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So i need to prove this function is constant and find the constant value

$$f:[2,\infty] \rightarrow \mathbb{R}\\ f(x)=2\arctan\left(\frac{x}{2}\right)-\arcsin\left(\frac{4x}{x^2+4}\right)$$

So i normally used those two theorems in the title for proving similar statements. But i'm not sure how to use it, so possibly my title is wrong, but i am open for suggestions. Any help solving this would be appreciated, thank you in advance.

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  • $\begingroup$ It should be[-2,2] rather than [2,+\infty) $\endgroup$ – Zack Ni Aug 2 '16 at 11:22
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HINT

One approach is to take the derivative and show it is zero. $$ f'(x) = \frac{2}{1+(x/2)^2} \times \frac{1}{2} - \frac{1}{\sqrt{1-\left(\frac{4x}{x^2+4}\right)^2}} \times \frac{d}{dx} \left[ \frac{4x}{x^2+4} \right] $$ which does not involve any trig...

Another is to note that $f(0)=f(2)=0$, so you want to show that $f(x)=0$ or in other words, $$ 2\arctan(x/2) = \arcsin\left(\frac{4x}{x^2+4}\right) $$ So take sines of both sides. RHS is easy and LHS is $$ \sin \left(2\arctan(x/2)\right) = 2 \sin (\arctan(x/2)) \cos(\arctan(x/2)) = 2 \frac{2}{\sqrt{x^2+4}} \frac{x}{\sqrt{x^2+4}} $$ and you are done

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