Question

in triangle abc prove that cos square A by 2 + cos square B by 2 + cos square C by 2 equal to 2 + 2 sin a by 2 Sin B by 2 Sin C by 2

Answer 1

Hope you understand

Answer 2

Answer:

Given: Triangle ABC ⇒ A + B + C = 180° = π

To prove: $cos^\,\frac{A}{2}+cos^2\,\frac{B}{2}+cos^2\,\frac{C}{2}=2+2\:sin\,\frac{A}{2}\:sin\,\frac{B}{2}\:sin\,\frac{C}{2}$

We use the following formula to prove it,

1 .cos² x - sin² y = cos( x + y ) . cos( x - y )

2. sin² x + cos² y = 1

3. cos ( π - x ) = -cos x

4. cos x - cos y  = 2 . sin (x+y/2) . sin (x-y/2)

5. cos (-x) = cos x

Consider,

LHS

$cos^2\,\frac{A}{2}+cos^2\,\frac{B}{2}+cos^2\,\frac{C}{2}$

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

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

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

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

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

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

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

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

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

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

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

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

$=2+2\:sin\,\frac{C}{2}\:(sin\,\frac{A}{2}\:sin\,\frac{B}{2})$

$=2+2\:sin\,\frac{A}{2}\:sin\,\frac{B}{2}\:sin\,\frac{C}{2}$

= RHS

Hence Proved.