If $${\Delta _1} = \left| {\matrix{ x & {\sin \theta } & {\cos \theta } \cr { - \sin \theta } & { - x} & 1 \cr {\cos \theta } & 1 & x \cr } } \right|$$ and $${\Delta _2} = \left| {\matrix{ x & {\sin 2\theta } & {\cos 2\theta } \cr { - \sin 2\theta } & { - x} & 1 \cr {\cos 2\theta } & 1 & x \cr } } \right|$, $x \ne 0$$ ; then for all $\theta \in \left( {0,{\pi \over 2}} \right)$ :

Question

If $${\Delta _1} = \left| {\matrix{ x & {\sin \theta } & {\cos \theta } \cr { - \sin \theta } & { - x} & 1 \cr {\cos \theta } & 1 & x \cr } } \right|$$ and
$${\Delta _2} = \left| {\matrix{ x & {\sin 2\theta } & {\cos 2\theta } \cr { - \sin 2\theta } & { - x} & 1 \cr {\cos 2\theta } & 1 & x \cr } } \right|$, $x \ne 0$$ ;

then for all $\theta \in \left( {0,{\pi \over 2}} \right)$ :

${\Delta _1} - {\Delta _2}$ = x (cos 2$\theta $ – cos 4$\theta $)

${\Delta _1} + {\Delta _2}$ = - 2x³

${\Delta _1} + {\Delta _2}$ = – 2(x³ + x –1)

${\Delta _1} - {\Delta _2}$ = - 2x³

Solution

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