IIT JEE , AIEEE Forum - Mathematics , Trignometry

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180 < a-b < 540

90 < (a-b)/2 < 270

In this range cosine is negative, hence cos{(a-b)/2} is negative.

sina + sinb = -21/65

2sin{(a+b)/2}cos{(a-b)/2} = -21/65.........(1)

cosa + cosb = -27/65

2cos{(a+b)/2}cos{(a-b)/2} = -27/65 ........(2)

Dividing these two : tan{(a+b)/2} = 21/27

Hence cos{(a+b)/2} = 27/

(21²+27²) = 27/

(1170) = 27/3(

130) = 9/(

130)

Hence cos{(a-b)/2} = (-27/65)/[ 2cos{(a+b)/2}] = (-27/65)/{18/(

130)}

= -3/(

130))

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