ɪғ ᴀ, ʙ ᴀɴᴅ ᴄ ᴀʀᴇ ɪɴᴛᴇʀɪᴏʀ ᴀɴɢʟᴇs ᴏғ ᴀ ᴛʀɪᴀɴɢʟᴇ ᴀʙᴄ, ᴛʜᴇɴ sʜᴏᴡ ᴛʜᴀᴛ sɪɴ [(ʙ + ᴄ)/2] = ᴄᴏs ᴀ/2.
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➪Given△ABC
➪We know that sum of three angles of a triangle is 180
➪Hence ∠A+∠B+∠C=180°
➪or A+B+C=180°
➪B+C=180° −A
➪Multiply both sides by 1/2
➪1/2 (B+C) = 1/2 (180° − A)
➪1/2 (B + C) = 90° - A/2...(1)
➪Now 1/2 (B+C)
➪Taking sine of this angle
➪sin (B + C /2)
➪[B + C /2 = 90° - A/2]
➪sin(90°-A/2)
➪cos A/2 [sin(90° - θ ) = cos θ]
➪Hence sin( B + C/2 )=cos A/2 proved
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