Question

in any triangle ABC,prove that (a-b)^2 cos^2c/2 + (a+b)^2 sin^c/2=c^2

Answer 1

We have in a triangle $\Delta ABC$ ,

[tex]\cos^2(\frac{C}{2} )+\sin^2(\frac{C}{2} )=1\\ \cos^2(\frac{C}{2} )-\sin^2(\frac{C}{2} )=\cos C\\ \cos C=\frac{a^2+b^2-c^2}{2ab}[/tex]

[tex]LHS=(a-b)^2\cos^2(\frac{C}{2} )+(a+b)^2\sin^2(\frac{C}{2} )\\ LHS=(a^2+b^2-2ab)\cos^2(\frac{C}{2} )+(a^2+b^2+2ab)\sin^2(\frac{C}{2} )\\ LHS=(a^2+b^2)(\cos^2(\frac{C}{2} )+\sin^2(\frac{C}{2} ))-2ab(\cos^2(\frac{C}{2} )-\sin^2(\frac{C}{2} ))[/tex].

Simplifying,

[tex]LHS=(a^2+b^2)(1)-2ab(\cos C)\\ LHS=a^2+b^2-2ab(\frac{a^2+b^2-c^2}{2ab} )\\ LHS=a^2+b^2-(a^2+b^2-c^2)\\ LHS=c^2=RHS[/tex].

The proof is complete.

Answer 2

Answer:

c² is the answer

Step-by-step explanation:

please look at the attachment