Find the greatest value of the function $$f\left( x \right) = \frac{{\sin \,2x}}{{\sin \left( {x + \frac{\pi }{4}} \right)}}$$       on the interval $$\left[ {0,\,\frac{\pi }{2}} \right]$$

Solution :
Let $$f\left( x \right) = \frac{{\sin \,2x}}{{\sin \left( {x + \frac{\pi }{4}} \right)}} = \sqrt 2 \left\{ {\frac{{{{\left( {\sin \,x + \cos \,x} \right)}^2} - 1}}{{\sin \,x + \cos \,x}}} \right\} = \sqrt 2 \left( {\frac{{{y^2} - 1}}{y}} \right),$$              where $$y = \sin \,x + \cos \,x$$
Let $$\phi \left( y \right) = \sqrt 2 \left( {\frac{{{y^2} - 1}}{y}} \right){\text{ and }}g\left( x \right) = \sin \,x + \cos \,x$$
We have, $$g'\left( x \right) = \cos \,x - \sin \,x$$
For max. or min. $$g'\left( x \right) = 0 \Rightarrow \tan \,x = 1 \Rightarrow x = \frac{\pi }{4}$$
For this value of $$x.$$
$$g''\left( x \right) < 0.$$   Thus, $$g\left( x \right)$$  is max. at $$x = \frac{\pi }{4}$$  and hence the domain of $$g\left( x \right)$$  is $$\left[ {1,\,\sqrt 2 } \right]$$  i.e. $$y$$ lies between $$1$$ and $$\sqrt 2 $$
Now, $$\phi '\left( y \right) = \sqrt 2 \left( {1 + \frac{1}{{{y^2}}}} \right) > 0{\text{ for all }}y\, \in \left[ {1,\,\sqrt 2 } \right]$$
That is $$\phi \left( y \right)$$  is increasing for all $$y\, \in \left[ {1,\,\sqrt 2 } \right]$$
Thus it attains the greatest value at $$\sqrt 2 $$ and is equal to $$\sqrt 2 \left( {\frac{{{{\left( {\sqrt 2 } \right)}^2} - 1}}{{\sqrt 2 }}} \right) = 1$$
Hence, greatest value of $$f\left( x \right)$$  on $$\left[ {0,\,\frac{\pi }{2}} \right] = $$   greatest value of $$\phi \left( y \right)\,{\text{on}}\,\left[ {1,\,\sqrt 2 } \right] = 1.$$