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Let $$f(x):=a_1+a_2 \sin(x)+a_3 \sin(x)^2$$ $$g(x):=b_1+b_2 \sin(x)+b_3 \sin(x)^2$$ $$h(x):=c_1+c_2 \sin(x)+c_3 \sin(x)^2$$

then Mathematica shows that the determinant of the matrix

$$\left(\begin{matrix} f(x) & g(x)& h(x) \\ f'(x) & g'(x) & h'(x) \\ f''(x) & g''(x) & h''(x) \end{matrix} \right)$$ is just

$$2 (-a_3 b_2 c_1 + a_2 b_3 c_1 + a_3 b_1 c_2 - a_1 b_3 c_2 - a_2 b_1 c_3 + a_1 b_2 c_3) \cos(x)^2$$

which is a suprisingly simple result given that the expressions from the chain-rule may become rather cumbersome.

I would like to know, is there a smart way to conclude this result without explicitly calculating everything and regrouping terms in order to see the result?

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    $\begingroup$ Perhaps replacing each $\sin^nx$ with $e^{int}$ will make computing derivatives easier. Then omit the real part of the result. $\endgroup$ – user170231 Apr 1 '17 at 17:33
  • $\begingroup$ @user170231good point, but still it looks as if there must be another significant simplification hidden somewhere to get this neat result. $\endgroup$ – user331288 Apr 1 '17 at 17:35
  • $\begingroup$ I think the coefficient is $2\cos^3(x)$, rather than $2\cos^2(x)$ $\endgroup$ – user125932 Apr 1 '17 at 19:01
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You have $$f(x)=p(\sin x),\ \ g(x)=q(\sin x),\ \ h(x)=r(\sin x)$$ for three degree-two polynomials. Then $$ f'(x)=p'(\sin x)\cos x,\ \ f''(x)=p''(\sin x)\cos x-p'(x)\sin x. $$ So the determinant is \begin{align} \begin{vmatrix} p(\sin x)&q(\sin x)& r(\sin x)\\ p'(\sin x)\cos x&q'(\sin x)\cos x& r'(\sin x)\cos x\\ p''(\sin x)\cos^2 x-p'(\sin x)\sin x&q''(\sin x)\cos^2 x-q'(\sin x)\sin x&r''(\sin x)\cos^2 x-r'(\sin x)\sin x \end{vmatrix}\\ \end{align} We can decompose this as the sum of $$ \begin{vmatrix} p(\sin x)&q(\sin x)& r(\sin x)\\ p'(\sin x)\cos x&q'(\sin x)\cos x& r'(\sin x)\cos x\\ p''(\sin x)\cos^2 x &q''(\sin x)\cos^2 x &r''(\sin x)\cos^2 x \end{vmatrix} = c^3\,\begin{vmatrix} p(\sin x)&q(\sin x)& r(\sin x)\\ p'(\sin x) &q'(\sin x) & r'(\sin x) \\ p''(\sin x) &q''(\sin x) &r''(\sin x) \end{vmatrix}, $$ where $c=\cos x$, and \begin{align} \begin{vmatrix} p(\sin x)&q(\sin x)& r(\sin x)\\ p'(\sin x)\cos x&q'(\sin x)\cos x& r'(\sin x)\cos x\\ -p'(\sin x)\sin x& -q'(\sin x)\sin x& -r'(\sin x)\sin x \end{vmatrix}=0 \end{align} For the determinant with the polynomials, we have $$ \begin{vmatrix} a_1+a_2s+a_3s^2& b_1+b_2s+b_3s^2& c_1+c_2s+c_3s^2\\ a_2+2a_3s&b_2+2b_3s&c_2+2c_3s\\ 2a_3&2b_3&2b_3 \end{vmatrix} $$ Doing row reduction (first row minus $s$ times the second, and second row minus $s$ times the third), this last determinant equals $$ \begin{vmatrix} a_1& b_1& c_1\\ a_2&b_2&c_2\\ 2a_3&2b_3&2b_3 \end{vmatrix} $$ So the determinant is equal to $$ \begin{vmatrix} a_1& b_1& c_1\\ a_2&b_2&c_2\\ 2a_3&2b_3&2b_3 \end{vmatrix} \,\cos^3x $$

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  • $\begingroup$ I think the third line of your first determinant should read $p''(\sin x)\cos^2 x - p'(\sin x) \sin x$ (and so on) rather than $p''(\sin x)\cos x - p'(\sin x) \sin x$, which gives a factor of $\cos^3 x$ $\endgroup$ – user125932 Apr 1 '17 at 18:58
  • $\begingroup$ You are so right. Edited, thanks. $\endgroup$ – Martin Argerami Apr 1 '17 at 19:00
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This becomes simpler if we let

$$u(x) = 1 \qquad v(x) = \sin(x) \qquad w(x) = \sin^2(x)$$

so since differentiation is linear,

$$\left(\begin{matrix} f(x) \\ f'(x) \\ f''(x) \end{matrix} \right) = a_1 \left(\begin{matrix} u(x) \\ u'(x) \\ u''(x) \end{matrix} \right) + a_2 \left(\begin{matrix} v(x) \\ v'(x) \\ v''(x) \end{matrix} \right) + a_3 \left(\begin{matrix} w(x) \\ w'(x) \\ w''(x) \end{matrix} \right)$$

and similar formulas hold for $g$ in terms of $b_i$'s and $h$ in terms of $c_i$'s. From here we can write

$$\left(\begin{matrix} f(x) & g(x)& h(x) \\ f'(x) & g'(x) & h'(x) \\ f''(x) & g''(x) & h''(x) \end{matrix} \right) = \left(\begin{matrix} u(x) & v(x)& w(x) \\ u'(x) & v'(x) & w'(x) \\ u''(x) & v''(x) & w''(x) \end{matrix} \right) \left(\begin{matrix} a_1 & b_1& c_1 \\ a_2 & b_2 & c_2 \\ a_3 & b_3 & c_3\end{matrix} \right). $$

Call these matrices $F$, $U$, $A$, so this equation reads $F = UA$. We conclude $\det F = \det U \det A$.

We can easily compute

$$ \det U = \det \left(\begin{matrix} 1 & \sin(x) & \sin^2(x) \\ 0 & \cos(x) & 2\cos(x)\sin(x) \\ 0 & -\sin(x) & 2(\cos^2(x) - \sin^2(x)) \end{matrix} \right) = 2\cos^3(x)$$

so $\det F = 2 \cos^3(x) \det A$.

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The critical step to doing this 'smartly' is to realize that you can write your determinant as the product of two matrices as follows. We also know that $\det(A\cdot B)=\det(A)\cdot \det(B)$, so we can write the determinant as $$ \begin{vmatrix} 1&\sin{x}&\sin^2{x}\\ 0&\cos{x}&\sin{2x}\\ 0&-\sin{x}&2\cos{2x} \end{vmatrix} \cdot \begin{vmatrix} a_1& b_1& c_1\\ a_2&b_2&c_2\\ a_3&b_3&b_3 \end{vmatrix} $$ Now we just need to find the determinant of the matrix on the left, which is much simpler than what we were doing before, and turns out just to be $2\cos^2{x}$

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