Why does $\sin(45)=\frac{\sqrt2}{2}$ when $45^{\circ}= \frac{\pi}{4}$ radians?

Question

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

Answer 1

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.

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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!