Finding angles in a parallelogram without trigonometry

Solution 1:

Let me rephrase your question first: Can the angle $x$ in the diagram be expressed in terms of angles $a$ and $b$ using only addition, subtraction, multiplication, division, exponentiation, $\pi$, and auxiliary rational numbers, and no trig or inverse trig functions? (For example, is $x$ an expression like $(a+2b)^{3/2}$?)

Here is strong evidence that the answer is no. If $a=\frac{\pi}{2}$ and $b=\frac{\pi}{3}$ (the two "simplest" angles to play with) then allowing ourselves access to trigonometry, we can show that $$x=2\arctan\left(\sqrt{13}-2\sqrt{3}\right)=\arctan\left(\frac{\sqrt{3}}{6}\right)$$

So you would have to believe that this expression is equal to a combination of $\frac{\pi}{2}$ and $\frac{\pi}{3}$ using only addition, subtraction, multiplication, division, exponentiation, $\pi$, and auxiliary rational numbers. I highly doubt this. (The transcendence of $\pi$ and $\arctan$ might be used to even prove that it is not possible - but that's just my conjecture.)

Solution 2:

The example by alex.jordan does finish the matter, and similar ones may be constructed. We have an angle $$ \theta = \arctan \left( \frac{1}{\sqrt{12}} \right) $$ and we wish to know whether $ x = \frac{\theta}{\pi} $ is the root of an equation with rational coefficients.

Well, $$ e^{i \theta} = \sqrt{\frac{12}{13}} + i \sqrt{\frac{1}{13}} $$ Next, $\cos 2 \theta = 2 \cos^2 \theta - 1 = \frac{11}{13}.$ So, by Corollary 3.12 on page 41 of NIVEN we know that $2 \theta$ is not a rational multiple of $\pi.$ So, neither is $\theta,$ and $$ x = \frac{\theta}{\pi} $$ is irrational.

Now, the logarithm is multivalued in the complex plane. We may choose $$ \log(-1) = \pi i. $$ With real $x,$ we have chosen $$ (-1)^x = \exp(x \log(-1)) = \exp(x\pi i) = \cos \pi x + i \sin \pi x. $$ With our $ x = \frac{\theta}{\pi}, $ we have $$ (-1)^x = e^{i \pi x} = e^{i \theta} = \sqrt{\frac{12}{13}} + i \sqrt{\frac{1}{13}} $$ The right hand side is algebraic.

The Gelfond-Schneider Theorem, Niven page 134, says that if $\alpha,\beta$ are nonzero algebraic numbers, with $\alpha \neq 1$ and $\beta$ not a real rational number, then any value of $\alpha^\beta$ is transcendental.

Taking $\alpha = -1$ and $\beta = x,$ which is real but irrational. We are ASSUMING that $x$ is algebraic over $\mathbb Q.$ The assumption, together with Gelfond-Schneider, says that $ (-1)^x$ is transcendental. However, we already know that $ (-1)^x = \sqrt{\frac{12}{13}} + i \sqrt{\frac{1}{13}} $ is algebraic. This contradicts the assumption. So $x = \theta / \pi$ is transcendental, with $ \theta = \arctan \left( \frac{1}{\sqrt{12}} \right) $

Solution 3:

In the case $a = \pi/2$, $x = \arccos \left( 2\,{\frac {\sin \left( b \right) }{\sqrt {4-3\, \cos^2 \left( b \right) }}} \right)$. This is not an algebraic function of $b$, because its derivative is $\frac{dx}{db} = \frac{2}{3 \cos^2 b - 4}$ for $-\pi/2 < b < \pi/2$, and $\cos(b)$ is not an algebraic function.