Question

Question #5f168

Answer 1

`sin(x-y) = -56/65` , `cos 2x= -7/25`,
`sin2y =120/169`

Explanation:

In 2nd quadrant `90^0 < x >180^0 ,sinx=p/h=4/5 ; p=4 , h=5 :. b=- sqrt(5^2-4^2)= -3 :.cosx= b/h=-3/5`where `p` is perpedicular , `b` is base , `h` is hypotenuse.

In 3rd quadrant `180^0 < x >270^0 , cosy=b/h= -5/13 ; b= -5 , h=13 :. p=- sqrt(13^2-5^2)= -12 :.siny= p/h=-12/13`

`sin(x-y) =sinx*cosy-cosx*siny = 4/cancel5* (-cancel5/13) - (-3/5)* (-12/13) = -4/13 - 36/65 = -56/65`

`cos 2x= cos^2x- sin^2x= (-3/5)^2 - (4/5)^2 =9/25 -16/25 = -7/25`
`sin2y = 2*siny*cosy = 2* (-12/13) * (-5/13) =120/169` [Ans]