Question

A, b, c are angles of triangle abc then cos(3a+2b+c/2)+cos(a-c/2)

Answer 1

HELLO DEAR,

GIVEN :- a,b,c are the angles of a triangle and

as, we know:- cosX + cosY = 2cos{(x + y)/2}cos{(x - y)/2}

therefore, cos(3a + 2b + c)/2 + cos(a - c)/2

=> $2\cos\{ \frac{\frac{3a \: + 2b \: + c}{2} \: + \frac{a \: - \: c}{2} } {2} \}\ cos\{ \frac{\frac{3a \: + 2b \: + c}{2} \: - \frac{a \: - \: c}{2} } {2} \}$

=>$2\cos\{\frac{4a + 2b}{4}\}\cos\{\frac{2a + 2b + 2c}{4}\}$

=> $2\cos\{\frac{2a + b}{2}\}\cos\{\frac{a + b + c}{2}\}$

as, we know in any triangle sum of all the angles of a triangle is 180°

so, a + b + c = 180°

=> $2\cos\frac{2a + b}{2}\cos\frac{180{\degree}}{2}$

$[\therefore, \cos90\degree = 0]$

hence, cos(3a + 2b + c)/2 + cos(a - c)/2 = 0

I HOPE IT'S HELP YOU DEAR,

THANKS

Answer 2

0. hope it helps you.

happy to help