If (A + B + C) = 90°, then what is the value of sin (A / 2) sin [(180 - B - C) / 2] + cos (A / 2) sin (B + C) / 2?
1 / √2
Step 1 — Use A+B+C=90°: \(B+C=90^\circ-A\), so \(\tfrac{B+C}{2}=45^\circ-\tfrac{A}{2}\) and \(\tfrac{180^\circ-B-C}{2}=45^\circ+\tfrac{A}{2}\).
Step 2 — Let \(\alpha=\tfrac{A}{2}\): expression becomes \(\sin\alpha\sin(45^\circ+\alpha)+\cos\alpha\sin(45^\circ-\alpha)\).
Step 3 — Expand: the \(\sin\alpha\cos\alpha\) cross terms cancel, leaving \(\tfrac{1}{\sqrt 2}(\sin^2\alpha+\cos^2\alpha)=\dfrac{1}{\sqrt 2}\).
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