If `sinalpha+cosbeta=2,(0^(@) le beta lt alpha90^(@))`, then `sin((2alpha+beta)/(3))=?`

Options

A. `sin((alpha)/(2))`
B. `cos((alpha)/(3))`
C. `sin((alpha)/(3))`
D. `cos((2alpha)/(3))`

Answer

b

Explanation

`sinalpha+cos beta= 2`
shortest method
put, `alpha=90^(@),beta=0^(@)`
`sin90^(@)+cos0^(@)=2`
`1+1=2`
2=2matched
So, `alpha=90^(@),beta=0^(@)`
`rArrsin((2alpha+beta)/(3))^(2)`
`=sin((2xx90+0)/(3))^(2)`
`=sin((180)/(3))^(2)`
`=sin60^(@)=cos30^(@)=sqrt(3)/(2)`
Take `cos(alpha)/(3)=cos(90^(@))/(2)`
`=cos30^(@)`
So, this is answer