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To be clearer, $45^{\circ} = \frac{\pi}{4}$ radians which estimates to $0.78539816$ radians; whereas, $\sin(45^{\circ}) = \frac{\sqrt2}{2}$ estimates to $0.70710678$ which is a ratio and has no units.

Why does $\sin(0.78539816) = 0.70710678$? Or why does the $\arcsin(0.70710678) = 0.78539816$ radians?

I want to be able to convert ratios of sides to angles without a trigonometric calculator or trigonometric tables.

I´d like to be able to go from knowing that $\sin(45^{\circ}) = \frac{\sqrt2}{2}$and then be able to tell how many radians and degrees it is without a trigonometric calculator or trigonometric tables. Is this possible?

Or considering the 3-4-5 right triangle, I´d like to be able to know that the $\arcsin(\frac{4}{5}) = 0.92729522$ radians or $53.13010235^{\circ}$ without a trigonometric calculator or trigonometric tables. I understand how to convert between radians and degrees using $\pi = 180^{\circ}$ , but I don´t understand how to go from $\arcsin(\frac{4}{5})$ to $0.92729522$ radians.

I was thinking that it related to the domain $[-1,1]$ of the sin function, but radians are the range units on the sine wave curve. Is it somehow related to polar coordinates where $x=r*\cos(θ)$ and $y=r*\sin(θ)$?

Also, this website seems to hint at an extension of the radius to calculate tangent so might that be the difference in the numbers? Link

Glorfindel
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Chris W Bates
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  • Are you asking how to compute sine and arcsine by hand? – Misha Lavrov Mar 18 '17 at 04:37
  • I guess so if it will give me the number of radians so I can convert to degrees, but not if I have to measure using a protractor; that would not be very accurate. Does it require integrals and summation like in this example: http://math.stackexchange.com/questions/421892/how-to-calc-arc-sine-without-a-calculator? I´d need a refresher on how to go through that calculation if for example x = 4/5. – Chris W Bates Mar 18 '17 at 04:50
  • I assume you know the SohCahToa definition of sine and how to use it to show that $\sin(\pi/4) = \sqrt{2}/2$? – littleO Mar 18 '17 at 04:50
  • @littleO Yes and I understand the common right triangle conversions (1-1-sqrt(2) and 1-sqrt(3)-2). My question is related to calculating any ratio of sides to radians and thus degrees. How do you you convert arcsin(4/5) to radians in terms of pi without a trig calculator or trig table? – Chris W Bates Mar 18 '17 at 04:56
  • You can use power series for arcsin:http://math.stackexchange.com/questions/105024/finding-the-power-series-of-arcsin-x, and approximate by truncating first few terms – user160738 Mar 18 '17 at 05:04
  • @user1607838 So here´s my first few terms for the power series for arcsin if x=4/5: The integral of sqrt(4/5+4/5^2+4/5^4) = The integral of 0.983056458. So that doesn´t seem right. How do I integrate a summation? So then I tried using the bionomial expansion (1-(4/5)^2)^(-0.5) = 1.66666667, and that isn´t right either. So then I see this notation 2k!! and don´t know how to apply that since 2! is factorial. Any ideas? – Chris W Bates Mar 18 '17 at 05:27
  • The conversion from degrees to radians is simply a scale conversion. What seems to be confusing you is the conversion factor is purely irrational. There is utterly no significance or reason to view radians in decimal as their only "meaning" is in relation to pi. You know why sin .785 =.707. Because .785 = pi/4 =360 degrees = the base angle of an isosceles right triangle and .707 is 1/sqrt 2. That is the reason and there is nothing more to it. The decimal value appearence of these numbers have no signifance other than they estimate certain irrational numbers. – fleablood Mar 18 '17 at 05:29
  • " I want to be able to convert ratios of sides to angles without a trigonometric calculator or trigonometric tables. Is this possible." Not really. You can recreate the tables entirely by hand by yourself using trig identities for specific values, estimation, and Taylor series. But all you would be doing would be recreating the trig tables. – fleablood Mar 18 '17 at 05:34
  • So I´m guessing that this notation is the factorial of a factorial. So arcsin(4/5) is approx. = 4/5+(1/2)((4/5)^3)/3+(1/4)((4/5)^5)/5 + (1/6)*((4/5)^7)/7 = 0.906710552 but this is still an estimate of arcsin(4/5) estimated to 0.92729522 radians. This would be difficult to remember and it´s still an estimate. So how do I calculate it in terms of pi so that it´s exact and then I can easily convert it to degrees? – Chris W Bates Mar 18 '17 at 05:45
  • @fleablood I fully understand that 1/sqrt 2 = sqrt 2/2 which is estimated to 0.707 because it is irrational due to the sqrt 2. My question is related to actually calculating the conversion and not just memorizing it. I also get that the Taylor series only estimates trig functions for trig tables. But my question is how does it do that so that I can learn a shorthand method in terms of pi and not just a never ending decimal estimation in radians. – Chris W Bates Mar 18 '17 at 05:56
  • Calculating what conversion? There is no algebraic to calculate trig values. If you understand Taylor series I don't understand your question. – fleablood Mar 18 '17 at 06:03

1 Answers1

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Typically, when computing inverse trig functions, people switch to arctangent: instead of $\arcsin \frac45$, compute $\arctan \frac43$. I'm not entirely sure why; maybe we don't like factorials?

There is a reasonably easy to remember Taylor series for arctangent: $$\arctan x = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots.$$ This converges quickly when $x$ is small, and badly or not at all when $x$ is large. So our first step is to switch to $\arctan \frac34$ instead of computing $\arctan \frac43$: this will give us the complementary angle, the one that's approximately $37^\circ$ instead of the one that's approximately $53^\circ$.

But the series for $\arctan \frac34$ doesn't converge particularly quickly either: if we take $\frac34 - \frac{(3/4)^3}{3} + \frac{(3/4)^5}{5} - \frac{(3/4)^7}{7}$, we get $0.6377\dots$, which isn't very close to $\arctan \frac34 = 0.6435\dots$.

To speed things up, we try to make clever use of the identity $$\arctan x + \arctan y = \arctan \frac{x + y}{1 - xy}.$$ Specifically, if we want to compute $\arctan x$, we try to pick $y$ such that both $y$ and $\frac{x+y}{1-xy}$ are smaller than $x$.

In the case of the angles for Pythagorean triples, there is a built-in special case of this identity that makes our lives easier: in an $(a,b,c)$ right triangle, $$\arctan \frac ab = 2 \arctan \frac{c-b}{a}.$$ So instead of computing $\arctan \frac34$, we can compute $2 \arctan \frac13$ instead.

This works much better: even with just the first $3$ terms, we get $$2 \left(\frac13 - \frac{(1/3)^3}{3} + \frac{(1/3)^5}{5}\right) = \frac{782}{1215} \approx 0.6436\dots$$ which gets us three correct decimal digits and most of the way to a fourth.

Even better, when you want to compute $\arctan x$, is to approximate $x$ by a close but simpler $y$, and use a variant of the identity above: $$\arctan x = \arctan y + \arctan \frac{x-y}{1+xy}.$$ In the case of $\arctan \frac34$, we can approximate $\frac34$ by $1$, and get $$\arctan \frac34 = \arctan 1 + \arctan \frac{\frac34 - 1}{1 + \frac34 \cdot 1} = \frac\pi4 - \arctan \frac17.$$ Assuming you know lots of digits of $\pi$, we now have $$\arctan \frac34 \approx \frac\pi4 - \frac17 + \frac{(1/7)^3}{3} = 0.64351\dots$$ which is four digits of accuracy, after only two terms of the Taylor series!

Misha Lavrov
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