Question

If A,B,C are the interor angles of triangle ABC, show that cos^2A/2+cos^2(B+C/2)=1

Answer 1

Given ∆ABC;

By angle sum property;

• /_A + /_B + /_C = 180°
• ( /_A + /_B + /_C )/2 = 180°
• A/2 + (B+C)/2 = 90°
• A/2 = 90° - (B+C)/2 ................ (1)

Using sin²θ + cos²θ = 1;

• cos²(A/2) + sin²(A/2) = 1
• cos²(A/2) + sin²(90 - (B+C)/2) = 1
• cos²(A/2) + cos²( (B+C)/2 ) = 1. {using sin(90-X) = cosX}
• HENCE PROVED.

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