simplify sin(pi/6 +x)+sin(pi/6 -x)?

sin(pi/6 +x)+sin(pi/6 -x)

sin(pi/6) cos x+cos (pi/6) sin(x) + sin(pi/6) cos x - cos(pi/6) sin x

2 sin (pi/6) cos x

2 sin (30) cos x

2 (1/2) cons x

cos x

Identities :
sin(A ± B) = sinA cosB ± cos A sinB

sin[(π/6) + x] + sin[(π/6) - x]
= [sin(π/6) cosx + cos(π/6) sinx] + [sin(π/6) cosx - cos(π/6) sinx]
= sin(π/6) cosx + cos(π/6) sinx + sin(π/6) cosx - cos(π/6) sinx
= 2 sin(π/6) cosx
= 2 × (1/2) × cosx
= cosx

Well,

sin(pi/6 +x) + sin(pi/6 -x) = sin(pi/6) cosx + cos(pi/6) sinx + sin(pi/6) cos(-x) - cos(pi/6) sinx
= sin(pi/6) cosx + cos(pi/6) sinx + sin(pi/6) cos(x) - cos(pi/6) sinx <--- because : cos(-x) = cosx
= 2 sin(pi/6) cosx
= 2 * 1/2 * cosx
= cosx

sin(pi/6 +x) + sin(pi/6 -x) = cosx

qed

hope it' ll help !!

sin(pi/6 +x) is the same thing as cos(pi/6). so cos pi /6 + -cos pi/6 so 0

sin(pi/6 +x)+sin(pi/6 -x)=[sin(pi/6)cosx+sinxcos(pi/6)]+[sin(pi/6)cosx-sinxcos(pi/6)] = 2sin(pi/6)cosx =
cosx.

Do you know this identity?

sin(a + b) = sin(a).cos(b) + cos(a).sin(b) → suppose that: a = (π/6) and b = x

sin[(π/6) + x] = sin(π/6).cos(x) + cos(π/6).sin(x) ← memorize this result as (1)


Do you know this identity?

sin(a - b) = sin(a).cos(b) - cos(a).sin(b) → suppose that: a = (π/6) and b = x

sin[(π/6) - x] = sin(π/6).cos(x) - cos(π/6).sin(x) ← memorize this result as (2)



= sin[(π/6) + x] + sin[(π/6) - x] → recall (1)

= sin(π/6).cos(x) + cos(π/6).sin(x) + sin[(π/6) - x] → recall (2)

= sin(π/6).cos(x) + cos(π/6).sin(x) + sin(π/6).cos(x) - cos(π/6).sin(x)

= 2.sin(π/6).cos(x) → you know that: sin(π/6) = 1/2

= 2.(1/2).cos(x)

= cos(x)