Question

In ΔABC, show that $sin^{2}\ \frac{A}{2} + sin^{2}\ \frac{B}{2} + sin^{2}\ \frac{C}{2} = 1\ -\ \frac{r}{2R}$.

Answer 1

Solution :

Step-by-step explanation:

LHS = sin²A/2+sin²B/2+sin²C/2

= (1-cosA)/2+(1-cosB)/2+(1-cosC)/2

= 3/2-1/2[cosA+cosB+cosC]

=3/2-1/2[2cos(A+B)/2cos(A-B)/2+1-2sin²C/2

=3/2-1/2[1+2sinC/2cos(A-B)/2-2sin²C/2]

=3/2-1/2[1+2sinC/2{cos(A-B)/2-sinC/2}

=3/2-1/2[1+2sinC/2{cos(A-B)/2-cos(A+B)/2}

= 3/2-1/2[1+2sinC/2(2sinA/2sinB/2)]

= 3/2-1/2[1+4sinA/2sinB/2sinC/2]

= 3/2-1/2[1+(4RsinA/2sinB/2sunC/2)/R]

= 3/2-1/2[1 + r/R]

= 3/2 - 1/2 - r/2R

= 1 - r/2R

= RHS

•••••

Answer 2

HELLO DEAR,

Answer:

Step-by-step explanation:

sin²A/2 + sin²B/2 + sin²C/2

= (1 - cosA)/2 + (1-cosB)/2 + (1 - cosC)/2

= 3/2 - 1/2[cosA + cosB + cosC]

=3/2 - 1/2[2cos(A+B)/2cos(A-B)/2+1-2sin²C/2

=3/2 - 1/2[1+2sinC/2cos(A-B)/2-2sin²C/2]

=3/2 - 1/2[1+2sinC/2{cos(A-B)/2-sinC/2}

=3/2 - 1/2[1+2sinC/2{cos(A-B)/2-cos(A+B)/2}

= 3/2 - 1/2[1 + 2sinC/2(2sinA/2sinB/2)]

= 3/2 - 1/2[1+4sinA/2sinB/2sinC/2]

= 3/2 - 1/2[1 + (4RsinA/2sinB/2sunC/2)/R]

= 3/2 - 1/2[1 + r/R]

= 3/2 - 1/2 - r/2R

= 1 - r/2R

I HOPE IT'S HELP YOU DEAR,

THANKS