Assuming that the question is as follows:
![\arccos (\frac{\sqrt[]{2}}{2})=\cos ^(-1)_{}(\frac{\sqrt[]{2}}{2}),0<em>The question is asking for the function arccos (or inverse cosine) of the value, that is the </em><strong>angle</strong><em>, theta, that gives us a cosine(theta) = (sqrt(2)/2)</em>. Then, we have that this value is, in degrees, as follows:<p>If we represented this angle as a right triangle (in fact, a right-angled isosceles triangle) with sides (legs) equal to one, then, we have that (<em>for this case, the triangle has two angles that equal 45 degrees</em>):</p>[tex]\cos (\theta)=\cos (45)=(adj)/(hyp)=\frac{1}{\sqrt[]{2}}\cdot\frac{\sqrt[]{2}}{\sqrt[]{2}}=\frac{\sqrt[]{2}}{\sqrt[]{2^2}}=\frac{\sqrt[]{2}}{2}\Rightarrow cos(45)=\frac{\sqrt[]{2}}{2}]()
We need to multiply both, the numerator and the denominator by the square root of 2 to have no irrational number in the denominator.
Therefore, the value of the inverse cosine of sqrt(2)/2 is the angle 45 (the correct answer is option D).