How to prove $\sin(x)\arccos(x) + x\cos(x) > 0$

Question

How to solve this: $\sin(x)\arccos(x) + x\cos(x) > 0$? I have drawn some graphs and investigated that $0<x\leq1$. But how to prove it?

Answer 1

Not a sufficient proof.

Because of the $\cos ^{-1}(x)$, the domain is limited to $-1 \leq x \leq 1$. Using a Taylor expansion built around $x=0$, we have $$f(x)=\sin (x) \cos ^{-1}(x)+x \cos (x)=\left(1+\frac{\pi }{2}\right) x-x^2-\left(\frac{1}{2}+\frac{\pi }{12}\right) x^3+O\left(x^5\right)$$ and the rhs does not show, in the considered interval, any root except the trivial $x=0$. The rhs goes through a maximum at $$x_*=\frac{\sqrt{160+64 \pi +8 \pi ^2}-8}{12+2 \pi }\approx 0.709757 \implies f(x_*) \approx 1.04772$$ and $f(1)=\cos(1) >0$.

For the fun of it, plot the function and the Taylor series on the same graph.

Answer 2

The domain here is $[-1,1]$. Clearly, when $x = 0$, the left hand side evaluates to $0$, so it doesn't hold.

When $x\in[-1,0)$, we have $\sin(x) < 0$ and $\arccos(x) > 0$, so the first term is negative; similarly, $x < 0$ and $\cos(x) > 0$, so the second term is negative. Thus the left hand side is negative, and so it doesn't satisfy the inequality.

Finally, when $x\in(0,1]$, we have $\sin(x),\cos(x),x > 0$ and $\arccos(x) \geq 0$, so the left hand side must be positive, thus satisfying the inequality.