# proof

• Jun 18th 2006, 06:37 PM
ling_c_0202
proof
i have got the question...

in triangle ABC, the lengths of the sides a, b, c have the relation 2b equals to a + c.

i. prove that 2 sin B equals sin A + sin C.
this part i have done...
but i can t handle the next part, which is

ii hence, show that cos {（A - C）/2 } = 2 sin（B / 2）
• Jun 18th 2006, 06:45 PM
malaygoel
Quote:

Originally Posted by ling_c_0202
i have got the question...

in triangle ABC, the lengths of the sides a, b, c have the relation 2b equals to a + c.

i. prove that 2 sin B equals sin A + sin C.
this part i have done...
but i can t handle the next part, which is

ii hence, show that cos {（A - C）/2 } = 2 sin（B / 2）

$\displaystyle 2sinB = sinA + sinC$
Use
$\displaystyle sinB =2sin\frac{B}{2}cos\frac{B}{2}$
$\displaystyle sinA + sinC = 2sin\frac{A+C}{2}cos\frac{A-C}{2}$
$\displaystyle sin\frac{A+C}{2}=sin\frac{\pi - B}{2}=sin\frac{B}{2}$
Keep smiling
• Jun 18th 2006, 07:15 PM
ling_c_0202
thank you
$\displaystyle sin\frac{A+C}{2}=sin\frac{\pi - B}{2}=cos\frac{B}{2}$