Question

Maximum and minimum values of sin 2x - cos 2x​

Answer 1

$\begin{gathered}\begin{gathered}\bf Given -  \begin{cases} &\sf{f(x) = sin2x - cos2x} \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\begin{gathered}\bf  To \:  Find :-  \begin{cases} &\sf{maximum \: and \: minimum \: value \: of \: f(x)}  \end{cases}\end{gathered}\end{gathered}$

$\begin{gathered}\Large{\bold{{\underline{Formula \: Used \::}}}} \end{gathered}$

$\begin{gathered}\bf\red{\tt \:  \boxed{\tt \:  (1). \: - 1 \leqslant sinx \leqslant 1}}\end{gathered}$

$\begin{gathered}\bf\green{\tt \:  \boxed{\tt \:  (2). \: sin(x - y) =sinxcosy - sinycosx }}\end{gathered}$

$\large\underline\purple{\bold{Solution :- }}$

$\tt \:  \longrightarrow \: f(x) = sin2x - cos2x$

$\tt \:  \longrightarrow \: multiply \: and \: divide \: by \: \sqrt{2}$

$\tt \:  \longrightarrow \: f(x) = \sqrt{2} \bigg( \dfrac{1}{ \sqrt{2} } \: sin2x - \dfrac{1}{ \sqrt{2} } \: cos2x\bigg)$

$\tt \:  \longrightarrow \: f(x) = \sqrt{2} \bigg(cos \dfrac{\pi}{ 4 } sin2x - sin\dfrac{\pi}{ 4} cos2x\bigg)$

$\tt \:  \longrightarrow \: f(x) = \sqrt{2} sin \bigg(2x - \dfrac{\pi}{4} \bigg )$

$\begin{gathered}\bf\red{Now,}\end{gathered}$

$\begin{gathered}\bf\red{We \: know,}\end{gathered}$

$\tt \:  \longrightarrow \: - 1 \leqslant sin \bigg(2x - \dfrac{\pi}{4} \bigg ) \leqslant 1$

$\tt \:  \longrightarrow \: multiply \: by \: \sqrt{2}$

$\tt \:  \longrightarrow \: \purple{ - \sqrt{2} \leqslant \: \sqrt{2} \: sin \bigg(2x - \dfrac{\pi}{4} \bigg ) \leqslant \sqrt{2} }$

$\green{\begin{gathered}\begin{gathered}\bf Hence - \begin{cases} &\tt{minimum \: value \: = - \: \sqrt{2} } \\ \\ &\tt{maximum \: value \: = \: \sqrt{2} } \end{cases}\end{gathered}\end{gathered}}$