Question

if A+B+C=90 then sinA/2×sin{(180-B-C)/2} + cosA/2×sin(B+C)/2. = ?​

Answer 1

Answer:

A+B+C = 90

sinA/2 sin((180-B-C)/2) + cosA/2 sin(B+C)/2

sinA/2 sin(90-(B+C)/2) + cosA/2 sin(B+C)/2

sinA/2 cos(B+C)/2 + cosA/2 sin(B+C)/2

sin((A+B+C)/2)

sin(90/2)

sin45°

1/√2

Answer 2

Answer:

Step-by-step explanation:

SinA/2 * Sin [(180-B-C)/2] + CosA/2 * Sin(B+C)/2

Sin [(180-B-C)/2]                     |                          Sin(B+C/2)

A + B + C = 180                       |   A + B + C = 180

A = 180 - (B+C)                        |   B + C = 180 - A

A/2 = [180 - (B+C)] / 2             |   B+C/2 = (180 - A)/2

Taking Sine on both sides,    | Taking Sine on both sides,  

SinA/2 = Sin[ (180 - B - C)/2]   | Sin(B+C)/2 = Sin(180 - A)/2 = Sin(90 - A/2) =    

                                                  CosA/2

Now,

SinA/2 * Sin [(180-B-C)/2] + CosA/2 * Sin(B+C)/2

=> SinA/2 * SinA/2 + CosA/2 * CosA/2

=> Sin²A/2 + Cos²A/2

=> 1.