Question

For the triangle ABC, show that `sin^2  A/2 + sin^2  (B + C)/2 = 1`.

Answer 1

Given:

Triangle ABC with angles A, B, and C

Formula:

Sum of angles in a triangle is 180°, complementary angle relation for sine, and the fundamental trigonometric identity

Solution:

A + B + C = 180°  [Sum of interior angles of a triangle is 180°]

B + C = 180° − A  [Transposing A to the right side]

\[\frac{B + C}{2} = 90^\circ - \frac{A}{2}\]  [Dividing both sides by 2]

\[\sin\left(\frac{B + C}{2}\right) = \sin\left(90^\circ - \frac{A}{2}\right)\]  [Taking sine on both sides}]

\[\sin\left(\frac{B + C}{2}\right) = \cos\left(\frac{A}{2}\right)\]  [Using the trigonometric relation sin(90° − θ) = cos(θ)]

\[\sin^2\left(\frac{A}{2}\right) + \sin^2\left(\frac{B + C}{2}\right) = \sin^2\left(\frac{A}{2}\right) + \cos^2\left(\frac{A}{2}\right)\]  [Substituting \[\sin\left(\frac{B + C}{2}\right)\] into the expression]

\[\sin^2\left(\frac{A}{2}\right) + \sin^2\left(\frac{B + C}{2}\right) = 1\]  [Using the trigonometric identity sin²(θ) + cos²(θ) = 1]

∴ \[\sin^2 \frac{A}{2} + \sin^2 \frac{B + C}{2} = 1\]