Why is tan(a) not equal to sin(a)/cos(a)

trying to simplify sin(alpha)/cos(alpha) Matlab does not give tan(alpha), why?

also isolating alpha from b == sin(alpha)/cos(alpha) does not give alpha = atan(b), why?

>> syms alpha b

equ1=b==sin(alpha)/cos(alpha)

simplify(equ1)

equ1=isolate(equ1,alpha)

equ1 =

b == sin(alpha)/cos(alpha)

ans =

b == sin(alpha)/cos(alpha)

equ1 =

alpha == -log((- b^2 - 1)^(1/2)/(b + 1i))*1i

[David Goodmanson]

Hi cpistor,

to be honest, there are certain situations where Matlab symbolics is not all that bright. But the expression for alpha on your last line is a good expression for atan(b).

( For complex input, the expression does not always exactly agree with atan. Somtimes there is disagreement by pi, reflecting the pi ambiguity in the tan function, i.e. tan(theta) = tan(theta +-pi) )

Answer 1

[Paul]

It's kind of a pain, but ....

syms b alpha
eq1 = b == sin(alpha)/cos(alpha);
eq1 = rewrite(eq1,'tan')

eq1 = 

eq1 = lhs(eq1) == simplify(rhs(eq1))

eq1 = 

alpha_sol = solve(eq1,alpha)

alpha_sol = 

Actually, I'm kind of surprised the SMT yileds this result w/o some further assumptions on the range of alpha. Suppose alpha = 3*pi/4.

syms b alpha
alpha = 3*sym(pi)/4;
b = sin(alpha)/cos(alpha);
atan(b)

ans = 

OTOH, if we solve() eq1 w/o the simplifications, we get

syms b alpha
eq1 = b == sin(alpha)/cos(alpha);
alpha_sol = solve(eq1,alpha,'Real',true,'ReturnConditions',true);
alpha_sol.alpha

ans = 

alpha_sol.conditions

ans = 

I think those solutions are basically atan(b) and atan(b) +- pi.