I'm trying to solve for a certain set of trig identities in terms of other trig identities. I saw this and this, but in my example, I have two tries:
Solve[{al == Sin[b] Cos[a],
am == Sin[b] Sin[a], an == Cos[b]},
Cos[b], {a, b}]
Solve[{am == Sin[b] Sin[a], an == Cos[b]},
Cos[b], {a, b}]
In the first example, I get a null solution. In the second, Mathematica recognizes that the expression is just 'an'. Apparently I need to building the rule set up from scratch for solve to use, any suggestions here? The actual expressions I plan to use will be more complex, so it failing here isn't giving me much confidence... In the meantime, I'll be trying to generalize the Weierstrass approach.
Edit:
A fuller example might be:
Solve[{al == Sin[b] Cos[a],
am == Sin[b] Sin[a], an == Cos[b]},
Sin[b] Cos[a] + Cos[b]^2, {a, b}]
Giving the result
al+an^2
Reduce[{al == Sin[b] Cos[a], am == Sin[b] Sin[a], an == Cos[b]}, {a, b}]
? $\endgroup$Solve
. Why do you includeCos[b]
andSin[b] Cos[a] + Cos[b]^2
as the 2nd argument? $\endgroup$Solve
documentation but don't see it. But that's my problem, not yours.) $\endgroup$