It’s not particularly efficient, but you can work the problem using very elementary analytic geometry and no trig identities.
Let $P=\left\langle\frac35,\frac45\right\rangle$, $Q=\langle -1,0\rangle$, $R=\langle 1,0\rangle$, and $O=\langle 0,0\rangle$. Clearly angle $POR$ is $\cos^{-1}\frac35$, so angle $POQ$ is $\pi-\cos^{-1}\frac35$. Let $S=\langle x_0,y_0\rangle$ be the point on the unit circle in the upper half plane such that $OS$ bisects angle $POQ$. Then $y_0$ is the perpendicular distance from $S$ to $OQ$, which must be equal to the perpendicular distance from $S$ to $OP$. The latter is the distance from $S$ to the line $y=\frac43x$.
The perpendicular from $S$ to that line has slope $-\frac34$, so its equation is $y-y_0=-\frac34(x-x_0)$, and it intersects the line $y=\frac43x$ at the point $\langle x_1,y_1\rangle$ where $y_0-\frac34(x_1-x_0)=\frac43x_1$, i.e., where $x_1=\frac{12}{25}\left(y_0+\frac34x_0\right)$ and $y_1=\frac43x_1=\frac{16}{25}\left(y_0+\frac34x_0\right)$. The distance from $\langle x_1,y_1\rangle$ to $S$ is
$$\begin{align*}&\sqrt{\left(\frac{12}{25}\left(y_0+\frac34x_0\right)-x_0\right)^2+\left(\frac{16}{25}\left(y_0+\frac34x_0\right)-y_0\right)^2}\\
&\qquad\qquad=\sqrt{\left(\frac{12}{25}y_0-\frac{16}{25}x_0\right)^2+\left(\frac{12}{25}x_0-\frac9{25}y_0\right)^2}\\
&\qquad\qquad=\frac1{25}\sqrt{(12y_0-16x_0)^2+(12x_0-9y_0)^2}\\
&\qquad\qquad=\frac1{25}\sqrt{225y_0^2-600x_0y_0+400x_0^2}\\
&\qquad\qquad=\frac15\sqrt{9y_0^2-24x_0y_0+16x_0^2}\\
&\qquad\qquad=\frac15\sqrt{(3y_0-4x_0)^2}\\
&\qquad\qquad=\frac15|3y_0-4x_0|\\
&\qquad\qquad=\frac15(3y_0-4x_0)\;,
\end{align*}$$
since clearly $y_0>0$ and $x_0<0$.
This must be equal to $y_0$, so we have $\frac15(3y_0-4x_0)=y_0$, and hence $y_0=-2x_0$. Thus, the slope of the line $OS$ is $-2$, and it follows that the tangent of angle $SOQ$ is $2$ and hence that $2\tan^{-1}2$ equals angle $POQ$, which we already saw is $\pi-\cos^{-1}\frac35$.
