Quick trig question?

Question

How does cos(x) + cos(x + 2π/3) + cos(x + 4π/3) = 0?

This exercise requires us to use "sum to product" formulae.

I only know of the formula where there are 2 "parts" eg: cos(x) + cos(y) =...

I'm not sure how to utilize this formulae for the above question.

Any help would be appreciated. Thanks a lot.

Answer 1

cos x+cos(x+2pi/3)+cos(x+4pi/3)=sin(4pi/3)co...

solve for x equation

cos(x+2pi/3)=0

or use formula

cos x+cos(x+a)+cos(x+2a)+...+cos(x+na)=

sin((n+1)a/2)c0s(x+na/2)/sin(a/2)

Answer 2

cos(x) + cos(x+2pi/3) + cos(x + 4pi/3) = 0

cos(x) + cos(x + pi - pi/3) + cos(x + pi + pi/3) = 0

let u = x + pi and v = pi/3 then

cos(x) + cos(u - v) + cos (u + v) = 0

cos(x) + 2cos(u)cos(v) = 0 (from product-to-sum identities)

cos(x) + 2cos(x+pi)cos(pi/3) = 0

cos(x) + 2cos(x+pi)(1/2) = 0

cos(x) + cos(x + pi) = 0

cos(x) + (-cos(x)) = 0 (from phase angle identities)

cos(x) - cos(x) = 0

Answer 3

cos(x) + cos(x + 4π/3)

=2cos(x + 2π/3)cos(-4π/3)

=-cos(x + 2π/3) [since cos(-4π/3)=-0.5]

Moving the RHS term to the left completes the proof.

Answer 4

___ cos(x) + cos(x + 2π/3) + cos(x + 4π/3) = 0

cos(x) + cos(x + 2π/3) + cos(x + 4π/3)

= 2cos( (x+2π/3 + x +4π/3)/2 )*cos( (x + 2π/3 - x - 4π/3)/2 )

= 2cos(x+π)cos(π/3)

=cos(x+π)

Therefore:

cos(x) + cos(x + 2π/3) + cos(x + 4π/3) = 0

<=> cos(x)+ cos(x+π) =0

<=> cos(x) = -cos(x+π)

<=> cos(x) = cos(-x)

<=> cos(x)=cos(x)

x can be any real number.