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I know that sound a bit stupid, but is there a way to calculate basic trigonometric functions mentally like: $$ \sin(x) $$ $$ \cos(x) $$ $$ \tan(x) $$ $$ \arcsin(x) $$ $$ \arccos(x) $$ $$ \arctan(x) $$ I know it's a lot of functions, but if you can give me a simple and nice writen answer, that would be very nice!

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    $\begingroup$ You can for example calculate them with power series expansions or using geometry, looking at the unit circle or triangles. $\endgroup$ – mathreadler Jan 30 '17 at 5:55
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    $\begingroup$ There is no real point in being able to calculate these functions mentally, really. $\endgroup$ – Mariano Suárez-Álvarez Jan 30 '17 at 6:03
  • $\begingroup$ @MarianoSuárez-Álvarez I know, but I want just a simple method to approximate those functions $\endgroup$ – Arthur Guiot Jan 30 '17 at 6:04
  • $\begingroup$ You google "sin(23 degrees)" and you get an approximation. $\endgroup$ – Mariano Suárez-Álvarez Jan 30 '17 at 6:09
  • $\begingroup$ If you remove the "mentally" part then the question becomes one that has interest. $\endgroup$ – Mariano Suárez-Álvarez Jan 30 '17 at 6:09
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You can use the Taylor Series expansion. For example, $$\sin x= x - \frac{x^3}{3!}+\frac{x^5}{5!}+\dots$$ So as $x \approx 0$ $\sin x \approx x-\frac{x^3}{6}$. Since the error grows a bit larger as $x$ grows, you may also want to use the fact that $\sin x$ is a periodic function. Thus $$\sin(x+2 \pi)=\sin x$$ You can calculate $\cos$ and other trigonometric functions in similar ways. For $\cos$, the expansion is $$\cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}-\dots$$ And since $\tan x =\frac{\sin x}{\cos x}$, you can calculate the tangent as well. Other tools you could use include the double angle formula, and you can approximate using well known values such as $$\sin \frac{\pi}{4}=\frac{\sqrt{2}}{2}, \quad \sin \frac{\pi}{6}=\frac{1}{2}, \quad \sin \frac{\pi}{3}=\frac{\sqrt{3}}{2}$$ Or, somewhat less known but still famous $$\sin \frac{\pi}{12}=\frac{\sqrt{6}-\sqrt{2}}{4}$$ For the inverse trigonometric functions, the Taylor series for $\arcsin$ is as follows $$\arcsin x=x+\frac{1!!}{2!!}\frac{x^3}{3}+\frac{3!!}{4!!}\frac{x^5}{5}+\dots$$Where $n!!$ is the double factorial.Then, we have that $\arccos x=\frac{\pi}{2}-\arcsin x$. Using these, we can approximate the values required.

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  • $\begingroup$ Also double angle formulas can let you push closer to 0 giving better approximations. $\endgroup$ – mathreadler Jan 30 '17 at 5:58
  • $\begingroup$ @mathreadler Yes, they do. $\endgroup$ – S.C.B. Jan 30 '17 at 5:59
  • $\begingroup$ Yes, but, can you give me the "similar ways"? $\endgroup$ – Arthur Guiot Jan 30 '17 at 5:59
  • $\begingroup$ @ArthurGuiot See here and here $\endgroup$ – Stahl Jan 30 '17 at 6:01
  • $\begingroup$ @ArthurGuiot Does my edit address your question? $\endgroup$ – S.C.B. Jan 30 '17 at 6:04
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Convert the argument to a multiple of $\pi/2\approx1.57$, i.e. express the angle in "quadrants". It is an easy matter to reduce to the first quadrant.

Then

$$\cos x^*\approx 1-x^2$$ is a resonable approximation ($<5\%$ maximum error, green curve).

You can get better with an extra term,

$$\cos x^*\approx(1-x^2)\left(1-\frac{\pi^2-8}8x^2\right)\approx(1-x^2)\left(1-\frac7{30}x^2\right).$$

($<0.3\%$, blue curve.)

enter image description here

The sine is obtained from $\sin x^*=\cos(1-x^*)$ and the tangent from the ratio. The arc sine/cosine can be drawn form the first approximation, but require a square root extraction, which is uneasy.

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  • $\begingroup$ This looks remarkably similar to my own empirical approximation. $\endgroup$ – Lucian Jan 30 '17 at 8:27
  • $\begingroup$ @Lucian: no surprise, these are mere polynomial approximations with some tradeoff between accuracy and coefficient simplicity. $\endgroup$ – Yves Daoust Jan 30 '17 at 8:48
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Sorry, but it can't be done mentally. And I don't know about inverse functions but I can give you my simple method to calculate sin of any angle upto at least one digit of accuracy. First, convert your angle to radians if it's in degrees.

If you want to calculate $sin\theta$ where $\theta$ is in degrees, then first calculate $x=\frac{\pi\theta}{180}$

Now, if $\theta$ lies between 0 and 22.5 degrees, then $sin\theta=x$ approximately. The accuracy decreases as the angle increases.

If $\theta$ lies between 22.5 and 45 degrees, then $sin\theta=x\sqrt{1-(\frac{x}{2})^2}$ approximately. The closer $\theta$ is to 45 degrees, the lesser the accuracy.

If $\theta$ lies between 45 and 67.5 degrees, then use $sin\theta=\frac{1}{\sqrt{2}}(x-\frac{\pi}{4}+\sqrt{1-(x-\frac{\pi}{4})^2})$. Accuracy decreases as $\theta$ gets closer to 67.5 degrees.

If $\theta$ lies between 67.5 and 90 degrees, use $sin\theta=\sqrt{1-(\frac{\pi}{2}-x)^2}$. Accuracy increases as $\theta$ gets closer to 90 degrees.

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