Question

If A, B, C are angles in a triangle, then prove that sin A + sin B - sin C = $4sin\frac{A}{2}sin\frac{B}{2}cos\frac{C}{2}$.

Answer 1

Answer:

Step-by-step explanation:

If A, B, C are angles in a triangle, then prove that

$sin\:A+sin\:B-sin\:C=4sin\:\frac{A}{2}sin\:\frac{B}{2}cos\:\frac{C}{2}$

we have

$(sin\:A+sin\:B)-sin\:C\\\\\\=2\sin(\frac{A+B}{2} )\cos(\frac{A-B}{2} )-2sin\frac{C}{2}\:cos\frac{C}{2}\\\\=2\sin(\frac{\pi}{2} -\frac{C}{2} )\cos(\frac{A-B}{2} )-2sin\frac{C}{2}\:cos\frac{C}{2}$

∵

$\frac{A+B}{2}=\frac{\pi}{2} -\frac{C}{2}\\\\=2cos\frac{C}{2}cos\frac{A-B}{2}-2sin\frac{C}{2}\:cos\frac{C}{2}$

$=2cos\frac{C}{2}[cos\frac{A-B}{2}-sin\frac{C}{2}]\\\\=2cos\frac{C}{2}[cos\frac{A-B}{2}-sin(\frac{\pi}{2}-\frac{A+B}{2})]\\\\\\=2cos\frac{C}{2}[cos\frac{A-B}{2}-cos\frac{A+B}{2}]\\\\\\=2cos\frac{C}{2}[2sin\frac{A}{2}sin\frac{B}{2}]\\\\\\=4cos\frac{C}{2}sin\frac{A}{2}sin\frac{B}{2}$

=RHS