If A+B+C=(π/3) then sin ((π-6 A/6))+ sin ((π-6 B/6))+ sin C=

Question

If A+B+C=(π/3) then sin ((π-6 A/6))+ sin ((π-6 B/6))+ sin C= (A) -1+4 cos ((π-6 A/12)) cos ((π-6 B/12)) sin (C/2) (B) 4 sin ((π+6 A/12)) sin ((π+6 B/12)) cos (C/2) (C) 1-4 cos ((π-6 A/12)) cos ((π-6 B/12)) cos (π-6 C/12) (D) 4 cos ((π-6 A/12)) cos ((π-6 B/12)) sin (C/2)

Tardigrade Admin · Accepted Answer

Given, A+B+C=(π/3) sin ((π-6 A/6))+ sin ((π-6 B/6))+ sin C =2 sin ((π-6 A+π-6 B/12)) cos ((π-6 A-π+6 B/12))+ sin C =2 sin ((π/6)-((A+B/2))) cos ((A-B/2))+ sin C =2 sin ((π/6)-(π/6)+(C/2)) cos ((A-B/2))+2 sin (C/2) cos (C/2) =2 sin (C/2) cos (A-B/2)+2 sin (C/2) cos (C/2) =2 sin (C/2)( cos (A-B/2)+ cos (C/2)) =2 sin (C/2)(2 cos (((A-B/2)+(C/2)/2)) cos (((A-B/2)-(C/2)/2))) =4 sin (C/2) cos ((A+C-B/4)) cos ((A-(B+C)/4)) =4 sin (C/2) cos ((π-6 B/12)) cos ((π-6 A/12)) =4 cos ((π-6 A/12)) cos ((π-6 B/12)) sin (C/2)

Q. If $A + B + C = \frac{π}{3}$ then
$sin (\frac{π - 6 A}{6}) + sin (\frac{π - 6 B}{6}) + sin C =$

$- 1 + 4 cos (\frac{π - 6 A}{12}) cos (\frac{π - 6 B}{12}) sin \frac{C}{2}$

$4 sin (\frac{π + 6 A}{12}) sin (\frac{π + 6 B}{12}) cos \frac{C}{2}$

$1 - 4 cos (\frac{π - 6 A}{12}) cos (\frac{π - 6 B}{12}) cos \frac{π - 6 C}{12}$

$4 cos (\frac{π - 6 A}{12}) cos (\frac{π - 6 B}{12}) sin \frac{C}{2}$

Q. If A+B+C=3π​ then sin(6π−6A​)+sin(6π−6B​)+sinC=

−1+4cos(12π−6A​)cos(12π−6B​)sin2C​

4sin(12π+6A​)sin(12π+6B​)cos2C​

1−4cos(12π−6A​)cos(12π−6B​)cos12π−6C​

4cos(12π−6A​)cos(12π−6B​)sin2C​

Q. If $A + B + C = \frac{π}{3}$ then
$sin (\frac{π - 6 A}{6}) + sin (\frac{π - 6 B}{6}) + sin C =$

$- 1 + 4 cos (\frac{π - 6 A}{12}) cos (\frac{π - 6 B}{12}) sin \frac{C}{2}$

$4 sin (\frac{π + 6 A}{12}) sin (\frac{π + 6 B}{12}) cos \frac{C}{2}$

$1 - 4 cos (\frac{π - 6 A}{12}) cos (\frac{π - 6 B}{12}) cos \frac{π - 6 C}{12}$

$4 cos (\frac{π - 6 A}{12}) cos (\frac{π - 6 B}{12}) sin \frac{C}{2}$