Let varphi = arctan(3/4).
This means tan varphi = 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.
sinvarphi/(+-sqrt(1-sin^2varphi)) = 3/4
=> sinvarphi = +-3/4 sqrt(1-sin^2varphi)
Square both sides .
sin^2 varphi = 9/16 (1-sin^2 varphi)
1*sin^2 varphi = 9/16 - 9/16 sin^2varphi
(1+9/16)sin^2varphi = 9/16
25/16 sin^2varphi = 9/16
Now we can the square root of both :
+-5/4 sin varphi = +-3/4 => sin varphi = +-(3/4 * 4/5)
sin varphi = +-3/5.
Calculating cos varphi, you'll get
cos varphi = +-4/5
color(red)("Important Note"): sin varphi and cos varphi must have the same sign such that tan varphi > 0.
=> varphi = arcsin(+-3/5) = arcsin (+-0.6)
In order to simplify things, let's take only the positive value and assume varphi in Q_I.
varphi^+ = arcsin(0.6)
Notice how 0.6 is between 0.5 and 0.707 = sqrt2/2.
We can deduce varphi^+ to be in the interval (pi/6,pi/4).
Considering how sin varphi^+ ~~ (sin(pi/6) + sin(pi/4))/2,
this implies that varphi^+ is close to being in "between" pi/6 and pi/4. In other words,
varphi^+ ~~ (pi/6 +pi/4)/2
varphi^+ ~~ (5pi)/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((5pi)/24) is ~0.608.
And finally, since the tan function has period npi, n in ZZ, this means that
color(red)(arctan 0.75 in {(5pi)/24 + npi, n in ZZ}.
