Question

If A, B, C are angles in a triangle, then prove that cos A + cos B - cos C = -1 + 4$cos\frac{A}{2}cos\frac{B}{2}sin\frac{C}{2}$.

Answer 1

Answer:

Step-by-step explanation:

If A, B, C are angles in a triangle,than A+B+C=π

as we have to prove

$cos\:A+cos\:B-cos\:C=-1+4cos\:\frac{A}{2}cos\:\frac{B}{2}cos\:\frac{C}{2}$

we have

$=cos\:A+cos\:B-cos\:C\\\\\\=2cos(\frac{A+B}{2})cos(\frac{A-B}{2})-(1-2sin^{2}\frac{C}{2})$

since cos 2A=1-2sin²A

so

$=2cos(\frac{A+B}{2})cos(\frac{A-B}{2})-1+2sin^{2}\frac{C}{2})\\\\=2cos(\frac{\pi}{2}-\frac{C}{2})cos(\frac{A-B}{2})-1+2sin^{2}\frac{C}{2})\\\\=2sin(\frac{C}{2})cos(\frac{A-B}{2})-1+2sin^{2}\frac{C}{2})\\\\=2sin(\frac{C}{2})[cos(\frac{A-B}{2})+sin\frac{C}{2})]-1\\\\\\=2sin(\frac{C}{2})[cos(\frac{A-B}{2})+sin(\frac{\pi}{2}-\frac{A+B}{2})]-1\\\\\\=2sin(\frac{C}{2})[cos(\frac{A-B}{2})+cos\frac{A+B}{2})]-1$

$=2sin(\frac{C}{2})[2cos(\frac{A}{2})+cos\frac{B}{2})]-1\\\\\\=4sin(\frac{C}{2})cos(\frac{A}{2})+cos\frac{B}{2})-1\\\\\\=-1+4sin(\frac{C}{2})cos(\frac{A}{2})+cos\frac{B}{2})$

hence proved