In the book, Control Systems Engineering – frequency design, the author used the equality $$\phi_{max}=\arctan{\frac{1-\beta}{2\sqrt{\beta}}}=\arcsin{\frac{1-\beta}{1+\beta}}$$
Is this some famous identity?
Am I seriously missing out since I've never seen this formula before.
Edit:
Using the formula on the comments:
$$\sin{\arctan{\theta}}=\frac{x}{\sqrt{1+x^2}}$$
Let $\theta= \frac{1-\beta}{2\sqrt{\beta}}$
$$\arctan{\theta}=\arcsin{\frac{\theta}{\sqrt{1+\theta^2}}}
$$
$$\arctan{\theta}=\arcsin{\frac{1-\beta^2}{\beta^2+2\beta+1}}
$$
Thus
$$\arctan{\frac{1-\beta}{2\sqrt{\beta}}}=\arcsin{\frac{1-\beta}{1+\beta}}
$$
Best Answer
Not really new.
Let $\tan\theta=(1-\beta)/2\sqrt{\beta}$ with $-\pi/2<\theta<\pi/2$. Then using $\sec^2\theta=1+\tan^2\theta$ with $\sec\theta>0$ we have
$\sec\theta=\sqrt{(1-\beta)^2+4\beta}/(2\sqrt{\beta})=\sqrt{1+2\beta+\beta^2}/(2\sqrt{\beta}=(1+\beta)/2\sqrt{\beta}$
The sign on the square toot is controlled by the requirement $\beta>0$ for the initial arctangent to be real. So then
$\sin\theta=(\tan\theta)(\cos\theta)=\dfrac{(1-\beta)/2\sqrt{\beta}}{(1+\beta)/2\sqrt{\beta}}$
$=(1-\beta)/(1+\beta)$
as claimed.