Question

For `DeltaABC` show that ?

`sin((A+B)/2) = cosC/2`

Answer 1

Given any `triangle ABC` with standard labelling:

We know that the sum of the interior angles is `180^o`, thus:

`A + B + C = 180^o`

`:. A + B = 180^o - C`
`:. (A + B)/2 = (180^o - C)/2`
`:. (A + B)/2 = 90^o - C/2`

Now we take the sine of both sides and use the sum of angle identity:

`sin(A+-B) -= sinAcosB +- cosAsinB`

Then we get:

`sin((A - B)/2) = sin(90^o - C/2)`
`" " = sin90^o cos (C/2) - cos90^o sin( C/2)`

And finally we have:

`sin 90^o = 1 \ \ \` and `\ \ \ cos 90^o = 0`

Giving the final result:

`sin((A - B)/2) = cos (C/2) \ \ \` QED