Powers of sin(x)

If we have the equality $f(x) = g(x)$ that implies $b^{\sin x} = x^{\sin x}$ . This can only occur in two cases:

Case 1: $\sin x = 0$ . Then equality is trivial because $1 = 1$ .

Case 2: $b = x$ . Equality is trivial because $b^{\sin b} = b^{\sin b}$ .

These cases may overlap, but we can prove they are exhaustive. To do this assume that $\sin x \neq 0$ and $b \neq x$ . Then $b^{\sin x} = x^{\sin x} \implies b = x$ because we can cancel the exponent, and this contradicts $b \neq x$ .

This means that when examining the second equality, $f'(x) = g'(x)$ all we have to do is check the two cases. To do this first compute $f'(x) = \log b (\cos x) b^{\sin x}$ and $g'(x) = x^{\sin x} \left[ \frac{\sin x}{x} + \log x \cos x \right]$ . For the first case, $\sin x = 0 \implies \cos x = \pm 1$ so the equality says $\log b = \log x \implies b = x$ . For the second case we have $\log b (\cos b) b^{\sin b} = b^{\sin b} \left( \frac{\sin b}{b} + \log b \cos b \right) \implies \log b \cos b = \frac{\sin b}{b} + \log b \cos b \implies \frac{\sin b}{b} = 0 \implies \sin b = 0$ .

So what we basically found is that in either case we must have a $b$ such that $\sin b = 0$ and an $x$ such that $x = b$ . For the first equality, $b$ needs to be an integer multiple of $\pi$ and to minimize it we can choose $b = \pi$ . And then choose $x = \pi$ to get the desired equalities.