Question

A+B+C=pie then cos2A + cos2B+cos2c=?

Answer 1

Answer:

⇒ $-4 \ cos \ A \ cos \ B \ cos \ C \ - \ 1$

Step-by-step explanation:

Given :

A + B + C = π ,

To Find : The value of :

cos 2A + cos 2B + cos 2C,.

Solution :

cos 2A + cos 2B + cos 2C

⇒ cos 2A + (cos 2B + cos 2C)

We know that,

$cos \ A + cos \ B \ = \ 2 \ cos \ (\frac{A+B}{2}) \ cos \ (\frac{A-B}{2})$

also,

$cos \ 2A=2cos^2 \ A \ - \ 1$

So,

⇒ $cos \ 2A \ + \ 2 \ cos \ (\frac{2B+2C}{2}) \ cos \ (\frac{2B-2C}{2})$

⇒ $2 \ cos^2 \ A \ - \ 1 \ + \ 2 \ cos \ (B+C) \ cos \ (B-C)$

⇒ $2 \ cos^2 \ A \ - \ 1 \ + \ 2 \ cos \ (\pi - A) \ cos \ (B-C)$

∵ A + B + C = π ⇒ B + C = π - A..

∵ cos (π - A) = - cos A..

So,

⇒ $2 \ cos^2 \ A \ - \ 1 \ - \ 2 \ cos \ A \ cos \ (B-C)$

⇒ $2 \ cos \ A \ [cos \ A - cos \ (B-C)] \ - \ 1$

∵ cos (A) = cos [π - (B + C)] = - cos (B + C)

⇒ $2 \ cos \ A \ [-cos \ (B+C) - cos \ (B-C)] \ - \ 1$

⇒ $-2 \ cos \ A \ [cos \ (B+C) + cos \ (B-C)] \ - \ 1$

⇒ $-2 \ cos \ A \ [2cos(\frac{(B+C)+(B-C)}{2})cos\frac{(B+C)-(B-C)}{2}] \ - \ 1$

⇒ $-2 \ cos \ A \ [2 \ cos \ B \ cos \ C] \ - \ 1$

⇒ $-4 \ cos \ A \ cos \ B \ cos \ C \ - \ 1$

Answer 2

Answer:

Step-by-step explanation:

Given=a+b+c=pie

cos2a + cos2b + cos2c 

= cos2a + [ cos(2b) + cos(2c) ] 

= cos2a + 2*cos(b + c)*cos(b - c) 

= (2cos²a - 1) + 2cos(180 - a)*cos(b - c) 

= 2cos²(a) - 1 - 2cos(a)*cos(b - c) 

= - 1 + 2cos(a) * [cos(a) - cos(b - c)] 

= - 1 + 2cos(a) [ cos {180 - (b + c) } - cos(b - c) ] 

= - 1 + 2cos(a) [ - cos (b + c) - cos(b - c) ] 

= - 1 - 2cos(a) [ cos (b + c) - cos(b - c) ] 

= - 1 - 2cos(a) [ 2*cos(b)*cos(c) ] 

= - 1 - 4*cosa*cosb*cosc