Answer
0
Work Step by Step
Given expression
= $ \sin^{2} 45^{\circ}$ - 2 $ \sin 45^{\circ} \cos 45^{\circ}$ + $ \cos^{2} 45^{\circ}$
= $(\frac{1}{\sqrt 2})^{2}$ - (2$ \times \frac{1}{\sqrt 2} \times\frac{1}{\sqrt 2})$ + $ (\frac{1}{\sqrt 2})^{2}$
( substituting for exact values of trigonometric functions)
=$\frac{1}{2} - (2\times \frac{1}{2}) + \frac{1}{2}$
= $\frac{1}{2} + \frac{1}{2} - (2\times \frac{1}{2}) $
= 1 - 1
= 0
