=\sum_{n=0}^\infty (-1)^n \frac{x^{2n+1}}{(2n+1)!} $$

the negative signs are sort of important when you want to preserve the correctness of the identity $sin(x) \leq |x|$.

My second point is a more serious one, I would like to engage in a serious discussion about aesthetic and technical issues regarding units in the teaching and learning of mathematics. Is it some romantic attachment to the Babylonian number system that forces us to tolerate a situation in which the expression $sin^{-1}(1)$ cannot be evaluated without some specification of whether we are using degrees or radians?

Trigonometric functions seem to be the only area of mathematics where this issue can arise.

With regard to dimensional functions I think that the issue of suppressed coefficients and units is relevant. consider the equation for the distance travelled by an object having initial velocity $v_0$ experiencing uniform acceleration $a$. $$ d=v_0t+\frac{1}{2}at^2 $$ In the particular case where $v_0=1\frac{m}{s}$ and $a=4\frac{m}{s^2}$ I think it would be perfectly reasonable to write the above equation as $$ d=t+2t^2 $$ The suppression of coefficients and units is both aesthetically appealing and practical - if a student were to be required to factor the above expression, the units would just get in the way and make the task more difficult. At some point I developed the habit of rearranging formulas symbolically and evaluating only at the end, it is at this evaluation phase that I make choices about the correct units to use and some sort of dimensional analysis is often key to detecting errors in my algebra.

So any polynomial of finite degree can always be interpreted as having coefficients with suppressed units. It is only with transcendental functions like $sin(x)$ that the arguments really need to be pure numbers, it's just too bad that in the case of angles, it seems to require a separate statement of intent to say what pure number we are talking about.

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Thank you for the reply even though I had some time to jump in to this forum. Are you saying in your first statement that , one can actually calculate ( theoretically ) using Mr Taylor's equation series to find the sin(x) were x can be rads or degrees?. Please forgive me I can't follow the second part of your statement at the moment. – themhz Jun 23 '15 at 14:35
if you want the function to take degrees as its argument the expansion would be $sin(x)=\sum_{n=0}^\infty (-1)^n \frac{ (\frac{ 2 \pi x}{180}) ^{2n+1}}{(2n+1)!}$ – WW1 Jun 23 '15 at 22:38

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