If `cos3theta=cos3alpha,` then the value of `sintheta` can be given by `+-sinalpha` (b) `sin(pi/3+-alpha)` `sin((2pi)/3+alpha)` (d) `sin((2pi)/3-alpha)`

Options

A. `pm sin alpha`
B. `sin (pi/3 pm alpha)`
C. `sin ((2pi)/3+alpha)`
D. `sin ((2pi)/3- alpha)`

Answer

A, C, D

Explanation

`cos 3 theta= cos 3 alpha`
`3alpha = 2npi pm 3 alpha, n in Z`
Putting `n=0, 1`, we have
`3 theta=3 alpha or -3 alpha or 2pi +3 alpha or 2pi -3alpha`
`theta=alpha or -alpha or theta =(2pi)/3+alpha or (2pi)/3-alpha`
Hence, options (1), (3), (4) are correct
If `n=-1,` then `3 theta=-2 pi pm 3 alpha`
`rArr theta=-(2pi)/3 pm alpha`
`rArr sin theta= sin (-(2pi)/3 pm alpha)=- sin ((2pi)/3 pm alpha)`
`=- sin (pi-pi/3 pm alpha)=-sin (pi/3 pm alpha)`
Hence, (2) is not correct.