Trigonometry
Trigonometry
Allen Star Batch
Grade 11
Question:
The least value of $\sin^2\frac{A}{2}+\sin^2\frac{B}{2}+\sin^2\frac{C}{2}$ is: (Where $A, B, C$ are interior angles of a triangle)
$\frac{3}{2}$
$\frac{3}{4}$
$1$
None of these
Step-by-Step Solution
Key Concept: Sum-to-product formulas combined with constraint analysis yield the minimum value $\frac{3}{4}$ when angles are equal.
Starting from $\frac{1}{2}(1-\cos A + 1 - \cos B + 1 - \cos C) = \frac{3}{2} - \frac{1}{2}(\cos A + \cos B + \cos C)$, let $\cos A + \cos B + \cos C = \lambda$. Then $2\cos\frac{A+B}{2}\cos\frac{A-B}{2} + 1 - 2\sin^2\frac{C}{2} = \lambda$ leads to $4\cos^2\frac{A-B}{2} - 8(\lambda - 1) \geq 0$. Since $\sin\frac{C}{2}$ is real, $\lambda \leq \frac{2 + \cos^2\frac{A-B}{2}}{2} \leq \frac{3}{2}$. The minimum value of $\sin^2\frac{A}{2} + \sin^2\frac{B}{2} + \sin^2\frac{C}{2} = \frac{3}{4}$.
Correct Answer: 2