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

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Answer 1 · 2017-02-03T02:57:08

see explanation below

expand #sin(A+B)=sinAcosA+sin BcosA# and,
#sin(A-B)=sinAcosA-sin BcosA#

Therefore
#sin(pi/6+x) + sin(pi/6-x) # #= sin(pi/6)cosx + cancel (sinx cos(pi/6))+ sin(pi/6)cosx - cancel (sinx cos(pi/6))#

#= 2 sin(pi/6)cosx#, where #sin(pi/6) =1/2#

therefore,

#= 2 sin(pi/6)cosx=2(1/2)cos x = cos x#,

Answer 2 · 2017-02-03T03:06:02

Explained below.

Expand the left hand side,

#(sin (pi/6) cos x + cos (pi/6) sinx) +(sin (pi/6) cosx - cos (pi/6) sin x)#

= #2 sin (pi/6) cos x#

=#2 (1/2)cos x#

= cos x = right hand side