polynomial function and sub

$\begin{aligned} \frac{\sin^3 a + 6\sin^2 a + \sin a + 2\cos^2 a - 8}{\sin a - 1} &= \frac{\sin^3 a + 4\sin^2 a + \sin a + 2 - 8}{\sin a - 1} && \color{#3D99F6} \text{By the pythagorean identity } \sin^2 a + \cos^2 a = 1 \\ & = \frac{\sin^3 a + 4\sin^2 a + \sin a - 6}{\sin a - 1} \\ & = \frac{x^3 + 4x^2 + x - 6}{x - 1} && \color{#3D99F6} \text{Substituting }x=\sin a \\ &= \frac{(x-1)(x^2 + 5x + 6)}{x-1} \\ &= x^2 + 5x + 6 \end{aligned}$

Now, replace $x$ with $\sin a$ again. The initial expression is equal to $\sin^2 a + 5\sin a + 6$ . To find the minimum, differentiate and set equal to zero: $\begin{aligned} 0 &= 2\sin a\cos a + 5\cos a \\ &= \cos a(2\sin a + 5) \end{aligned}$ Either $\cos a = 0$ or $\sin a = -\frac{5}{2}$ . The latter is impossible, since $-1\leq\sin a \leq 1$ . Thus $\cos a = 0$ .

When $\cos a = 0$ , $\sin a = \pm 1$ . To determine our minimum, we must plug these values back into our original equation. Since $\sin a = 1$ makes the expression undefined (the denominator would be zero), we need only check $\sin a = -1$ : $\frac{(-1)^3 + 6(-1)^2 + (-1) + 2(0) - 8}{(-1) - 1} = \frac{-1 + 6 - 1}{-2} = \frac{-4}{-2} = 2$ We still need to verify that this is indeed a minimum. Remembering that our original expression is equal to $\sin^2 a + 5\sin a + 6$ , it's second derivative is $2\sin^2a - 2\cos^2 a - 5\sin a$ . For $\cos a = 0$ and $\sin a = 1$ , the value of this expression is $2 + 5 = 7 > 0$ Thus we have indeed found a minimum of $\boxed{2}$ .

$\begin{aligned} f(a) & = \frac {\sin^3 a+ 6\sin^2 a + \sin a + 2{\color{#3D99F6}\cos^2 a}-8}{\sin a -1} \\ & = \frac {\sin^3 a+ 6\sin^2 a + \sin a + 2{\color{#3D99F6}(1- \sin^2 a)}-8}{\sin a -1} \\ & = \frac {\sin^3 a+ 4\sin^2 a + \sin a -6}{\sin a -1} & \small \color{#3D99F6} \text{By long division} \\ & = \sin^2 a + 5 \sin a + 6 \\ & = {\color{#3D99F6}\left(\sin a + \frac 52\right)^2} - \frac {25}4 + 6 & \small \color{#3D99F6} \text{Since }\min \left( \left({\color{#D61F06}\sin a} + \frac 52\right)^2 \right) = \left({ \color{#D61F06}-1} + \frac 52\right)^2 \\ & = {\color{#3D99F6}\frac 94} - \frac {25}4 + 6 \\ & = \boxed 2 \end{aligned}$