Fun Trigonometric Equivalence

$\sqrt{2}\sin \left ( \frac{2x+1}{4}\pi \right )=\sqrt{2}\sin \left ( \frac{2x}{4}\pi+\frac{\pi}{4} \right )$

$=\sqrt{2}\sin \left ( \frac{\pi x}{2}+\frac{\pi}{4} \right )$

$=\sqrt{2}\left [\sin \left ( \frac{\pi x}{2}\right )\cos\frac{\pi}{4} +\cos \left ( \frac{\pi x}{2}\right )\sin\frac{\pi}{4} \right ]$

Using: $\sin (A+B)= \sin(A)\cos(B)+\cos(A)\sin(B)$

$=\sqrt{2}\left [\sin \left ( \frac{\pi x}{2}\right ).\frac{1}{\sqrt{2}} +\cos \left ( \frac{\pi x}{2}\right ).\frac{1}{\sqrt{2}} \right ]$

$=\boxed{\cos \left ( \frac{\pi x}{2}\right ) +\sin \left ( \frac{\pi x}{2}\right )}$

Since the question is a variable one and the ans options are also variable,u can put x=0 in the expression given in question and u will get the ans as 1, and then put x=0 in each of the ans option, and check whih option gives 1, using this method u will eliminate fisrt two ans options,but last two ans option gives same value for x=0 so u can put x=1 in the givn expression and match the value by putting x=1 in the last two expressions, so finally u get option d