Question

The most general values of $\theta $ for which $\sin \theta - \cos \,\theta \, = \,\mathop {\min }\limits_{{\text{a}} \in {\text{R}}} \,(1,\,{{\text{a}}^2}\, - \,6{\text{a}}\,{\text{ + }}\,{\text{11)}}$ are given by
A) \[{\text{n}}\pi \,{\text{ + }}\,{{\text{( - 1)}}^n}\dfrac{\pi }{4} - \dfrac{\pi }{4},\, \& \,{\text{n}}\, \in \,z\,\]
B) \[{\text{n}}\pi \,{\text{ + }}\,{{\text{( - 1)}}^n}\dfrac{\pi }{4} + \dfrac{\pi }{4},\, \&\, {\text{n}}\, \in \,z\,\]
C) \[2{\text{n}}\pi \, + \,\dfrac{\pi }{2}\, \& ,\,{\text{n}} \in {\text{z}}\]
D) \[{\text{n}}\pi \, + \,\dfrac{\pi }{2}\, \& ,\,{\text{n}} \in {\text{z}}\]

Answer 1

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