Question

cos(b+c-a)-cos(c+a-b)+cos(a+b-c)-cos(a+b+c)​

Answer 1

HELLO DEAR,

HELLO DEAR,as we know, a + b + c = 180°

HELLO DEAR,as we know, a + b + c = 180°now, cos(b + c - a) - cos(c + a - b) + cos(a + b - c) - cos(a + b + c)

HELLO DEAR,as we know, a + b + c = 180°now, cos(b + c - a) - cos(c + a - b) + cos(a + b - c) - cos(a + b + c)so, cos(180 - 2a) - cos(180 - 2b) + cos(180 - 2c) - cos(180)

=> - cos2a + cos2b - cos2c - ( -1)

=> -(2cos²a - 1) + (2cos²b - 1) - (2cos²c - 1) + 1

=> -2cos²a + 1 + 2cos²b - 1 - 2cos²c + 1 + 1

=> -2(cos²a - cos²b + cos²c)

I HOPE IT'S HELP DEAR,

I HOPE IT'S HELP DEAR,THANKS

Answer 2

Answer:

4 SinaCosbSinc

Step-by-step explanation:

cos(b+c-a)-cos(c+a-b)+cos(a+b-c)-cos(a+b+c)​

= cos(b+c-a) + +cos(a+b-c) - {cos(c+a-b) + cos(a+b+c)​}

Using CosX + CosY = 2Cos((X+Y)/2)Cos((X-Y)/2)

= 2Cos(b)Cos(c-a) - 2Cos(c+a)Cos(-b)

= 2Cos(b)Cos(c-a) - 2Cos(c+a)Cosb

= 2Cosb(Cos(c-a)  - Cos(c+a))

Using CosX - CosY = -2Sin((X+Y)/2)Sin((X-Y)/2)

= 2Cosb (-2SincSin(-a))

= 2Cosb2SincSina

= 4 SinaCosbSinc