Let `alpha, beta" such that "pi lt alpha-beta lt 3pi." If sin "alpha+sin beta=-21//65 cos alpha+cos beta=-27//65`, then find the value of `cos ""(alpha-beta)/(2)`.

Options

A. `-(3)/(sqrt(130))`
B. `(3)/(sqrt(130))`
C. `(6)/(65)`
D. `-(6)/(65`

Answer

A

Explanation

We have `sin alpha+sin beta=-(21)/(65)`
`cos alpha+cos beta=-(27)/(65)`
Squaring Eq. i, we get
`sin^(2)alpha+sin^(2)beta+2sin alpha sin beta=((21)/(65))^(2)`
Squaring Eq. ii, we get
`cos^(2)alpha+cos^(2)beta+2cos alpha cos beta=((27)/(65))^(2)`
Adding Eqs. iii and iv, we get
`2+2cos(alpha-beta)=(1)/((65)^(2))[(27)^(2)+(21)^(2)]`
`=(1)/((65)^(2))(729+441)`
`=(1)/((65)^(2))(1170)=(18)/(65)`
or `1+cos (alpha-beta)=(9)/(65)`
or `2cos^(2)""(alpha-beta)/(2)=(9)/(65)`
or `cos""(alpha-beta)/(2)=-(3)/sqrt(130)`