Question

if a b c are the interior angles of a triangle ABC then prove that sin (A+B/2) cos c/2 + cos(A+B/2) sin c/2 = 1​

Answer 1

$\underline{\large{\sf To Prove:}}$

$\sf sin(\frac{A+B}{2})cos(\frac{c}{2})+cos(\frac{A+B}{2})sin(\frac{c}{2})=1$

$\underline{\large{\sf Proof:}}$

$\sf LHS\:sin (\frac{A+B}{2}).cos(\frac{c}{2})+cos(\frac{A+B}{2}).sin(\frac{c}{2})$

using sin (A + B) formula,

sin(A+B)=sinA.cosB + cosA.sinB

$\implies \sf sin((\frac{A+B}{2})+(\frac{C}{2}))$

$\implies \sf sin(\frac{A+B+C}{2})$

we know the sum of interior angles of triangle is 180° or π

Therefore,

$\implies \sf sin(\frac{\pi}{2})$

$\implies\boxed{\large{\sf 1}}$

$\implies \sf RHS$

Hence proved .