Calculating \(\sin(75^\circ)\)

The objective is to determine the value of \(\sin(75^\circ)\) using trigonometric principles.

Applying the Half-Angle Formula

We utilize the half-angle identity for sine, which states:

\(\sin\left(\frac{x}{2}\right) = \pm \sqrt{\frac{1-\cos x}{2}}\)

To find \(\sin(75^\circ)\), we can set \(\frac{x}{2} = 75^\circ\), which implies \(x = 150^\circ\). Since \(75^\circ\) lies in the first quadrant, \(\sin(75^\circ)\) is positive. Therefore, we use the positive form of the identity:

\(\sin(75^\circ) = \sqrt{\frac{1-\cos(150^\circ)}{2}}\)

Determining \(\cos(150^\circ)\)

The angle \(150^\circ\) can be expressed as \(180^\circ - 30^\circ\). Using the cosine identity \(\cos(180^\circ - \theta) = -\cos(\theta)\), we get:

\(\cos(150^\circ) = \cos(180^\circ - 30^\circ) = -\cos(30^\circ)\)

We know that \(\cos(30^\circ) = \frac{\sqrt{3}}{2}\). Thus:

\(\cos(150^\circ) = -\frac{\sqrt{3}}{2}\)

Substituting and Simplifying

Substitute the value of \(\cos(150^\circ)\) back into the half-angle formula:

\(\sin(75^\circ) = \sqrt{\frac{1 - \left(-\frac{\sqrt{3}}{2}\right)}{2}}\)

Simplify the expression under the square root:

\(\sin(75^\circ) = \sqrt{\frac{1 + \frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{\frac{2}{2}+\frac{\sqrt{3}}{2}}{2}} = \sqrt{\frac{\frac{2+\sqrt{3}}{2}}{2}}\)

\(\sin(75^\circ) = \sqrt{\frac{2+\sqrt{3}}{4}}\)

Simplify the fraction inside the square root:

\(\sin(75^\circ) = \frac{\sqrt{2+\sqrt{3}}}{\sqrt{4}} = \frac{\sqrt{2+\sqrt{3}}}{2}\)

The resulting value matches the expression provided in Option D.