Question

sina+sinb+sinc=4cosa/2 cosb/2 cosc/2​

Answer 1

Answer:

$sinA+sinB+sinC=4\:cos\frac{A}{2}\:cos\frac{B}{2}\:cos\frac{C}{2}$

Step-by-step explanation:

In ΔABC , Show that

$sinA+sinB+sinC=4\:cos\frac{A}{2}\:cos\frac{B}{2}\:cos\frac{C}{2}$

$Formula\:used:\\\\sinC+sinD=2\:sin(\frac{C+D}{2})\:cos(\frac{C-D}{2})\\\\sinA=2\:sin\frac{A}{2}\:cos\frac{A}{2}$

In ΔABC , $A+B+C=\pi$

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

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

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

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

Answer 2

I hope this answer helps you