Can we show that $K=\tan(\frac{\theta_B}{2} + 45^\circ)$, given $\theta_B = \arctan(K) - \arctan(\frac{1}{K})$?

Question

I am studying two separate technical documents which are about the same topic. I would like to know if they are defining certain two variables exactly the same.

In the first document, it defines a $\theta_B$ variable in terms of a positive $K$ variable as

$\theta_B = \arctan\left(K\right) - \arctan\left(\dfrac{1}{K}\right).\quad\quad$ (source, page 11)

In the another document, it calculates $K$ back from $\theta_B$ as

$K=\tan\left(\dfrac{\theta_B}{2} + 45^o\right).\quad\quad$ (source, page 32)

Are these two conversions between $\theta_B$ and $K$ consistent with each other? If yes, how do you get the second formula from the first one?

Answer 1

Consider a right triangle with legs $1$ and $K$:

$\alpha=\arctan K$, $\beta=\arctan \frac{1}{K}$, $\alpha+\beta=\frac{\pi}{2}$.

Then $\theta_B+\alpha+\beta=\theta_B+\frac{\pi}{2}$. On the other hand, $\theta_B+\alpha+\beta=\alpha-\beta+\alpha+\beta=2\alpha$ , hence $K=\tan\left(\frac{\theta_B}{2}+\frac{\pi}{4}\right)$.

Answer 2

They are consistent. Eliminate K by putting the first equation into the second, the second equation becomes: $\theta=0.5\theta+45-\arctan(\frac{1}{\tan(0.5\theta+45)})$ This becomes $0.5\theta-45=-\arctan(\frac{1}{\tan(0.5\theta+45)})$ Change signs and take tangent on both sides, we get $\tan(45-0.5\theta)=\frac{1}{\tan(0.5\theta+45)}$ Cross multiply we arrive at $\tan(45-0.5\theta)\tan(0.5\theta+45)=1$ With the tangent sum formulas, the last line indeed can be shown to be an identitiy. Please excuse my poor formatting....

Answer 3

Recall the identity $$\arctan x + \arctan\frac1x = \begin{cases}\frac\pi2&\qquad x > 0\\ -\frac\pi2&\qquad x < 0 \end{cases}$$

Thus $$\theta_B = \arctan K - \left(\frac\pi2 - \arctan K\right) = 2\arctan K - \frac\pi2$$

Therefore $$\arctan K = \frac{\theta_B}2 + \frac\pi4 \implies K = \tan\left(\frac{\theta_B}2 + \frac\pi4\right)$$ as desired.

Answer 4

It's best to draw the right triangle.

Consider a right triangle with legs $1$ and $K$.

We know the acute angles are complementary.

Call them $45^\circ+\alpha$ and $45^\circ-\alpha$, where $\alpha>0$.

We know their tangents are reciprocals.

So $\arctan(K)-\arctan\left(\frac1K\right)=\pm2\alpha$ depending whether $K>1$ or $K<1$.

If $K>1$, $2\alpha=\theta_B$ so $\tan\left(\frac{\theta_B}2+45^\circ\right)=K$

If $K<1$, $K$ is opposite $45^\circ-\alpha$ and $-2\alpha=\theta_B$ so $\tan\left(\frac{\theta_B}2+45^\circ\right)=K$