Trigonometric Identities

The central tool is the product-to-sum identity $2\sin A\cos B=\sin (A+B)+\sin (A-B)$, which converts a product of a sine and a cosine into an additive form that is far easier to evaluate at standard angles. This identity is essentially the angle-addition formulas for sine read in reverse, packaging two terms into one product. Here the product already carries the convenient leading factor of two, so with $A=75^{\circ }$ and $B=15^{\circ }$ the expression becomes $\sin (75^{\circ }+15^{\circ })+\sin (75^{\circ }-15^{\circ })=\sin 90^{\circ }+\sin 60^{\circ }$. The first term equals one and the second equals $\frac{\sqrt{3}}{2}$, so the total is $1+\frac{\sqrt{3}}{2}=\frac{2+\sqrt{3}}{2}$; the clean appearance of $\sin 90^{\circ }$ reflects the deliberate choice of complementary angles. Option $\frac{\sqrt{3}}{2}$ drops the $\sin 90^{\circ }$ contribution. Option $\frac{1+\sqrt{3}}{2}$ mishandles the constant by halving it incorrectly. Option $1+\sqrt{3}$ forgets to halve the radical term. This matches the standard product-to-sum conversion pattern emphasized in JEE Advanced trigonometry. As a final numeric plausibility check, $\sin 75^{\circ }\approx 0.966$ and $\cos 15^{\circ }\approx 0.966$, so $2(0.966)(0.966)\approx 1.866$, which equals $\frac{2+\sqrt{3}}{2}\approx \frac{3.732}{2}=1.866$, confirming the result.