Question

cos(-a+b+c) +cos (a-b+c) + cos(a+b-c)+cos (a+b+c)=4cosa cosb cosc

Answer 1

Use cos C + cos D = 2 cos ((C+D)/2) cos ((C-D)/2)
Please comment if you don't understand what's written.

Answer 2

Answer:

To show: cos( -a + b + c ) + cos ( a - b + c ) + cos ( a + b - c ) + cos ( a + b + c ) = 4 . cos a . cos b . cos c

We use the following formula,

$cos\,x+cos\,y=2cos\,(\frac{x+y}{2})\:cos\,(\frac{x-y}{2})$

Consider,

LHS

$=cos\,(-a+b+c)+cos \,(a-b+c)+cos\,(a+b-c)+cos\,(a+b+c)$

$=2\:cos\,(\frac{-a+b+c+a-b+c}{2})\:cos\,(\frac{-a+b+c-(a-b+c)}{2})+2\:cos\,(\frac{a+b-c+a+b+c}{2})\:cos\,(\frac{a+b-c-(a+b+c)}{2})$

$=2\:cos\,c\:cos\,(b-a)+2\:cos\,(a+b)\:cos\,c$   ( cos(-c) = cos c )

$=2\:cos\,c(cos\,(b-a)+cos\,(a+b))$

$=2\:cos\,c(2\:cos\,(\frac{b-a+a+b}{2})\:cos\,(\frac{b-a-(a+b)}{2})$

$=2\:cos\,c(2\:cos\,b\:cos\,a)$    ( cos(-a) = cos a )

$=4\:cos\,c\:cos\,b\:cos\,a$

$=4\:cos\,a\:cos\,b\:cos\,c$

$=RHS$

Hence Proved.