How to calculate $\arccos (x)$?

Question

how do i calculate for example $\arccos (2/\sqrt [2] 5)$ without calculator? I had some exercises with calculator and i could not find any good explanation for calculating per hand.

@YvesDaoust He doesn't talk about rounding or decimals, if someone asks "Calculate $arccos0.5$ without calculator, do you expect $1.04$ or $\pi/3$? — imranfat, Jan 31 '17 at 16:49
@imranfat: it depends if the quantity has a closed-form expression or not. — , Jan 31 '17 at 16:51
That's true, but if the OP assumes that there is a closed form, whereas in fact there may not be (as far as I know), then.... — imranfat, Jan 31 '17 at 16:52
@imranfat: I don't see any assumption about a closed form in the post. — , Jan 31 '17 at 16:52
You can lookup this amazing table: http://intmstat.com/blog/2011/06/exact-values-sin-degrees.pdf; you can find more by using the angle bisection formula. But in most cases, (like your example I guess) there is no closed-form formula. — , Jan 31 '17 at 17:04

score 1 · Accepted Answer · answered Jan 31 '17 at 17:13

When there is no closed-formula and no calculator is allowed, about the only resort is the Taylor formula.

In your case, you are lucky, as

$$\arccos\frac2{\sqrt 5}=\arctan\frac12,$$ and you can use the Gregory's series

$$\arctan\frac12=\frac12-\frac1{3\cdot2^3}+\frac1{5\cdot2^5}-\frac1{7\cdot2^7}+\cdots$$ which gives you at least two more bits of accuracy on every term with not too painful by-hand computation.

score 0 · Answer 2 · answered Jan 31 '17 at 16:33

0

You cannot calculate it's value without calculator.

You can only calculate values of inverse trigonometric functions for common angles viz. 0, 30, 45, 60, 90, 180, etc

answered Jan 31 '17 at 16:33

Raknos13

572

There are more acute angles $x$ such that $x/\pi$ and $\cos x$ are radical numbers. ($x$ is in radians). – ajotatxe Jan 31 '17 at 16:40
How do you think that the first trigonometric tables were established ? – Jan 31 '17 at 16:47
@Yves the first trigonometric table were made by calulating the values of sin and cos from right angled triangles with one angle 0,30,45,.......respectively. A different triangle was drawn for each angle and the values of sin and cos were measured physically (since, sinx = altitude/hypo. and cosx = base/hypo) – Raknos13 Feb 03 '17 at 15:49
@mikealise: I assume this was done at a time the calculators were still made of stones... Anyway, very early tables was obtained numerically, by the half-angle fomulas. – Feb 03 '17 at 16:06
@Yves They didn't had calculators at that time (around 500bc), instead they used abacus. – Raknos13 Feb 03 '17 at 16:10
@mikealise: I meant electronic calculators (by the way still made of silicon). – Feb 03 '17 at 16:12

score 0 · Answer 3 · answered Jan 31 '17 at 17:12

$\theta = \arccos(2/\sqrt{5}) = \arctan(1/2)$ is not a rational multiple of $\pi$. This can be seen from the fact that $\exp(i\theta) = (2+i)/\sqrt{5}$ is not an algebraic integer (its minimal polynomial is $z^4-(6/5) z^2+1$), but $\exp(i \pi m/n)$ is (since it's a root of $z^{2n}-1$).

How to calculate $\arccos (x)$?

3 Answers3