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

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Answer 1 · 2017-03-31T15:20:58

See below.

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 · 2017-03-31T15:22:34

See proof below

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#