This may sound dumb, but does such a way exist to mentally (and quickly) determine the values of trigonometric functions such as $\sin(47^\circ)$ and so forth--quickly being a mere matter of seconds? My physics teacher suggested to our class that it is in fact possible, though I see no other ways apart from memorization and the standard methods involving triangles or trig identities (which are decidedly not quick).

I'm on my high school's 'mathletes' team, and though it's unnecessary minutiae I thought that it would be a fun thing to share with my teammates. That, and calculating logarithms mentally, though it appears that the means for doing so has already been answered on this site. (Feel free to comment on the logarithm issue as well, however, if you do have a particularly clever method that you'd be willing to share.)

Parameters: to 2-3 accurate decimal places is ideal. I'm not quite sure how that corresponds %-wise.

  • $\begingroup$ This looks closely related to math.stackexchange.com/questions/446076/…, though that one asks only about tangent, not sine or cosine. That was a relatively highly-ranked question, by the way. $\endgroup$ – David K Feb 13 '16 at 22:28
  • $\begingroup$ The way calculators evaluate $\sin$, $\cos$, $\tan$, $\log$, $\ln$, and other complicated functions is with a Taylor series which is basically an approximation of these functions using a polynomial. This is relatively easy for a calculator, but really hard for mental math. $\endgroup$ – Noble Mushtak Feb 13 '16 at 22:30
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    $\begingroup$ I believe it is within your ability to improve this Question by quantifying how much accuracy is to be obtained by what mixture of memorization and mental computation. In fact you might benefit from attempting the problem yourself once the parameters are better specified. $\endgroup$ – hardmath Feb 13 '16 at 22:55
  • $\begingroup$ I'd prefer to use only limited memorization as possible, apart from fundamentals of trigonometry (ex. the most common unit circle trig values). The relative usages does not matter to me as long as it is relatively quick. $\endgroup$ – Avant Guard Feb 14 '16 at 1:00

For something like $47^\circ$, you know that $\sin45^\circ =\dfrac1{\sqrt2}\approx0.7071$ and this is $2^\circ$ more than that. The rate of change of the sine-in-radians at that point is the cosine at that point, and the cosine of $45^\circ$ is also $\frac 1 {\sqrt 2}\approx0.7071$, so multiply by $\pi/180^\circ$ to get the rate of change in degrees. If you want a really crude approximation, you can say $\pi/180\approx3/180=1/60$, and $0.7\times1/60\approx0.0116666\times2$, so the answer is about $0.7+(2/\times0.0116666)=0.73333$.

But to be really accurate you'd probably want a piece of paper at least.


For sine in degrees, you can get an estimation using this formula (where x is a number of degrees from 0 to 90):


Simplified, this is just the following steps:

  1. Subtract the number of degrees from 100
  2. Square that number
  3. Subtract the result from 10,000.
  4. Divide step 3's result by 10,000.

Note that, for step 2, you should be familiar with quick methods for squaring 2-digit numbers quickly.

Let's try sin(47°) with this method:

  1. $100 - 47 = 53$
  2. $53^2 = 2,809$
  3. $10,000 - 2,809 = 7,191$
  4. $\frac{7,191}{10,000} = 0.7191$

sin(47°), in actuality, is roughly 0.7314, so the above method isn't perfect. Here's the error margin (red=sin(x), blue=approximation, green=error) on this calculation, ranging from -0.019 to +0.03.

Similarly, for cos in degrees from 0 to 90, you can use a similar formula:


This breaks down into:

  1. Add 10 to the number of degrees
  2. Square that number
  3. Subtract the result from 10,000.
  4. Divide step 3's result by 10,000.

Let's estimate cos(47°):

  1. $47 + 10 = 57$
  2. $57^2 = 3,249$
  3. $10,000 - 3,249 = 6,751$
  4. $\frac{6,751}{10,000} = 0.6751$

cos(47°) is roughly 0.682, so an estimate of 0.6751 isn't that bad. This shortcut has the same margin of error, from -0.019 to +0.03.

Obviously, these approaches put more emphasis on speed than accuracy, but you'll find that most schortcut methods have the same challenge.

  • $\begingroup$ I think this should be the best answer, but unfortunately it is late. This one helped me a lot thank you. $\endgroup$ – KKZiomek May 13 '17 at 1:38

When you have to do things like that, I suggest you find the closest angle for which you know the value of the function you need to compute (name $a$ this reference angle). At this point, use a very limited Taylor expansion to get the result.

For an example with the sine function, around $x=a$, you have $$\sin(x)=\sin (a)+ \cos (a)(x-a) +O\left((x-a)^2\right)$$ So, for your example (sorry for working using radians), $$\sin(\frac{47\pi}{180})=\sin(\frac{\pi}{4}+\frac{\pi}{90})\approx \frac{\sqrt 2}2+\frac{\sqrt 2}2\frac\pi {90}$$ which gives as an approximation $0.73179$ instead of $0.73135$.

I give you below similar formulae for some other trigonometric functions $$\cos(x)=\cos (a)- \sin (a)(x-a)+O\left((x-a)^2\right)$$ $$\tan(x)=\tan (a)+ \left(\tan ^2(a)+1\right)(x-a)+O\left((x-a)^2\right)$$ $$\cot(x)=\cot (a)- \left(\cot ^2(a)+1\right)(x-a)+O\left((x-a)^2\right)$$ $$\sec(x)=\sec (a)+\tan (a) \sec (a)(x-a) +O\left((x-a)^2\right)$$ $$csc(x)=\csc (a)- \cot (a) \csc (a)(x-a)+O\left((x-a)^2\right)$$


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