If sinC – sinD = x, then the value of x is

$2 sin left[frac{(mathrm{C}+mathrm{D})}{2} ight] cos left[frac{(mathrm{C}-mathrm{D})}{2} ight]$

$2 sin left[frac{(mathrm{C}+mathrm{D})}{2} ight] sin left[frac{(mathrm{D}-mathrm{C})}{2} ight]$

$2 cos left[frac{(mathrm{C}+mathrm{D})}{2} ight] cos left[frac{(mathrm{C}-mathrm{D})}{2} ight]$

$2 cos left[frac{(mathrm{C}+mathrm{D})}{2} ight] sin left[frac{(mathrm{C}-mathrm{D})}{2} ight]$

Answer : Option D

Explanation :

$sin C-sin D=2 cos frac{C+D}{2} cdot sin frac{C-D}{2}=x$

Illustration

sin (A + B)

= sinA . cosB + cosA . sinB

sin (A – B)

= sinA . cosB – cosA . sinB

∴ sin (A + B) – sin (A – B)

= 2 cosA . sinB

Let, A + B = C ; A – B = D

∴ On adding,

$A=frac{C+D}{2}$

On subtracting,

$B=frac{C-D}{2}$

$herefore sin C-sin D=2 cos frac{C+D}{2} cdot sin frac{C-D}{2}$