Question

How do you verify the identity `sin(pi/6+x)=1/2(cosx+sqrt3sinx)`?

Answer 1

See below.

Explanation:

We use the `sin` sum identity.

`sin(\theta+\alpha)=sin(\theta)cos(alpha)+sin(alpha)cos(theta)`

So, in this case, `theta=pi/6` and `alpha=x`

`sin(pi/6+x)=sin(pi/6)cos(x)+sin(x)cos(pi/6)`

where `sin(pi/6)=1/2` and `cos(pi/6)=sqrt(3)/2`

Plugging these back in,
`sin(pi/6+x)=1/2cos(x)+sqrt(3)/2sin(x)`

Taking out `1/2`, we find that
`sin(pi/6+x)=1/2(cosx+sqrt3sinx)`.

Answer 2

See proof below

Explanation:

We apply

`sin(a+b)=sinacosb+sincosa`

We need

`sin(1/6pi)=1/2`

`cos(1/6pi)=sqrt3/2`

Therefore,

`LHS=sin(pi/6+x)`

`=sin(pi/6)cosx+cos(pi/6)sinx`

`=1/2cosx+sqrt3/2cosx`

`=1/2(cosx+sqrt3sinx)`

`=RHS`

`QED`