If `sin theta+sin2theta+sin3theta=sin alpha` and `cos theta+cos 2theta+cos3theta=cos alpha`, then `theta` is equal to

Options

A. `(alpha)/(2)`
B. `alpha`
C. `2alpha`
D. `(alpha)/(6)`

Answer

A

Explanation

`sintheta+sin3theta+sin2theta=sinalpha`
`implies2sin2thetacostheta+sin2theta=sinalpha`
`implies sin2theta(2costheta+1)=sinalpha" ... (i)"`
Now, `costheta+cos3theta+cos2theta=cosalpha`
`2cos2thetacostheta+cos2theta=cosalpha`
`cos2theta(2costheta+1)=cosalpha" ... (ii)"`
From Eqs. (i) and (ii), `tan2theta=tanalpha`
`implies 2theta=alpha`
`implies theta=(alpha)/(2)`