Question

If A + B + C = 0, then prove that sin A + sin B - sin C = - 4 $cos\frac{A}{2}cos\frac{B}{2}sin\frac{C}{2}$.

Answer 1

Hey there! Thanks for the question!

Given,

A + B + C = 0

A + B = - C

Then,

sinA + sinB - sinC

= 2 sin(A+B/2) cos( A - B/2) - sinC

= 2 sin( - C /2) cos( A - B/2) - sinC

= -2 sinC/2 cos(A-B) /2 - 2sinC/2cosC/2

= -2sinC/2 [ cos(A-B) /2 + cos -(A+B)/2 ]

= -2sinC/2 [ 2cos(A/2)*cos(B/2)]

= -4sinA/2*sinB/2*sinC/2

Hence proved.

We used,

sin2A = 2sinA*cosA
sin(-x) = - sinx
sinA + sinB = Sin(A+B/2) * cos(A-B/2)
cos(A-B)+ cos(A+B) = 2cosAcosB.

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Latex version of the proof :
$sinA + sinB - sinC \: \: \\ \\ = 2 sin( \frac{ A+B}{2})cos( \frac{A - B}{2}) - sinC \\ \\ \\ \\ = 2 sin( \frac{- C}{ 2} ) cos( \frac{ A - B}{2}) - sinC \\ \\ = -2 sin (\frac{ C}{2} )cos (\frac{ A-B }{2}) - 2sin \frac{ C}{2}cos \frac{C}2 \\ \\ = -2sin \frac{C}2 [ cos (\frac {A-B}{2})+ cos( \frac{-(A+B)}2) ] \\ \\ -2sin \frac{C}2 [ cos (\frac {A-B}{2})+ cos( \frac{(A+B)}2) ] \\ \\ = -2sin \frac{C}2 [ 2cos (\frac{ A}2) \times cos \frac{B}2)] \\ \\ = -4cos \frac{A}2 \times cos \frac{B} 2 \times sin \frac{ C}2$

Answer 2

Step-by-step explanation:

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