Assuming that the question is as follows:
[tex]\arccos (\frac{\sqrt[]{2}}{2})=\cos ^{-1}_{}(\frac{\sqrt[]{2}}{2}),0The question is asking for the function arccos (or inverse cosine) of the value, that is the angle, theta, that gives us a cosine(theta) = (sqrt(2)/2). Then, we have that this value is, in degrees, as follows: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 (for this case, the triangle has two angles that equal 45 degrees):
[tex]\cos (\theta)=\cos (45)=\frac{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}[/tex]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).
