The maximum value of `sin(theta+pi/6)+cos(theta+pi/6)` is attained at `theta in (0,pi/2)`

Options

A. `(pi)/(12)`
B. `(pi)/(6)`
C. `(pi)/(3)`
D. `(pi)/(2)`

Answer

A

Explanation

`sin(x+(pi)/(6))+cos(x+(pi)/(6))`
`=sqrt(2)[(1)/sqrt(2)sin(x+(pi)/(6))+1/sqrt(2)(x+(pi)/(6))]`
`=sqrt(2)[sin((x+(pi)/(6))cos (pi)/(4)+cos((x+(pi)/(6))sin(pi)/(4)]`
`=sqrt(2)[sin(x+(pi)/(6)+(pi)/(4))]`
`=sqrt(2)[sin(x+(5 pi)/(12))]`
Given interval is `(0,(pi)/(2))`
for maximum vlaue `x+(5pi)/(12)=(pi)/(2)`
for maximum value `x+(5 pi)/(12)=(pi)/(2)`
`rarr =(pi)/(2)-(5pi)/(12)=(6 pi- 5pi)/(12)=(pi)/(12)`