If `xsina+ysin2a+zsin3a=sin4a` `xsinb+ysin2b+zsin3b=sin4b` `xsinc+ysin2c+zsin3c=sin4c` then the roots of the equation `t^3-(z/2)t^2-((y+2)/4)t+((z-x)/8)=0,a , b , c ,!=npi,` are `sina ,sinb ,sinc` (b) `cosa ,cosb ,cosc` `sin2a ,sin2b ,sin2c` (d) `cos2a ,cos2bcos2c`

Options

A. `sin a, sin b, sinc`
B. `cos a, cos b,cos c`
C. `sin 2a, sin 2b,sin2c`
D. `cos 2a,cos 2b cos 2c`

Answer

B

Explanation

`x sin a+y sin 2a+z sin 3a=sin4alpha`
`rArr x sin a+y=2sin a cos a+z x sin a (3-4sin^(2)a)`
`rArr x sin a+yx 2 sin a cos a+z x sin a(3-4sin^(2)a)`
`=2xx2 sina cos a cos 2a`
`rArr x+2ycos a+z(3+4cos^(2)a-4)`
`=4cos a(2cos^(2)a-1)` [as `sin alphane0`]
`rArr 8cos^(3)a-4z cos^(2)a-(2y+4)cosa+(z-x)=0`
`rArr cos^(3)a-((z)/(2))cos^(2)a-((y+2)/(4))cosa+((z-x)/(8))=0`
Which shows that `cos alpha` is root of the equation
`t^(3)=((z)/(2))t^(2)-((y+2)/(4))t+((z-x)/(8))=0`
Similarly, from second and thrid equations we can slow that `cos b ` and `cos c` are the roots of the given equation.