I have posted in Math rather than Physics as the problem is mainly abuse(?) of trig identities rather than physics.
This comes from an exercise within some lecture notes on Feynman Path integrals and I am trying to find the classical path for the Harmonic Oscillator, $\mathcal{L}(x, \dot{x}, t) = \frac{1}{2}m\dot{x}^2 + \frac{1}{2}m\omega^2x^2$
By satisfying the Euler-Lagrange equation, $\frac{d}{dt}\left(\frac{\partial \mathcal{L}}{\partial \dot{x}}\right) - \frac{\partial \mathcal{L}}{\partial x}=0$, we obtain the following ODE,
$$ \tag{1} \ddot{x} + \omega^2 x = 0 $$
The boundary conditions are, $$ \tag{2} x(t_a) = x_a \qquad\text{and}\qquad x(t_b) = x_b $$ The general solution is, $$ \tag{3} x(t) = A\sin(\omega t) + B\cos(\omega t) $$ From now on I will use the notation, $$ s_a = \sin(\omega t_a) \qquad\text{and}\qquad c_b = \cos(\omega t_b)\\ (\text{e.g.}\quad s_{a+b} = \sin(\omega [t_a + t_b])) $$
Solving for $A$ using $x(t_a) = x_a$, $$ \tag{4} A = \frac{x_a - Bc_a}{s_a} $$ Using $A$, I solve for $B$ using $x(t_b) = x_b$, $$ \tag{5} B = \frac{x_bs_a - x_as_b}{c_bs_a-c_as_b} = x_b\frac{s_a}{c_bs_a-c_as_b} - x_a\frac{s_b}{c_bs_a-c_as_b} $$ now I sub (5) into (4) to get, $$ A = \frac{x_a}{s_a} - \frac{c_a}{s_a}x_b\frac{s_a}{c_bs_a-c_as_b} + \frac{c_a}{s_a}x_a\frac{s_b}{c_bs_a-c_as_b}\\ $$
$$ \require{cancel} A = x_a\frac{(c_bs_a-\cancel{c_as_b})}{s_a(c_bs_a-c_as_b)} - x_b\frac{c_as_a}{s_a(c_bs_a-c_as_b)} + x_a\frac{\cancel{c_as_b}}{s_a(c_bs_a-c_as_b)} $$ so we get an expression for $A$ as, $$ \tag{6} A = x_a\frac{c_bs_a}{s_a(c_bs_a-c_as_b)} - x_b\frac{c_as_a}{s_a(c_bs_a-c_as_b)} $$ Using (5) and (6) in (3) gives, $$ x_t = x_a\frac{c_bs_a}{s_a(c_bs_a-c_as_b)}s_t - x_b\frac{c_as_a}{s_a(c_bs_a-c_as_b)}s_t + x_b\frac{s_a^2}{s_a(c_bs_a-c_as_b)}c_t - x_a\frac{s_bs_a}{s_a(c_bs_a-c_as_b)}c_t $$ $$ \tag{7} (c_bs_a-c_as_b) x_t = x_b(c_ts_a-c_as_t) + x_a(c_bs_t- c_ts_b) $$
I know the solution is, $$ x(t) = x_b \frac{\sin{\omega (t-t_a)}}{\sin{\omega T}} +x_a\frac{\sin{\omega (t_b-t)}}{\sin{\omega T}} $$ Where, $T = t_b-t_a$ but I can't see how I will get there from this mess. I have wrote about 4-5 pages of failed trig identities without luck.
The answer seemed to be in a recurring error I must have had in my paperwork as copying and pasting the MathJax forced me to retain terms carefully... This meta post suggested I should still post my answer from the point of my error.
From (7), using trig identity, $$ \cos(u)\sin(v) = \frac{1}{2}\sin(u+v) - \sin(u-v)\\ $$ we get, $$ \require{cancel} c_bs_a-c_as_b = \frac{1}{2}\left(\cancel{s_{b+a}} - s_{b-a} - \cancel{s_{b+a}} + s_{a-b}\right) $$ Using $\sin(a-b) = -\sin(b-a)$, $$ \tag{8} c_bs_a-c_as_b=-s_{b-a} = -s_T $$ where I define $T = t_a = t_b$. Thus, we can apply the identity derived in (8) to the numerators of $x_a, x_b$, $$ \tag{9} c_ts_a-c_as_t=-s_{t-a}\\ c_bs_t-c_ts_b=-s_{b-t} $$ Using (8),(9) in (7) we get, $$ s_Tx_t = x_bs_{t-a} +x_as_{b-t} $$ giving the final solution of, $$ x(t) = x_b \frac{\sin{\omega (t-t_a)}}{\sin{\omega T}} +x_a\frac{\sin{\omega (t_b-t)}}{\sin{\omega T}} $$