Question

In ΔABC, show that $cos^{2}\ \frac{A}{2} + cos^{2}\ \frac{B}{2} + cos^{2}\ \frac{C}{2} = 2\ +\ \frac{r}{2R}$.

Answer 1

Solution :

LHS= cos²A/2 + cos²B/2 + cos²C/2

=cos²A/2+(1-sin²B/2)+cos²C/2

=1+cos²A/2-sin²B/2+cos²C/2

=1+(cos²A/2-sin²B/2)+cos²C/2

=1+cos[(A+B)/2]cos[(A-B)/2]+cos²C/2

=1+sinC/2cos[(A-B)/2]+1-sin²C/2

=2+sinC/2[cos{(A-B)/2} -sinC/2]

= 2+sinC/2{cos[(A-B)/2-cos[(A+B)/2]}

= 2+sinC/2{(cosA/2cosB/2+sinA/2sinB/2)

-(cosA/2cosB/2-sinA/2sinB/2)}

= 2+sinC/2[2sinA/2sinB/2]

= 2+ 2sinA/2sinB/2sinC/2

= 2+[ (4RsinA/2sinB/2sinC/2)/2R ]

= 2 + r/2R

= RHS

•••

Answer 2

HELLO DEAR,

Answer:

Step-by-step explanation:

Solution :

LHS= cos²A/2 + cos²B/2 + cos²C/2

= cos²A/2 + (1 - sin²B/2) + cos²C/2

= 1 + cos²A/2 - sin²B/2 + cos²C/2

= 1 + (cos²A/2 - sin²B/2) + cos²C/2

= 1 + cos[(A+B)/2]cos[(A-B)/2] + cos²C/2

= 1 + sinC/2cos[(A-B)/2] + 1 - sin²C/2

= 2 + sinC/2[cos{(A-B)/2} - sinC/2]

= 2 + sinC/2{cos[(A-B)/2 - cos[(A+B)/2]}

= 2 + sinC/2{(cosA/2cosB/2 + sinA/2 sinB/2) - (cosA/2cosB/2-sinA/2sinB/2)}

= 2 + sinC/2[2sinA/2sinB/2]

= 2 + 2sinA/2sinB/2sinC/2

= 2 + [ (4RsinA/2sinB/2sinC/2)/2R ]

= 2 + r/2R

I HOPE IT'S HELP YOU DEAR,

THANKS