Trigonometric function problem

Question

Given: $f(x)=2\arctan(x) +\arcsin(2x/(1+x^2))$ prove that for every $x \ge 1, f(x)=\pi.$

any idea how to approach this question? thanks

Answer 1

Let $\arctan{x}=\theta$. Then $2 x/(1+x^2) = 2 \sin{\theta} \cos{\theta} = \sin{2 \theta} = \sin{(\pi-2 \theta)}$

Why have I made that last statement, even if true? Because $x \ge 1$, so that $2 x/(1+x^2)$ decreases from $1$ to $0$ for increasing $x$. In this case, the argument of the sine increases from $\pi/2$ to $\pi$. Thus,

$$f(x) = 2 \theta + \pi-2 \theta = \pi$$

Answer 2

Compute the derivative of $f$. You should find $f'(x)=0$. And therefore $f$ which is defined when the argument of $\arcsin$ is in the right domain is constant. $$f(x)=f(1)=\pi$$ Indeed $$f'(x)=\frac{2}{1+x^2}+(\frac{2}{1+x^2}-\frac{4x^2}{(1+x^2)^2})\frac{1}{\sqrt{1-(\frac{2x}{1+x^2})^2}}$$ and this simplifies into $$f'(x)=\frac{2}{1+x^2}+\frac{2-2x^2}{(1+x^2)^2}\frac{1+x^2}{x^2-1}=0$$

Answer 3

Hint : $arctan(x)=arcsin(x/\sqrt{1+x^2})$ $arcsin(x)+\arcsin(y)=\pi-arcsin(x\sqrt{1-y^2}+y\sqrt{1-x^2})$(if $x^2+y^2>1$ , $a>0$ and $b>0$)

Answer 4

Let $S^1$ be the unit circle in $\mathbb{R}^2$ and define $\theta: S^1 \to (-\pi, \pi]$ as the unique value satisfying $(\cos \theta(x), \sin \theta(x)) = x$. Using $\arcsin$, we can find an expression for $\theta$ as $\theta(x) = \begin{cases} \arcsin x_2, & x_1 \ge 0 \\ \pi - \arcsin x_2, & x_1 <0, x_2 \ge 0 \\ -\pi - \arcsin x_2, & x_1 < 0, x_2 < 0\end{cases}$.

Now let us find an expression for $2\arctan x$ in terms of $\theta$.

Using the double angle formulae, we have $\cos (2\arctan x) = 1 -2 \sin^2 (\arctan x)$, $\sin (2\arctan x) = 2 \sin (\arctan x)\cos (\arctan x)$. Letting $\alpha = \arctan x$, we have $\sin(\alpha) = {x \over \sqrt{1+x^2}}$, $\cos(\alpha) = {1 \over \sqrt{1+x^2}}$, and so $\cos (2\arctan x) = {1 - x^2 \over 1+x^2}$, $\sin (2\arctan x) = { 2x \over 1+x^2}$.

Then we have $2 \arctan x = \theta(({1 - x^2 \over 1+x^2},{ 2x \over 1+x^2})) = \begin{cases} \arcsin { 2x \over 1+x^2}, & |x| \le 1 \\ \pi - \arcsin { 2x \over 1+x^2}, & x > 1 \\ -\pi - \arcsin { 2x \over 1+x^2}, & x <-1\end{cases}$

Putting this together, we have $f(1) = \arcsin 1 + \arcsin 1 = \pi$, and for $x>1$, we have $f(x) = \pi - \arcsin { 2x \over 1+x^2} + \arcsin { 2x \over 1+x^2} = \pi$.