Solving for $a$ in this equation $\sin^a(a) = b$ $$\sin^{a}a=b$$
Do you know how to solve for $a$ here algebraicaly? This is a little bit too abstract for my understanding. Thank you very much.
 A: Looking at the plot for this function you probably want $b$ to be in the range $[0,1]$ and $a\in[0,\pi]$.
The best you can do in the general case is to use Newton's method. Set $$f(x) = \sin^x(x) - b,$$then $$f'(x) = \sin^x(x)(\cot(x) + \log(\sin(x)) )$$
If, given $b$, I had to numerically approximate this, I would 
1) Graph $f(x)$ and look for a zero.
2) Select a starting input to Newton's method that is close to your zero.
3) Iterate until you have desired convergence keeping track so your solution does not diverge. But it looks like a nice enough function that you probably needn't worry in this range.
A: We look at it this way: To find $a$ for given value of $b$,instead  we can see behavior of the function $y = (\sin x )^{x}  $ treating them as variables. Much can be learnt by graphing it.
e.g., $ y = 1 $ at $ x = ( 4 k -1) \pi/2 $ if $ x >0$
at all  $ x = k \pi, y \rightarrow  \infty , x<0 $ as at poles ( zero in denominator), $ x=  k \pi $ are vertical asymptotes.
A: This equation can't be solved using algebra. You can find the solution using the Newton's method that gives you a very good approximation in few steps.
A: You have to use power reducing trig identities. Wikipedia has a list of them: https://en.wikipedia.org/wiki/List_of_trigonometric_identities. Go to the power reducing section and use either the one for odd sine powers or the one for even sine powers. The formula is rather messy, which is why it isn't typically a problem discussed much. Even after reducing the power, you'll probably have to use a fair amount of algebra, to massage things into a nice format.
