Question

For a triangle ABC show that
Sin (A + B)/2 = cos C/2

Answer 1

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

A+B+C=180 (sum of angles of a triangle=180)

A+B=180-C

A+B/2=180-C/2

=180/2-C/2

=(90-C/2)

SUBSTITUTION....

sin(90-C/2) since,sin(90-A)=cos A

=cosC/2

LHS=RHS

Answer 2

Given:

Triangle ABC.

To prove:

$\sin \left(\dfrac{A+B}{2}\right)=\cos \dfrac{c}{2}$

Proof:

In triangle ABC,

$A+B+C=180^\circ$      (Angle sum property)

$A+B=180^\circ-C$            ...(i)

Taking LHS of $\sin \left(\dfrac{A+B}{2}\right)=\cos \dfrac{c}{2}$, we get

$LHS=\sin \left(\dfrac{A+B}{2}\right)$

$LHS=\sin \left(\dfrac{180^\circ -C}{2}\right)$     [Using (i)]

$LHS=\sin \left(90^\circ -\dfrac{C}{2}\right)$

$LHS=\cos \dfrac{C}{2}$     $[\because \sin(90^\circ -\theta)=\cos \theta]$

$LHS=RHS$

Hence proved.