If `sqrt(2) cos A=cosB+cos^(3)B,` and `sqrt(2)sinA=sinB-sin^(3)B ` then `sin(A-B)=`

Options

A. `+-1`
B. `+-(1)/(2)`
C. `+-(1)/(3)`
D. `+-(1)/(4)`

Answer

C

Explanation

`sqrt(2)cos A=cosB+cos^(3)B`
and `sqrt(2)sinA=sinB -sin^(3)B`
`rArr sqrt(2)sinA cos B-sqrt(2)cosA sinB`
`=(sin B-sin^(3)B)cosB-(cos B+cos^(3)B)sinB`
`=-sin B cos B `
`rArr sin(A-B)=-(1)/(2sqrt(2))sin 2B`
Now squaring and adding Eqs. i and ii, we get
`2=cos^(2)B+sin^(2)B+cos(6)B+sin^(6)B+2(cos^(4)B-sin^(4)B)`
`rArr 1=(cos^(2)A+sin^(2)A)^(3)-3cos^(2)Asin^(2)A(cos^(2)A+sin^(2)A)+2cos 2B`
`rArr 1=1-(3//4)sin^(2)AB+2cos 2B`
`rArr -3sin^(2)2B+8cos2B=0`
`rArr 3cos^(2)2B+8cos 2B-3=0`
`rArr cos 2B=(1)/(3)`
`rArr sin 2B=+-(2sqrt(2))/(3)therefore sin(A-B)=+-(1)/(3)`.