# How do you solve arctan(0.75)?

Mar 17, 2018

$\arctan \left(0.75\right) = \frac{5 \pi}{24} + n \pi , \forall n \in \mathbb{Z}$.

#### Explanation:

Let $\varphi = \arctan \left(\frac{3}{4}\right)$.

This means $\tan \varphi = \frac{3}{4}$.

tan varphi = 3/4 = sin varphi/cosvarphi = sin varphi/(+-sqrt(1-sin^2varphi)

It is not specified which quadrant does the angle $\varphi$ is in, so we'll take the general case.

$\sin \frac{\varphi}{\pm \sqrt{1 - {\sin}^{2} \varphi}} = \frac{3}{4}$

$\implies \sin \varphi = \pm \frac{3}{4} \sqrt{1 - {\sin}^{2} \varphi}$

Square both sides .

${\sin}^{2} \varphi = \frac{9}{16} \left(1 - {\sin}^{2} \varphi\right)$

$1 \cdot {\sin}^{2} \varphi = \frac{9}{16} - \frac{9}{16} {\sin}^{2} \varphi$

$\left(1 + \frac{9}{16}\right) {\sin}^{2} \varphi = \frac{9}{16}$

$\frac{25}{16} {\sin}^{2} \varphi = \frac{9}{16}$

Now we can the square root of both :

$\pm \frac{5}{4} \sin \varphi = \pm \frac{3}{4} \implies \sin \varphi = \pm \left(\frac{3}{4} \cdot \frac{4}{5}\right)$

$\sin \varphi = \pm \frac{3}{5}$.

Calculating $\cos \varphi$, you'll get

$\cos \varphi = \pm \frac{4}{5}$

$\textcolor{red}{\text{Important Note}}$: $\sin \varphi$ and $\cos \varphi$ must have the same sign such that $\tan \varphi > 0$.

$\implies \varphi = \arcsin \left(\pm \frac{3}{5}\right) = \arcsin \left(\pm 0.6\right)$

In order to simplify things, let's take only the positive value and assume $\varphi \in {Q}_{I}$.

${\varphi}^{+} = \arcsin \left(0.6\right)$

Notice how $0.6$ is between $0.5$ and $0.707 = \frac{\sqrt{2}}{2}$.

We can deduce ${\varphi}^{+}$ to be in the interval $\left(\frac{\pi}{6} , \frac{\pi}{4}\right)$.

Considering how $\sin {\varphi}^{+} \approx \frac{\sin \left(\frac{\pi}{6}\right) + \sin \left(\frac{\pi}{4}\right)}{2}$,

this implies that ${\varphi}^{+}$ is close to being in "between" $\frac{\pi}{6}$ and $\frac{\pi}{4}$. In other words,

${\varphi}^{+} \approx \frac{\frac{\pi}{6} + \frac{\pi}{4}}{2}$

${\varphi}^{+} \approx \frac{5 \pi}{24}$.

I know this answer has a lot of implied info, but it's as good as you can get without using a calculator or doing calculus.

The actual value of $\sin \left(\frac{5 \pi}{24}\right)$ is ~$0.608$.

And finally, since the $\tan$ function has period $n \pi , n \in \mathbb{Z}$, this means that

color(red)(arctan 0.75 in {(5pi)/24 + npi, n in ZZ}.