Identity $\arctan{\frac{1-\beta}{2\sqrt{\beta}}}=\arcsin{\frac{1-\beta}{1+\beta}}$

Question

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

$$\arctan{\frac{1-\beta}{2\sqrt{\beta}}}=\arcsin{\frac{1-\beta}{1+\beta}} $$

Answer 1

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.

Answer 2

HINT

Let use trigonometric functions and inverse trigonometric functions identity:

$$\tan(\arcsin x)=\frac x{\sqrt{1-x^2}}$$

which can be easily estabilshed by the following scheme:

(credit Wikipedia)

Refer also to the related:

• Show that $ \tan(\arcsin x)= \frac{x}{\sqrt{1-x^2}} $

Answer 3

Another way to show two functions are equal.

Let:

$$f(x)=\arctan{\frac{1-x}{2\sqrt{x}}}$$ $$g(x)=\arcsin{\frac{1-x}{1+x}}$$

Both are defined and $C^1$ for $x>0$.

Now,

$$\arctan'(x)=\frac{1}{1+x^2}$$ $$\arcsin'(x)=\frac{1}{\sqrt{1-x^2}}$$

Therefore

$$f'(x)=\left(\frac{1-x}{2\sqrt{x}}\right)'\frac{1}{1+\left(\dfrac{1-x}{2\sqrt{x}}\right)^2}=\frac{-2\sqrt{x}-(1-x)\frac{1}{\sqrt{x}}}{4x}\cdot\frac{4x}{4x+1-2x+x^2}\\ =\frac{1}{\sqrt x}\cdot\frac{-x-1}{(1+x)^2}=-\frac{1}{(1+x)\sqrt x}$$

$$g'(x)=\left(\frac{1-x}{1+x}\right)'\frac{1}{\sqrt{1-\left(\frac{1-x}{1+x}\right)^2}}=\frac{-(1+x)-(1-x)}{(1+x)^2}\cdot\frac{1+x}{\sqrt{(1+x)^2-(1-x)^2}}\\ =-\frac{2}{1+x}\cdot\frac{1}{\sqrt{4x}}=-\frac{1}{(1+x)\sqrt x}$$

Hence $f'(x)=g'(x)$ for $x>0$. Since $f(1)=g(1)=0$, we have $f(x)=g(x)$ for all $x>0$.