SSC CHSL 7 Jan 2017 Shift 1 Paper

Question : 64 of 100

Marks: +1, -0

If cosC - cosD = x, then value of x is

$2sin[(C+D)/2]cos[(C-D)/2]$
$2cos[(C+D)/2]sin[(C-D)/2]$
$2sin[(C+D)/2]sin[(D-C)/2]$
$2cos[(C+D)/2]cos[(C-D)/2]$

Solution:

We know that, cos (x+y) = cosx cosy - sinx siny ----(i)
and cos (x-y) = cosx cosy + sinx siny -----(ii)
Subtracting equations (ii) from (i), we get :
cos (x+y) - cos (x-y) = - 2 sinx siny ----------(iii)
Let x+y = C and x-y = D
⇒

x={C+D}/2

⇒ and

y= {C-D}/2

Substituting above values in equation (iii)
⇒

cosC - cosD

= -2sin({C+D}/2)sin({C-D}/2)

⇒

cosC - cosD

= 2sin({C+D}/2)sin({D-C }/2)

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