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**Hint:**Differentiating the equation ${{\text{a}}^{\text{2}}}{\text{ - 6a + 11}}$ and finding the minimum values of a gives us the minimum value of $(1,\,{{\text{a}}^2} - 6{\text{a}} + 11)$ using sin(A-B) formula, we can get the general values of $\theta $.

**Complete step by step solution:**For the $\,\mathop {\min }\limits_{{\text{a}} \in {\text{R}}} \,(1,\,{{\text{a}}^2}\, - \,6{\text{a}}\,{\text{ + }}\,{\text{11)}}$

Let ${\text{f}}({\text{a}})\, = \,{{\text{a}}^2} - 6{\text{a}} + 11$

${{\text{f}}^1}{\text{(a)}}\,{\text{ = }}\,{\text{2a - 6}}$

For minimum ${{\text{f}}^1}{\text{(a)}}\,{\text{ = }}\,{\text{0}}$

\[\begin{gathered}

\Rightarrow \,2{\text{a - 6}}\,{\text{ = }}\,{\text{0}} \\

\Rightarrow \,{\text{a}}\,{\text{ = }}\,{\text{3}} \\

\end{gathered} \]

At ${\text{a}}\,{\text{ = }}\,{\text{3}}$ we get minimum for f(a)

$\therefore \,{\text{f}}(3)\, = \,9 - 18 + 11\, = \,2$

Then among {1,2} minimum is|

So $\sin \,\theta - \cos \,\theta \, = \,1$

Let us divide both the sides by $\sqrt 2 $.

(Because $\sin \,{45^ \circ }\, = \,\cos \,{45^ \circ }\, = \,\dfrac{1}{{\sqrt 2 }}$ )

\[ \Rightarrow \,\dfrac{{\sin \theta }}{{\sqrt 2 }} - \dfrac{{\cos \,\theta }}{{\sqrt 2 }} = \dfrac{1}{{\sqrt 2 }}\]

We know $\sin \,({\text{A}} - {\text{B}})\, = \,\sin \,{\text{A cos B - cos A sin B}}$.

Hence \[\begin{gathered}

{\text{A}}\, \to \,{\text{0}} \\

{\text{B}}\, \to \,\dfrac{\pi }{4} \\

\end{gathered} \]

Then we get

\[\sin \,\left( {\theta - \dfrac{\pi }{4}} \right)\, = \,f1\dfrac{1}{{\sqrt 2 }}\, = \,\sin \,\left( {\dfrac{\pi }{4}} \right)\]

For general solution

When

$\begin{gathered}

\sin \,{\theta ^{’}}\, = \,\sin \,{\text{x}} \\

{\theta ^{’}}\, = \,{\text{n}}\pi {\text{ + ( - 1}}{{\text{)}}^n}{\text{x for n}} \in \,{\text{z}} \\

{\text{Here }}{\theta ^{’}}\, = \,\theta - \dfrac{\pi }{4} \\

{\text{and x = }}\dfrac{\pi }{4} \\

\Rightarrow \,\left( {\theta - \dfrac{\pi }{4}} \right)\, = \,n\pi \, + \,{( - 1)^n}\dfrac{\pi }{4} \\

\Rightarrow \,\theta = \,n\pi \, + \,{( - 1)^n}\dfrac{\pi }{4} \\

\end{gathered}$

**Note:**For a quadratic equation whenever the highest degree coefficient is positive there is always a global minima .$\sin \theta - \cos \theta = 1$; This equation can also be solved by squaring on both the sides and using the property

${\sin ^2}\theta + {\cos ^2}\theta = 1$

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