Question

ABC are interior angle of triangle ABC prove that sin (B+C) ➗ 2=cosA/2

Answer 1

hey!

in a triangle ABC;

A+B+C = 180°

=) B+C = 180°-A

Divide both sides by 2;

=) (B+C) /2 = 180°/2 - A/2

=) (B+C) /2 = 90°- A/2

take sin in both sides;

=) sin(B+C) /2 = sin {90° - (A/2)}

=) sin(B+C) /2 = cos A/2

hope it helps you!!

Answer 2

A+B+C=180°A+B+C=180° . . . Sum of all interior angles of ∆, and

sinX≡cos(90°−X)sinX≡cos(90°−X) . . . for all X

Now since

A+B+C=180°A+B+C=180° . . . Fact 1

(A+B+C)/2=90°(A+B+C)/2=90°

(A+B)/2=90°−C/2(A+B)/2=90°−C/2

sin(A+B)/2=sin(90°−C/2)sin(A+B)/2=sin(90°−C/2) . . . Taking sines on both sides

sin(A+B)/2=cos(C/2)sin(A+B)/2=cos(C/2) . . . Fact 2